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Given an $m\times n$-matrix over some field $F$, e.g. $F=\mathbb{R}$, take row rank to denote the dimension of the row subspace of $F^n$ and column rank to be the dimension of the column subspace of $F^m$.

Is there an argument for row rank = column rank for the class of all matrices, possibly for $m=n$ only, which reduces the proof to that for some class of symmetric matrices (for which it is trivial)?

I.e. something along the lines: Let $A\in F^{m\times n}$. We want to show row rank(A)= column rank(A). Since [...], it suffices to prove the equality for the (suitable) associated symmetric matrix $S_A\in F^{k\times k}$, where $k=k(m, n)$ (some suitable function). However, this is clear from $S_A^T=S_A$.

That would be quite elegant. Just wondering.

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  • $\begingroup$ You can prove that the elementary operations on rows (columns) don't change either the row or the column rank. Thus, you can reduce the matrix to the diagonal form (going e.g. through row echelon form), where it is obvious that the rank is the count of non-zero elements on the diagonal. This happens to work for rectangular matrices which are not square, whereas I assume any proof using symmetric matrices would work only for square mateices. $\endgroup$ – user491874 Dec 29 '17 at 18:50
  • $\begingroup$ Yes. But that proof reduces the equality to a particular matrix, for which the equality is trivial, not to the (abstract) property of symmetry. That's the difference I had in mind. $\endgroup$ – Damian Reding Dec 29 '17 at 18:52
  • $\begingroup$ Similar to proving existence of Jordan normal form by reducing the statement to nilpotent matrices/maps. $\endgroup$ – Damian Reding Dec 29 '17 at 18:53

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