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My problem:

Proof that in every matrix $A \in M_{mn}(\Bbb F)$ exists a square matrix with the degree $r(A)$ (a rank) which is invertible, that matrix is $B$.

My attempt:

I am not sure how to prove it properly, my proof is like a circle I think, so I will be happy for another suggestions.

For $A$ we have: $\;$ $r(A) + n(A) = n$ (rank-nullity theorem)

  • $n(A)$ is a nullity.

For $B$ we have: $\:$ $r(B) + n(B) = r(A)$, because $B$ has the degree of $r(A)$.

Then we have $n - n(A) = r(B) + n(B)$ and from this equation we see that if $r(A) = r(B)$, then $n(B) = 0$ so $B$ is regular matrix.

I think I didn't prove anything.

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    What is the "degree" of a matrix? Is that the size? – 2017-02-11
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    I think it is.. – 2017-02-11
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    What is $B$ in your proof? You don't define it. – 2017-02-11

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If $A$ is an $m\times n$ matrix of rank $r$, there exist $r$ columns that form a linearly independent set. So consider the submatrix $B$ formed by those $r$ columns. Since it has rank $r$, there must exist $r$ rows that form a linearly independent set. The required square submatrix is the one consisting of those linearly independent rows.

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    I was thinking about this too, but is it enough? – 2017-02-11
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    @Leif By construction, the obtained matrix has linearly independent rows, so it is invertible, being square. – 2017-02-12