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MATH 33AH Lecture Notes - Lecture 26: Invertible Matrix, Linear Map, Solution SetExam


Department
Mathematics
Course Code
MATH 33AH
Professor
All
Study Guide
Final

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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:XYis a function from Xto Y, then
im(f) = {f(x) : xX}.
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:XRnis a function from Xto
Rn, then
ker(f) = {xX: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 bRn.
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 xRmfor which ABx= 0, and similarly
that ker(B) is the set of vectors xRmfor 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 xRm. 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 vR2. 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= (A1A)A=A1A2=A1A=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 wker(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 xR2and let w=xAx. Notice that
Aw=AxA2x= 0, so wker(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 xker(A3). Then A3x=0, so
A4x=A(A3x) = A0=0,
meaning that xker(A4). So ker(A3) is contained in ker(A4). On the other hand, suppose
xker(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 xker(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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