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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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    $\begingroup$ Here are two: en.wikipedia.org/wiki/… $\endgroup$ – vadim123 May 6 '13 at 13:44
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    $\begingroup$ 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 $A$. $\endgroup$ – Ted Shifrin May 6 '13 at 14:42
  • $\begingroup$ @robjohn: You're right. Sorry about that, and thanks for pointing it out. $\endgroup$ – kjo Jun 9 '13 at 12:31
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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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    $\begingroup$ Im sorry, when you say "Linear translate" and "linear translation"... do you mean linear transformation? Perhaps I'm missing something. $\endgroup$ – Merkh Aug 10 '16 at 13:48

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