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Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity xx3.1{3.3: Introducing . . . Vector Spaces Math 214-001/002 W10 Day 6 M. Cap Khoury 2 February 2010 M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Combinations 1 Recall a very important de nition. Given some vectors ~v1;:::;~vk in Rn, the linear combinations of the ~vi are all the vectors of the form c1~v1 + c2~v2 + + ck 1~vk 1 + ck~vk. That is, we add with coe cients. The set of all linear combinations of a set of vectors is called their linear span. If V is the span of some set ~v1;~v2;:::;~vk, then we say ~v1;~v2;:::;~vk is a spanning set for V or just that they span V. The span of 2 4 1 0 2 3 5 and 2 4 0 1 1 3 5 is all vectors of the form 2 4 s t 2s + t 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Combinations 1 Recall a very important de nition. Given some vectors ~v1;:::;~vk in Rn, the linear combinations of the ~vi are all the vectors of the form c1~v1 + c2~v2 + + ck 1~vk 1 + ck~vk. That is, we add with coe cients. The set of all linear combinations of a set of vectors is called their linear span. If V is the span of some set ~v1;~v2;:::;~vk, then we say ~v1;~v2;:::;~vk is a spanning set for V or just that they span V. The span of 2 4 1 0 2 3 5 and 2 4 0 1 1 3 5 is all vectors of the form 2 4 s t 2s + t 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Combinations 1 Recall a very important de nition. Given some vectors ~v1;:::;~vk in Rn, the linear combinations of the ~vi are all the vectors of the form c1~v1 + c2~v2 + + ck 1~vk 1 + ck~vk. That is, we add with coe cients. The set of all linear combinations of a set of vectors is called their linear span. If V is the span of some set ~v1;~v2;:::;~vk, then we say ~v1;~v2;:::;~vk is a spanning set for V or just that they span V. The span of 2 4 1 0 2 3 5 and 2 4 0 1 1 3 5 is all vectors of the form 2 4 s t 2s + t 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Combinations 1 Recall a very important de nition. Given some vectors ~v1;:::;~vk in Rn, the linear combinations of the ~vi are all the vectors of the form c1~v1 + c2~v2 + + ck 1~vk 1 + ck~vk. That is, we add with coe cients. The set of all linear combinations of a set of vectors is called their linear span. If V is the span of some set ~v1;~v2;:::;~vk, then we say ~v1;~v2;:::;~vk is a spanning set for V or just that they span V. The span of 2 4 1 0 2 3 5 and 2 4 0 1 1 3 5 is all vectors of the form 2 4 s t 2s + t 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Spans 2 Suppose you have a list of vectors ~v1;~v2;:::;~vk, and you want to know whether ~w is in their linear span. This is the same as asking whether there are ci such that c1~v1 + c2~v2 + + ck~vk = ~w. This, in turn, is the same as asking whether ~v 1 ~v2 ~vk 2 66 64 c1 c2 ... ck 3 77 75 = ~w is consistent. Thus we can always answer such questions by row-reducing. ~v1 ~v2 ~vk ~w . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Spans 2 Suppose you have a list of vectors ~v1;~v2;:::;~vk, and you want to know whether ~w is in their linear span. This is the same as asking whether there are ci such that c1~v1 + c2~v2 + + ck~vk = ~w. This, in turn, is the same as asking whether ~v 1 ~v2 ~vk 2 66 64 c1 c2 ... ck 3 77 75 = ~w is consistent. Thus we can always answer such questions by row-reducing. ~v1 ~v2 ~vk ~w . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Spans 2 Suppose you have a list of vectors ~v1;~v2;:::;~vk, and you want to know whether ~w is in their linear span. This is the same as asking whether there are ci such that c1~v1 + c2~v2 + + ck~vk = ~w. This, in turn, is the same as asking whether ~v 1 ~v2 ~vk 2 66 64 c1 c2 ... ck 3 77 75 = ~w is consistent. Thus we can always answer such questions by row-reducing. ~v1 ~v2 ~vk ~w . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Spans 2 Suppose you have a list of vectors ~v1;~v2;:::;~vk, and you want to know whether ~w is in their linear span. This is the same as asking whether there are ci such that c1~v1 + c2~v2 + + ck~vk = ~w. This, in turn, is the same as asking whether ~v 1 ~v2 ~vk 2 66 64 c1 c2 ... ck 3 77 75 = ~w is consistent. Thus we can always answer such questions by row-reducing. ~v1 ~v2 ~vk ~w . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Vector Spaces 3 De nition A subspace of Rn is a nonempty subset V of Rn which is closed under linear combinations. That is, if ~v1;~v2 2V, then ~v1 + ~v2 2V and k~v1 2V. The zero vector is always a subspace by itself. In R3, the lines through the origin and the planes through the origin are subspaces. So is R3 itself. The set of vectors ~x 2R4 such that x1 + x2 + x3 + x4 = 0 is a subspace of R4. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Vector Spaces 3 De nition A subspace of Rn is a nonempty subset V of Rn which is closed under linear combinations. That is, if ~v1;~v2 2V, then ~v1 + ~v2 2V and k~v1 2V. The zero vector is always a subspace by itself. In R3, the lines through the origin and the planes through the origin are subspaces. So is R3 itself. The set of vectors ~x 2R4 such that x1 + x2 + x3 + x4 = 0 is a subspace of R4. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Vector Spaces 3 De nition A subspace of Rn is a nonempty subset V of Rn which is closed under linear combinations. That is, if ~v1;~v2 2V, then ~v1 + ~v2 2V and k~v1 2V. The zero vector is always a subspace by itself. In R3, the lines through the origin and the planes through the origin are subspaces. So is R3 itself. The set of vectors ~x 2R4 such that x1 + x2 + x3 + x4 = 0 is a subspace of R4. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Vector Spaces 3 De nition A subspace of Rn is a nonempty subset V of Rn which is closed under linear combinations. That is, if ~v1;~v2 2V, then ~v1 + ~v2 2V and k~v1 2V. The zero vector is always a subspace by itself. In R3, the lines through the origin and the planes through the origin are subspaces. So is R3 itself. The set of vectors ~x 2R4 such that x1 + x2 + x3 + x4 = 0 is a subspace of R4. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nitions 4 Let T : Rn !Rm represented by an m n matrix A. De nition The image of T (or A), written imT or imA, is the set of all ~w 2Rm such that ~w = T(~v) for some ~v. The image of a matrix sometimes called the column space. De nition The kernel of T (or A), written ker T or ker A, is the set of all ~v 2Rn such that T(~v) = ~0. The kernel of a matrix sometimes called the nullspace. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nitions 4 Let T : Rn !Rm represented by an m n matrix A. De nition The image of T (or A), written imT or imA, is the set of all ~w 2Rm such that ~w = T(~v) for some ~v. The image of a matrix sometimes called the column space. De nition The kernel of T (or A), written ker T or ker A, is the set of all ~v 2Rn such that T(~v) = ~0. The kernel of a matrix sometimes called the nullspace. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nitions 4 Let T : Rn !Rm represented by an m n matrix A. De nition The image of T (or A), written imT or imA, is the set of all ~w 2Rm such that ~w = T(~v) for some ~v. The image of a matrix sometimes called the column space. De nition The kernel of T (or A), written ker T or ker A, is the set of all ~v 2Rn such that T(~v) = ~0. The kernel of a matrix sometimes called the nullspace. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Examples 5 Consider the orthogonal projection T1 of R2 onto the line y = x. The image of T1 is the line y = x itself; the kernel is the line y = x. Now consider T2, the rotation by 45 degrees counterclockwise. The image of T2 is all of R2; the kernel is just the zero vector. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Examples 5 Consider the orthogonal projection T1 of R2 onto the line y = x. The image of T1 is the line y = x itself; the kernel is the line y = x. Now consider T2, the rotation by 45 degrees counterclockwise. The image of T2 is all of R2; the kernel is just the zero vector. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Examples 5 Consider the orthogonal projection T1 of R2 onto the line y = x. The image of T1 is the line y = x itself; the kernel is the line y = x. Now consider T2, the rotation by 45 degrees counterclockwise. The image of T2 is all of R2; the kernel is just the zero vector. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Examples 5 Consider the orthogonal projection T1 of R2 onto the line y = x. The image of T1 is the line y = x itself; the kernel is the line y = x. Now consider T2, the rotation by 45 degrees counterclockwise. The image of T2 is all of R2; the kernel is just the zero vector. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 6 Key Fact The image of T : Rn !Rm is a subspace of Rm. Proof If ~w1 = T(~v1);~w2 = T(~v2), then T(~v1 + ~v2) = ~w1 + ~w2 and T(k~v1) = k~w1. Also, the image of a matrix is the space spanned by its columns (hence the term column space). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 6 Key Fact The image of T : Rn !Rm is a subspace of Rm. Proof If ~w1 = T(~v1);~w2 = T(~v2), then T(~v1 + ~v2) = ~w1 + ~w2 and T(k~v1) = k~w1. Also, the image of a matrix is the space spanned by its columns (hence the term column space). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 6 Key Fact The image of T : Rn !Rm is a subspace of Rm. Proof If ~w1 = T(~v1);~w2 = T(~v2), then T(~v1 + ~v2) = ~w1 + ~w2 and T(k~v1) = k~w1. Also, the image of a matrix is the space spanned by its columns (hence the term column space). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 7 Key Fact The kernel of T : Rn !Rm is a subspace of Rn. Proof If T(~v1) = T(~v2) = ~0, then T(~v1 + ~v2) = ~0 +~0 = ~0 and T(k~v1) = k~0 = ~0. Notice that ker A =f~0g means that the system A~x = ~0 has a unique solution. Thus ker A =f~0g is another way of saying A is invertible. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 7 Key Fact The kernel of T : Rn !Rm is a subspace of Rn. Proof If T(~v1) = T(~v2) = ~0, then T(~v1 + ~v2) = ~0 +~0 = ~0 and T(k~v1) = k~0 = ~0. Notice that ker A =f~0g means that the system A~x = ~0 has a unique solution. Thus ker A =f~0g is another way of saying A is invertible. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 7 Key Fact The kernel of T : Rn !Rm is a subspace of Rn. Proof If T(~v1) = T(~v2) = ~0, then T(~v1 + ~v2) = ~0 +~0 = ~0 and T(k~v1) = k~0 = ~0. Notice that ker A =f~0g means that the system A~x = ~0 has a unique solution. Thus ker A =f~0g is another way of saying A is invertible. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivation 8 Suppose I tell you that we are studying the space spanned by the following following vectors. 2 4 0 0 0 3 5; 2 4 1 1 1 3 5; 2 4 2 2 6 3 5; 2 4 3 3 0 3 5; 2 4 1 1 2 3 5; 2 4 1 1 0 3 5; 2 4 9 9 19 3 5; 2 4 0 0 2 3 5; 2 4 3 3 1 3 5 I hope you would laugh at me. There is obviously redundancy in the list. It is desirable to give spanning sets which do not have this redundancy. The rst step is to make precise what is wrong with this list. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivation 8 Suppose I tell you that we are studying the space spanned by the following following vectors. 2 4 0 0 0 3 5; 2 4 1 1 1 3 5; 2 4 2 2 6 3 5; 2 4 3 3 0 3 5; 2 4 1 1 2 3 5; 2 4 1 1 0 3 5; 2 4 9 9 19 3 5; 2 4 0 0 2 3 5; 2 4 3 3 1 3 5 I hope you would laugh at me. There is obviously redundancy in the list. It is desirable to give spanning sets which do not have this redundancy. The rst step is to make precise what is wrong with this list. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivation 8 Suppose I tell you that we are studying the space spanned by the following following vectors. 2 4 0 0 0 3 5; 2 4 1 1 1 3 5; 2 4 2 2 6 3 5; 2 4 3 3 0 3 5; 2 4 1 1 2 3 5; 2 4 1 1 0 3 5; 2 4 9 9 19 3 5; 2 4 0 0 2 3 5; 2 4 3 3 1 3 5 I hope you would laugh at me. There is obviously redundancy in the list. It is desirable to give spanning sets which do not have this redundancy. The rst step is to make precise what is wrong with this list. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivation 8 Suppose I tell you that we are studying the space spanned by the following following vectors. 2 4 0 0 0 3 5; 2 4 1 1 1 3 5; 2 4 2 2 6 3 5; 2 4 3 3 0 3 5; 2 4 1 1 2 3 5; 2 4 1 1 0 3 5; 2 4 9 9 19 3 5; 2 4 0 0 2 3 5; 2 4 3 3 1 3 5 I hope you would laugh at me. There is obviously redundancy in the list. It is desirable to give spanning sets which do not have this redundancy. The rst step is to make precise what is wrong with this list. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivation 8 Suppose I tell you that we are studying the space spanned by the following following vectors. 2 4 0 0 0 3 5; 2 4 1 1 1 3 5; 2 4 2 2 6 3 5; 2 4 3 3 0 3 5; 2 4 1 1 2 3 5; 2 4 1 1 0 3 5; 2 4 9 9 19 3 5; 2 4 0 0 2 3 5; 2 4 3 3 1 3 5 I hope you would laugh at me. There is obviously redundancy in the list. It is desirable to give spanning sets which do not have this redundancy. The rst step is to make precise what is wrong with this list. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 1 9 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. A relation is an equation c1~v1 + c2~v2 + + cm~vm = ~0; that is, a relation a way to write the zero vector as a linear combination of the given vectors. There is always the trivial relation with ci = 0. If there are no other relations, we say the set is linearly independent. If there are any nontrivial relations, we say the set is linearly dependent. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 1 9 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. A relation is an equation c1~v1 + c2~v2 + + cm~vm = ~0; that is, a relation a way to write the zero vector as a linear combination of the given vectors. There is always the trivial relation with ci = 0. If there are no other relations, we say the set is linearly independent. If there are any nontrivial relations, we say the set is linearly dependent. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 1 9 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. A relation is an equation c1~v1 + c2~v2 + + cm~vm = ~0; that is, a relation a way to write the zero vector as a linear combination of the given vectors. There is always the trivial relation with ci = 0. If there are no other relations, we say the set is linearly independent. If there are any nontrivial relations, we say the set is linearly dependent. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 1 9 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. A relation is an equation c1~v1 + c2~v2 + + cm~vm = ~0; that is, a relation a way to write the zero vector as a linear combination of the given vectors. There is always the trivial relation with ci = 0. If there are no other relations, we say the set is linearly independent. If there are any nontrivial relations, we say the set is linearly dependent. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 1 9 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. A relation is an equation c1~v1 + c2~v2 + + cm~vm = ~0; that is, a relation a way to write the zero vector as a linear combination of the given vectors. There is always the trivial relation with ci = 0. If there are no other relations, we say the set is linearly independent. If there are any nontrivial relations, we say the set is linearly dependent. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 2 10 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. Let V0 =f~0g;V1 = sp(~v1);V2 = sp(~v1;~v2);V3 = sp(~v1;~v2;~v3);::: Evidently V0 V1 V2 Vm 1 Vm. If Vi 1 = Vi for some i, we say ~vi was redundant. A set is linearly independent if no vectors are redundant (and linearly dependent otherwise. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 2 10 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. Let V0 =f~0g;V1 = sp(~v1);V2 = sp(~v1;~v2);V3 = sp(~v1;~v2;~v3);::: Evidently V0 V1 V2 Vm 1 Vm. If Vi 1 = Vi for some i, we say ~vi was redundant. A set is linearly independent if no vectors are redundant (and linearly dependent otherwise. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 2 10 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. Let V0 =f~0g;V1 = sp(~v1);V2 = sp(~v1;~v2);V3 = sp(~v1;~v2;~v3);::: Evidently V0 V1 V2 Vm 1 Vm. If Vi 1 = Vi for some i, we say ~vi was redundant. A set is linearly independent if no vectors are redundant (and linearly dependent otherwise. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 2 10 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. Let V0 =f~0g;V1 = sp(~v1);V2 = sp(~v1;~v2);V3 = sp(~v1;~v2;~v3);::: Evidently V0 V1 V2 Vm 1 Vm. If Vi 1 = Vi for some i, we say ~vi was redundant. A set is linearly independent if no vectors are redundant (and linearly dependent otherwise. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linear Independence, Take 2 10 Consider a list of vectors ~v1;~v2;:::;~vm in Rn. Let V0 =f~0g;V1 = sp(~v1);V2 = sp(~v1;~v2);V3 = sp(~v1;~v2;~v3);::: Evidently V0 V1 V2 Vm 1 Vm. If Vi 1 = Vi for some i, we say ~vi was redundant. A set is linearly independent if no vectors are redundant (and linearly dependent otherwise. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity A Useful Algorithm 11 Suppose you have a list of vectors, and you want to know if it is independent (and if not, which vectors are redundant). Form the matrix whose columns are the listed vectors, in order. Use GJE to put this matrix in rref. The columns containing pivots are irredundant. The columns not containing pivots are redundant. If there are relations, you can use the rref to help you see them. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity A Useful Algorithm 11 Suppose you have a list of vectors, and you want to know if it is independent (and if not, which vectors are redundant). Form the matrix whose columns are the listed vectors, in order. Use GJE to put this matrix in rref. The columns containing pivots are irredundant. The columns not containing pivots are redundant. If there are relations, you can use the rref to help you see them. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity A Useful Algorithm 11 Suppose you have a list of vectors, and you want to know if it is independent (and if not, which vectors are redundant). Form the matrix whose columns are the listed vectors, in order. Use GJE to put this matrix in rref. The columns containing pivots are irredundant. The columns not containing pivots are redundant. If there are relations, you can use the rref to help you see them. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity A Useful Algorithm 11 Suppose you have a list of vectors, and you want to know if it is independent (and if not, which vectors are redundant). Form the matrix whose columns are the listed vectors, in order. Use GJE to put this matrix in rref. The columns containing pivots are irredundant. The columns not containing pivots are redundant. If there are relations, you can use the rref to help you see them. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity A Useful Algorithm 11 Suppose you have a list of vectors, and you want to know if it is independent (and if not, which vectors are redundant). Form the matrix whose columns are the listed vectors, in order. Use GJE to put this matrix in rref. The columns containing pivots are irredundant. The columns not containing pivots are redundant. If there are relations, you can use the rref to help you see them. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Linearly Independent Sets 12 Let ~v1;:::;~vm be a list of vectors in Rn. Then the following are equivalent. 1 the vectors are linearly independent 2 no ~vk is in the span of the previous ~v1;:::;~vk 1 3 no ~vk is in the span of the other vectors ~v1;:::;~vk 1;~vk+1;:::;~vm 4 the only relation c1~v1 + + cm~vm = ~0 is the trivial one 5 ker ~v 1 ~v2 ~vm =f~0g 6 ~v 1 ~v2 ~vm has rank m M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, De nitions 13 Let V be a vector space (that is, a subspace of some Rn). De nition Then a basis of V (plural bases, BAY-seas) is a spanning set of V which is linearly independent. Alternative De nition A basis of V is a set ~v1;:::;~vk of vectors in V such that every vector ~w in V can be written ~w = c1~v1 + + ck~vk for unique scalars ci. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, De nitions 13 Let V be a vector space (that is, a subspace of some Rn). De nition Then a basis of V (plural bases, BAY-seas) is a spanning set of V which is linearly independent. Alternative De nition A basis of V is a set ~v1;:::;~vk of vectors in V such that every vector ~w in V can be written ~w = c1~v1 + + ck~vk for unique scalars ci. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, De nitions 13 Let V be a vector space (that is, a subspace of some Rn). De nition Then a basis of V (plural bases, BAY-seas) is a spanning set of V which is linearly independent. Alternative De nition A basis of V is a set ~v1;:::;~vk of vectors in V such that every vector ~w in V can be written ~w = c1~v1 + + ck~vk for unique scalars ci. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, Properties 14 Fundamental Fact Every vector space has a basis. It?s not usually hard to nd a basis for a favorite space. Begin with a nonzero vector. If the vectors you have don?t span the space, add a vector not in their span. Repeat until you can?t nd any vectors not in the span. Then you have a basis. Every linearly independent set can be extended to a basis; every spanning set can be pruned to a basis. The standard basis vectors ~ei form the most convenient basis of Rn, hence the name. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, Properties 14 Fundamental Fact Every vector space has a basis. It?s not usually hard to nd a basis for a favorite space. Begin with a nonzero vector. If the vectors you have don?t span the space, add a vector not in their span. Repeat until you can?t nd any vectors not in the span. Then you have a basis. Every linearly independent set can be extended to a basis; every spanning set can be pruned to a basis. The standard basis vectors ~ei form the most convenient basis of Rn, hence the name. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, Properties 14 Fundamental Fact Every vector space has a basis. It?s not usually hard to nd a basis for a favorite space. Begin with a nonzero vector. If the vectors you have don?t span the space, add a vector not in their span. Repeat until you can?t nd any vectors not in the span. Then you have a basis. Every linearly independent set can be extended to a basis; every spanning set can be pruned to a basis. The standard basis vectors ~ei form the most convenient basis of Rn, hence the name. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Bases, Properties 14 Fundamental Fact Every vector space has a basis. It?s not usually hard to nd a basis for a favorite space. Begin with a nonzero vector. If the vectors you have don?t span the space, add a vector not in their span. Repeat until you can?t nd any vectors not in the span. Then you have a basis. Every linearly independent set can be extended to a basis; every spanning set can be pruned to a basis. The standard basis vectors ~ei form the most convenient basis of Rn, hence the name. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Intermission Take ve. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivating Question 15 We know that for any matrix A or map T, the image and the kernel are subspaces of some Rn. What if we turn the question around? Given a subspace V, can we always think of V as the image of something? Given a subspace V, can we always think of V as the kernel of something? M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivating Question 15 We know that for any matrix A or map T, the image and the kernel are subspaces of some Rn. What if we turn the question around? Given a subspace V, can we always think of V as the image of something? Given a subspace V, can we always think of V as the kernel of something? M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivating Question 15 We know that for any matrix A or map T, the image and the kernel are subspaces of some Rn. What if we turn the question around? Given a subspace V, can we always think of V as the image of something? Given a subspace V, can we always think of V as the kernel of something? M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Motivating Question 15 We know that for any matrix A or map T, the image and the kernel are subspaces of some Rn. What if we turn the question around? Given a subspace V, can we always think of V as the image of something? Given a subspace V, can we always think of V as the kernel of something? M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example 16 Consider the subspace V of R4 given by x1 = x2;x3 = x4. How to express it as a kernel? Or an image? First, V consists of all vectors ~x such that 1 1 0 0 ~x and 0 0 1 1 ~x = 0. Putting this together, V consists of all vectors such that 1 1 0 0 0 0 1 1 ~x = 0 0 . That is, V = ker 1 1 0 0 0 0 1 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example 16 Consider the subspace V of R4 given by x1 = x2;x3 = x4. How to express it as a kernel? Or an image? First, V consists of all vectors ~x such that 1 1 0 0 ~x and 0 0 1 1 ~x = 0. Putting this together, V consists of all vectors such that 1 1 0 0 0 0 1 1 ~x = 0 0 . That is, V = ker 1 1 0 0 0 0 1 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example 16 Consider the subspace V of R4 given by x1 = x2;x3 = x4. How to express it as a kernel? Or an image? First, V consists of all vectors ~x such that 1 1 0 0 ~x and 0 0 1 1 ~x = 0. Putting this together, V consists of all vectors such that 1 1 0 0 0 0 1 1 ~x = 0 0 . That is, V = ker 1 1 0 0 0 0 1 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example 16 Consider the subspace V of R4 given by x1 = x2;x3 = x4. How to express it as a kernel? Or an image? First, V consists of all vectors ~x such that 1 1 0 0 ~x and 0 0 1 1 ~x = 0. Putting this together, V consists of all vectors such that 1 1 0 0 0 0 1 1 ~x = 0 0 . That is, V = ker 1 1 0 0 0 0 1 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example, Continued 17 To express V as an image, rst write down a basis. (Actually any spanning set would work.) V is spanned by 2 66 4 1 1 0 0 3 77 5 and 2 66 4 0 0 1 1 3 77 5; each element of V has the form c1 2 66 4 1 1 0 0 3 77 5+ c2 2 66 4 0 0 1 1 3 77 5 = 2 66 4 1 0 1 0 0 1 0 1 3 77 5 c 1 c2 . That is, V = im 2 66 4 1 0 1 0 0 1 0 1 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example, Continued 17 To express V as an image, rst write down a basis. (Actually any spanning set would work.) V is spanned by 2 66 4 1 1 0 0 3 77 5 and 2 66 4 0 0 1 1 3 77 5; each element of V has the form c1 2 66 4 1 1 0 0 3 77 5+ c2 2 66 4 0 0 1 1 3 77 5 = 2 66 4 1 0 1 0 0 1 0 1 3 77 5 c 1 c2 . That is, V = im 2 66 4 1 0 1 0 0 1 0 1 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity First Example, Continued 17 To express V as an image, rst write down a basis. (Actually any spanning set would work.) V is spanned by 2 66 4 1 1 0 0 3 77 5 and 2 66 4 0 0 1 1 3 77 5; each element of V has the form c1 2 66 4 1 1 0 0 3 77 5+ c2 2 66 4 0 0 1 1 3 77 5 = 2 66 4 1 0 1 0 0 1 0 1 3 77 5 c 1 c2 . That is, V = im 2 66 4 1 0 1 0 0 1 0 1 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example 18 Now consider the subspace W of R4 spanned by 2 3 4 and 1 1 1 . How to express it as a kernel? Or an image? First, W consists of all vectors have the form c1 2 4 2 3 4 3 5+ c2 2 4 1 1 1 3 5 = 2 4 2 1 3 1 4 1 3 5 c 1 c2 . That is, V = im 2 4 2 1 3 1 4 1 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example 18 Now consider the subspace W of R4 spanned by 2 3 4 and 1 1 1 . How to express it as a kernel? Or an image? First, W consists of all vectors have the form c1 2 4 2 3 4 3 5+ c2 2 4 1 1 1 3 5 = 2 4 2 1 3 1 4 1 3 5 c 1 c2 . That is, V = im 2 4 2 1 3 1 4 1 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example 18 Now consider the subspace W of R4 spanned by 2 3 4 and 1 1 1 . How to express it as a kernel? Or an image? First, W consists of all vectors have the form c1 2 4 2 3 4 3 5+ c2 2 4 1 1 1 3 5 = 2 4 2 1 3 1 4 1 3 5 c 1 c2 . That is, V = im 2 4 2 1 3 1 4 1 3 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example, Continued 19 It?s a bit harder to see why W is a kernel. W consists of all the vectors ~w such that 2 4 2 1 3 1 4 1 3 5~x = ~w is consistent. Row-reducing 2 4 2 1 w1 3 1 w2 4 1 w3 3 5 gives 2 4 1 0 w1 + w2 0 1 3w1 2w2 0 0 w1 2w2 + w3 3 5 This can be consistent only if w1 2w2 + w3 = 0. W = ker 1 2 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example, Continued 19 It?s a bit harder to see why W is a kernel. W consists of all the vectors ~w such that 2 4 2 1 3 1 4 1 3 5~x = ~w is consistent. Row-reducing 2 4 2 1 w1 3 1 w2 4 1 w3 3 5 gives 2 4 1 0 w1 + w2 0 1 3w1 2w2 0 0 w1 2w2 + w3 3 5 This can be consistent only if w1 2w2 + w3 = 0. W = ker 1 2 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example, Continued 19 It?s a bit harder to see why W is a kernel. W consists of all the vectors ~w such that 2 4 2 1 3 1 4 1 3 5~x = ~w is consistent. Row-reducing 2 4 2 1 w1 3 1 w2 4 1 w3 3 5 gives 2 4 1 0 w1 + w2 0 1 3w1 2w2 0 0 w1 2w2 + w3 3 5 This can be consistent only if w1 2w2 + w3 = 0. W = ker 1 2 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example, Continued 19 It?s a bit harder to see why W is a kernel. W consists of all the vectors ~w such that 2 4 2 1 3 1 4 1 3 5~x = ~w is consistent. Row-reducing 2 4 2 1 w1 3 1 w2 4 1 w3 3 5 gives 2 4 1 0 w1 + w2 0 1 3w1 2w2 0 0 w1 2w2 + w3 3 5 This can be consistent only if w1 2w2 + w3 = 0. W = ker 1 2 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Second Example, Continued 19 It?s a bit harder to see why W is a kernel. W consists of all the vectors ~w such that 2 4 2 1 3 1 4 1 3 5~x = ~w is consistent. Row-reducing 2 4 2 1 w1 3 1 w2 4 1 w3 3 5 gives 2 4 1 0 w1 + w2 0 1 3w1 2w2 0 0 w1 2w2 + w3 3 5 This can be consistent only if w1 2w2 + w3 = 0. W = ker 1 2 1 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 20 Generally, when you are given a subspace of Rn, it is given (implicitly) either as an image or a kernel. Whenever you are given a spanning set for a subspace, your subspace is really de ned as an image. Whenever you are given de ning relations, your subspace is really de ned as a kernel. In fact, every subspace V is the image of some matrix and the kernel of some matrix. Gauss-Jordan Elimination (of course) is the key to convert between the two perspectives. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 20 Generally, when you are given a subspace of Rn, it is given (implicitly) either as an image or a kernel. Whenever you are given a spanning set for a subspace, your subspace is really de ned as an image. Whenever you are given de ning relations, your subspace is really de ned as a kernel. In fact, every subspace V is the image of some matrix and the kernel of some matrix. Gauss-Jordan Elimination (of course) is the key to convert between the two perspectives. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 20 Generally, when you are given a subspace of Rn, it is given (implicitly) either as an image or a kernel. Whenever you are given a spanning set for a subspace, your subspace is really de ned as an image. Whenever you are given de ning relations, your subspace is really de ned as a kernel. In fact, every subspace V is the image of some matrix and the kernel of some matrix. Gauss-Jordan Elimination (of course) is the key to convert between the two perspectives. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 20 Generally, when you are given a subspace of Rn, it is given (implicitly) either as an image or a kernel. Whenever you are given a spanning set for a subspace, your subspace is really de ned as an image. Whenever you are given de ning relations, your subspace is really de ned as a kernel. In fact, every subspace V is the image of some matrix and the kernel of some matrix. Gauss-Jordan Elimination (of course) is the key to convert between the two perspectives. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 20 Generally, when you are given a subspace of Rn, it is given (implicitly) either as an image or a kernel. Whenever you are given a spanning set for a subspace, your subspace is really de ned as an image. Whenever you are given de ning relations, your subspace is really de ned as a kernel. In fact, every subspace V is the image of some matrix and the kernel of some matrix. Gauss-Jordan Elimination (of course) is the key to convert between the two perspectives. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Dimension of a Vector Space 21 You may have noticed that the same vector space can have lots of bases. For example, the subspace V of R3 de ned by x1 + x2 + x3 = 0 has all of the following as bases: B1 = 8< : 2 4 1 1 0 3 5; 2 4 1 0 1 3 5 9= ;, B2 = 8< : 2 4 3 2 5 3 5; 2 4 0 3 3 3 5 9= ;, B3 = 8< : 2 4 2 3 1 3 5; 2 4 1 3 2 3 5 9= ;. One important thing that all the bases of a vector space have in common is the number of elements in the basis. Every basis of the vector space V above has two elements in it. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Dimension of a Vector Space 21 You may have noticed that the same vector space can have lots of bases. For example, the subspace V of R3 de ned by x1 + x2 + x3 = 0 has all of the following as bases: B1 = 8< : 2 4 1 1 0 3 5; 2 4 1 0 1 3 5 9= ;, B2 = 8< : 2 4 3 2 5 3 5; 2 4 0 3 3 3 5 9= ;, B3 = 8< : 2 4 2 3 1 3 5; 2 4 1 3 2 3 5 9= ;. One important thing that all the bases of a vector space have in common is the number of elements in the basis. Every basis of the vector space V above has two elements in it. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Dimension of a Vector Space 21 You may have noticed that the same vector space can have lots of bases. For example, the subspace V of R3 de ned by x1 + x2 + x3 = 0 has all of the following as bases: B1 = 8< : 2 4 1 1 0 3 5; 2 4 1 0 1 3 5 9= ;, B2 = 8< : 2 4 3 2 5 3 5; 2 4 0 3 3 3 5 9= ;, B3 = 8< : 2 4 2 3 1 3 5; 2 4 1 3 2 3 5 9= ;. One important thing that all the bases of a vector space have in common is the number of elements in the basis. Every basis of the vector space V above has two elements in it. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Dimension of a Vector Space 21 You may have noticed that the same vector space can have lots of bases. For example, the subspace V of R3 de ned by x1 + x2 + x3 = 0 has all of the following as bases: B1 = 8< : 2 4 1 1 0 3 5; 2 4 1 0 1 3 5 9= ;, B2 = 8< : 2 4 3 2 5 3 5; 2 4 0 3 3 3 5 9= ;, B3 = 8< : 2 4 2 3 1 3 5; 2 4 1 3 2 3 5 9= ;. One important thing that all the bases of a vector space have in common is the number of elements in the basis. Every basis of the vector space V above has two elements in it. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nition: Dimension 22 Key De nition If V is a subspace of Rn the dimension of V, written dimV, is the number of objects in a basis of V. Key Theorem That makes sense. In R3, lines through the origin are 1-dimensional, planes through the origin are 2-dimensional, and the space itself is 3-dimensional. (The one-point space is 0-dimensional.) In general, Rn is n-dimensional. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nition: Dimension 22 Key De nition If V is a subspace of Rn the dimension of V, written dimV, is the number of objects in a basis of V. Key Theorem That makes sense. In R3, lines through the origin are 1-dimensional, planes through the origin are 2-dimensional, and the space itself is 3-dimensional. (The one-point space is 0-dimensional.) In general, Rn is n-dimensional. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nition: Dimension 22 Key De nition If V is a subspace of Rn the dimension of V, written dimV, is the number of objects in a basis of V. Key Theorem That makes sense. In R3, lines through the origin are 1-dimensional, planes through the origin are 2-dimensional, and the space itself is 3-dimensional. (The one-point space is 0-dimensional.) In general, Rn is n-dimensional. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity De nition: Dimension 22 Key De nition If V is a subspace of Rn the dimension of V, written dimV, is the number of objects in a basis of V. Key Theorem That makes sense. In R3, lines through the origin are 1-dimensional, planes through the origin are 2-dimensional, and the space itself is 3-dimensional. (The one-point space is 0-dimensional.) In general, Rn is n-dimensional. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Fundamental Properties 23 Suppose that V is a subspace of Rn and dimV = m. Then we know all of the following: ! No linearly independent subset of V can have more than m vectors. ! No spanning set of V can have fewer than m vectors. ! If a linearly independent subset of V has m vectors, then it is also spanning (i.e., it is a basis). ! If a spanning set of V has m vectors, then it is also linearly independent (i.e., it is a basis). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Fundamental Properties 23 Suppose that V is a subspace of Rn and dimV = m. Then we know all of the following: ! No linearly independent subset of V can have more than m vectors. ! No spanning set of V can have fewer than m vectors. ! If a linearly independent subset of V has m vectors, then it is also spanning (i.e., it is a basis). ! If a spanning set of V has m vectors, then it is also linearly independent (i.e., it is a basis). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Fundamental Properties 23 Suppose that V is a subspace of Rn and dimV = m. Then we know all of the following: ! No linearly independent subset of V can have more than m vectors. ! No spanning set of V can have fewer than m vectors. ! If a linearly independent subset of V has m vectors, then it is also spanning (i.e., it is a basis). ! If a spanning set of V has m vectors, then it is also linearly independent (i.e., it is a basis). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Fundamental Properties 23 Suppose that V is a subspace of Rn and dimV = m. Then we know all of the following: ! No linearly independent subset of V can have more than m vectors. ! No spanning set of V can have fewer than m vectors. ! If a linearly independent subset of V has m vectors, then it is also spanning (i.e., it is a basis). ! If a spanning set of V has m vectors, then it is also linearly independent (i.e., it is a basis). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Fundamental Properties 23 Suppose that V is a subspace of Rn and dimV = m. Then we know all of the following: ! No linearly independent subset of V can have more than m vectors. ! No spanning set of V can have fewer than m vectors. ! If a linearly independent subset of V has m vectors, then it is also spanning (i.e., it is a basis). ! If a spanning set of V has m vectors, then it is also linearly independent (i.e., it is a basis). M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity An Important Problem Type (Ex 3.3.25) 24 Find bases for the image and the kernel of the following matrix. A = 2 66 4 1 2 3 2 1 3 6 9 6 3 1 2 4 1 2 2 4 9 1 2 3 77 5 . First, row-reduce. rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity An Important Problem Type (Ex 3.3.25) 24 Find bases for the image and the kernel of the following matrix. A = 2 66 4 1 2 3 2 1 3 6 9 6 3 1 2 4 1 2 2 4 9 1 2 3 77 5 . First, row-reduce. rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity An Important Problem Type (Ex 3.3.25) 24 Find bases for the image and the kernel of the following matrix. A = 2 66 4 1 2 3 2 1 3 6 9 6 3 1 2 4 1 2 2 4 9 1 2 3 77 5 . First, row-reduce. rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 25 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 This says that the second, fourth, and fth columns are redundant. The column space is spanned by the rst and third columns imA has a basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 25 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 This says that the second, fourth, and fth columns are redundant. The column space is spanned by the rst and third columns imA has a basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 25 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 This says that the second, fourth, and fth columns are redundant. The column space is spanned by the rst and third columns imA has a basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Image 25 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 This says that the second, fourth, and fth columns are redundant. The column space is spanned by the rst and third columns imA has a basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 26 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 The kernel contains the solutions to the system A~x = ~0. The solutions are given by x1 = 2x2 5x4 + 2x5;x3 = x4 x5 A basis for the solution space 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 26 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 The kernel contains the solutions to the system A~x = ~0. The solutions are given by x1 = 2x2 5x4 + 2x5;x3 = x4 x5 A basis for the solution space 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 26 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 The kernel contains the solutions to the system A~x = ~0. The solutions are given by x1 = 2x2 5x4 + 2x5;x3 = x4 x5 A basis for the solution space 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity The Kernel 26 rref(A) = 2 66 4 1 2 0 5 2 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 3 77 5 The kernel contains the solutions to the system A~x = ~0. The solutions are given by x1 = 2x2 5x4 + 2x5;x3 = x4 x5 A basis for the solution space 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 27 ker A is a three-dimensional space with basis 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . imA is a two-dimensional space with basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. dim ker A is called the nullity of A. Here it is 3. dim imA is called the rank of A. Here it is 2. We?ll have much more to say about such things in Day 07. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 27 ker A is a three-dimensional space with basis 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . imA is a two-dimensional space with basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. dim ker A is called the nullity of A. Here it is 3. dim imA is called the rank of A. Here it is 2. We?ll have much more to say about such things in Day 07. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 27 ker A is a three-dimensional space with basis 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . imA is a two-dimensional space with basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. dim ker A is called the nullity of A. Here it is 3. dim imA is called the rank of A. Here it is 2. We?ll have much more to say about such things in Day 07. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 27 ker A is a three-dimensional space with basis 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . imA is a two-dimensional space with basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. dim ker A is called the nullity of A. Here it is 3. dim imA is called the rank of A. Here it is 2. We?ll have much more to say about such things in Day 07. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity Conclusion 27 ker A is a three-dimensional space with basis 2 66 66 4 2 1 0 0 0 3 77 77 5 ; 2 66 66 4 5 0 1 1 0 3 77 77 5 ; 2 66 66 4 2 0 1 0 1 3 77 77 5 . imA is a two-dimensional space with basis 2 66 4 1 3 1 2 3 77 5; 2 66 4 3 9 4 9 3 77 5. dim ker A is called the nullity of A. Here it is 3. dim imA is called the rank of A. Here it is 2. We?ll have much more to say about such things in Day 07. M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces Linear Spans Images and Kernels Bases and Linear Independence Expressing a Subspace as an Image or Kernel Introduction to Dimension Rank and Nullity xx3.1{3.3: Introducing . . . Vector Spaces Math 214-001/002 W10 Day 6 M. Cap Khoury 2 February 2010 M. Cap Khoury xx3.1{3.3: Introducing . . . Vector Spaces M. Cap Khoury §§3.1--3.3: Introducing Vector Spaces - Math 214-001/002 W10 Day 6

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Author: Prateek G.

Textbook: Linear Algebra with Applications (4th Edition)

Created: 2010-05-15

Updated: 2010-05-15

File Size: 115 page(s)

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Textbook: Linear Algebra with Applications (4th Edition)

Created: 2010-05-15

Updated: 2010-05-15

File Size: 115 page(s)

Keywords: flash card flashcards digital flashcards note sharing notes textbook wiki college dorm class classroom exam homework test quiz university college education learn student teachers tutors share, study blue studyblue studyblu

Views: 12

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