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, 2013 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 $A$. $\endgroup$ May 6, 2013 at 14:42

1 Answer 1


Let $\mathbb F$ be a field and let $T: \mathbb F^n \to \mathbb F^m$ be a linear transformation. Then there exists an $m$ by $n$ matrix $A$ such that $T(x)=Ax.$ Let $A_i \text{ (where }i = 1,...,n)$ be the $i$th column of $A$. Then

$T(x)=A_1x_1+A_2x_2+....+A_nx_n$, so that $\text{rank}(T) = $ dimension of the subspace spanned by the columns of $A$. On the other hand,$\text{rank}(T)+\text{null}(T)=n$ since $T$ is a linear transformation from $\mathbb F^n$ to $\mathbb F^m$.

By definition, the $\text{null}(T)$ is the dimension of the nullspace of $T$, where the nullspace of $T$ is $\{x \in \mathbb F^n : Ax=0\}$. Thus $\text{null}(T)= n- \text{dimension of row space of A}$. By rank nullity theorem, we thus have $\text{column space of A} + n - \text{row space of }A = n$. Thus, dimension of the column space of $A$ is equal to the dimension of the 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, 2016 at 13:48
  • $\begingroup$ It's unclear about what do you mean $null(T)$? null space or the nullity? $\endgroup$ Sep 20, 2020 at 12:07
  • $\begingroup$ You define $rank(T)$ as rank of column space of $A$ and you said $\dim(null(T))=n-\textrm{rank of row space of }A$, which implicitly use "row rank=column rank". $\endgroup$ Sep 20, 2020 at 12:46
  • $\begingroup$ You didn't prove it. $\endgroup$ Sep 20, 2020 at 12:51

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