Department

MathematicsCourse Code

MATH 33AHProfessor

AllStudy Guide

FinalThis

**preview**shows pages 1-2. to view the full**8 pages of the document.**4 Images, Kernels, and Subspaces

In our study of linear transformations we’ve examined some of the conditions under which a

transformation is invertible. Now we’re ready to investigate some ideas similar to invertibility.

Namely, we would like to measure the ways in which a transformation that is not invertible

fails to have an inverse.

4.1 The Image and Kernel of a Linear Transformation

Deﬁnition. The image of a function consists of all the values the function assumes. If

f:X→Yis a function from Xto Y, then

im(f) = {f(x) : x∈X}.

Notice that im(f) is a subset of Y.

Deﬁnition. The kernel of a function whose range is Rnconsists of all the values in its

domain at which the function assumes the value 0. If f:X→Rnis a function from Xto

Rn, then

ker(f) = {x∈X:f(x) = 0}.

Notice that ker(f) is a subset of X. Also, if T(x) = Axis a linear transformation from Rm

to Rn, then ker(T) (also denoted ker(A)) is the set of solutions to the equation Ax=0.

The kernel gives us some new ways to characterize invertible matrices.

Theorem 1. Let Abe an n×nmatrix. Then the following statements are equivalent.

1. Ais invertible.

2. The linear system Ax=bhas a unique solution xfor every b∈Rn.

3. rref(A) = In.

4. rank(A) = n.

5. im(A) = Rn.

6. ker(A) = {0}.

Example 13. (§3.1, Exercise 39 of [1]) Consider an n×pmatrix Aand a p×mmatrix B.

(a) What is the relationship between ker(AB) and ker(B)? Are they always equal? Is one

of them always contained in the other?

(b) What is the relationship between im(A) and im(AB)?

(Solution)

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(a) Recall that ker(AB) is the set of vectors x∈Rmfor which ABx= 0, and similarly

that ker(B) is the set of vectors x∈Rmfor which Bx= 0. Now if xis in ker(B), then

Bx= 0, so ABx= 0. This means that xis in ker(AB), so we see that ker(B) must

always be contained in ker(AB). On the other hand, ker(AB) might not be a subset

of ker(B). For instance, suppose that

A=0 0

0 0and B=1 0

0 1.

Then Bis the identity matrix, so ker(B) = {0}. But every vector has image zero under

AB, so ker(AB) = R2. Certainly ker(B) does not contain ker(AB) in this case.

(b) Suppose yis in the image of AB. Then y=ABxfor some x∈Rm. That is,

y=ABx=A(Bx),

so yis the image of Bxunder multiplication by A, and is thus in the image of A. So

im(A) contains im(AB). On the other hand, consider

A=1 0

0 0and B=0 0

0 1.

Then

AB =0 0

0 0,

so im(AB) = {0}, but the image of Ais the span of the vector 1

0. So im(AB) does

not necessarily contain im(A).

♦

Example 14. (§3.1, Exercise 48 of [1]) Consider a 2 ×2 matrix Awith A2=A.

(a) If wis in the image of A, what is the relationship between wand Aw?

(b) What can you say about Aif rank(A) = 2? What if rank(A) = 0?

(c) If rank(A) = 1, show that the linear transformation T(x) = Axis the projection onto

im(A) along ker(A).

(Solution)

(a) If wis in the image of A, then w=Avfor some v∈R2. Then

Aw=A(Av) = A2v=Av=w,

since A2=A. So Aw=w.

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(b) If rank(A) = 2, then Ais invertible. Since A2=A, we see that

A=I2A= (A−1A)A=A−1A2=A−1A=I2.

So the only rank 2 2 ×2 matrix with the property that A2=Ais the identity matrix.

On the other hand, if rank(A) = 0 then Amust be the zero matrix.

(c) If rank(A) = 1, then Ais not invertible, so ker(A)6={0}. But we also know that A

is not the zero matrix, so ker(A)6=R2. We conclude that ker(A) must be a line in

R2. Next, suppose we have w∈ker(A)∩im(A). Then Aw= 0 and, according to part

(a), Aw=w. So wis the zero vector, meaning that ker(A)∩im(A) = {0}. Since

im(A) is neither 0 nor all of R2, it also must be a line in R2. So ker(A) and im(A)

are non-parallel lines in R2. Now choose x∈R2and let w=x−Ax. Notice that

Aw=Ax−A2x= 0, so w∈ker(A). Then we may write xas the sum of an element

of im(A) and an element of ker(A):

x=Ax+w.

According to Exercise 2.2.33, the map T(x) = Axis then the projection onto im(A)

along ker(A).

♦

Example 15. (§3.1, Exercise 50 of [1]) Consider a square matrix Awith ker(A2) = ker(A3).

Is ker(A3) = ker(A4)? Justify your answer.

(Solution) Suppose x∈ker(A3). Then A3x=0, so

A4x=A(A3x) = A0=0,

meaning that x∈ker(A4). So ker(A3) is contained in ker(A4). On the other hand, suppose

x∈ker(A4). Then A4x=0, so A3(Ax) = 0. This means that Axis in the kernel of A3, and

thus in ker(A2). So

A3x=A2(Ax) = 0,

meaning that x∈ker(A3). So ker(A4) is contained in ker(A3). Since each set contains the

other, the two are equal: ker(A3) = ker(A4). ♦

4.2 Subspaces

Deﬁnition. A subset Wof the vector space Rnis called a subspace of Rnif it

(i) contains the zero vector;

(ii) is closed under vector addition;

(iii) is closed under scalar multiplication.

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