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Is it possible to give an intuitive/elementary proof of the theorem that says that the row rank of a (finite-dimensional) square matrix matrix equals its column rank?

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marked as duplicate by Douglas S. Stones, Johannes Kloos, Vedran Šego, Lord_Farin, azimut Nov 4 '13 at 19:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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Here are two: en.wikipedia.org/wiki/… –  vadim123 May 6 '13 at 13:44
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The right understanding, I think, is what's in the second proof at the link @vadim123 provided. The point is that multiplying by the matrix $A$ (applying the linear map) gives a one-to-one correspondence between the row space of $A$ and the column space of $A$. –  Ted Shifrin May 6 '13 at 14:42
    
@robjohn: You're right. Sorry about that, and thanks for pointing it out. –  kjo Jun 9 '13 at 12:31

1 Answer 1

suppose T be a linear translation such that $T(x)=Ax$ and A be a m*n matrix.

$T(x)=A_1x_1+A_2x_2+....+A_nx_n$so

rank(T)=rank column space of A

in other hand :

$Rank(T)+null(T)=n$ since (T is a linear translate form $F^n \to F^{m}$)

$null(T)=\{x, Ax=0\}$ so

$dim(null(T))=n-$ rank 's row space of A

with attention to :

$Rank(T)+null(T)=n$

we will have :

$rank column space of A+n -rank row space of A=n$

so : rank column space of A=rank row space of A

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