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Chapter 5: Q42E (page 249)

If Ais any matrix, show that the linear transformation L(x鈫�)=Ax鈫�from im(AT)to im(A)is an isomorphism. This provides yet another proof of the formula rank(A)=rank(AT).

Short Answer

Expert verified

L is an isomorphism.

Step by step solution

01

L is an isomorphism.

Let v鈭�im(AT)such that v鈭�ker(L). Then it has ATw=v for some w. Also Lv=0implies Av=0.

Thus, it has AATw=0.

If it multiplies both side of this equation by wTfrom the left, then it gets

wTA(ATw)=0

Which implies that (ATW)-(AT(W)=0.

Then it gets ATW2=0.

Thus, kernel is Ker(L)=0.

So, L is injective. now it knows thatimA鈯�=kerAIfor any matrix A. so if it applies this formula toATthen,(im(A))鈯�=ker(A).

Also, it knows that Rn=im(AT)鈯�(im(A))鈯�.

Which implies that Rn=im(AT)鈯�ker(A).

From the above terms it gets W=W1+W2.

Wherew1鈭�im(AT),w2鈭�ker(A)

Then, it is written as,

V=Aw=Aw1+Aw2=Aw1+0=Aw1

Thus, it gets v=AW1, where w1鈭�imAT. So,Aw1=L(w1) and v鈭�im(L).

Hence, L is subjective and an isomorphism.

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