(adsbygoogle = window.adsbygoogle || []).push({}); Theorem:Let A be an m x n matrix. If P and Q are invertible m x m and n x n matrices, respectively, then

(a.) rank(AQ) = rank(A)

(b.) rank(PA) = rank(A)

(c.) rank(PAQ) = rank(A)

Proof:

[tex]R(L_A_Q)[/tex] = [tex]R(L_AL_Q)[/tex] = [tex]L_AL_Q(F^n)[/tex] = [tex]L_A(L_Q(F^n)) [/tex]= [tex]L_A(F^n)[/tex] = [tex]R(L_A)[/tex]

since [tex]L_Q[/tex] is onto. Therefore,

rank(AQ) = dim(R([tex]L_A_Q[/tex])) = dim(R([tex]L_A[/tex])) = rank(A). (#1)

Question1:How is [tex]L_Q[/tex] onto?

Question2:How does the onto-ness imply (#1)?

Question3:Can anyone help me/supply ideas for the proof for parts (b.) and (c.) of the theorem?

NOTE: the symbol R denotes the terminology of images.

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# The rank of a matrix

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