When reading a text, I came across a statement saying
"the rank of an $m\times n$ matrix is $r$ if and only if all $(n-r+1)\times(n-r+1)$. minors vanish"
Could anyone explain what it means by a minor, and how to prove the statement?
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A minor is the determinant of a square submatrix. However the statement given is not valid.
Consider a $1\times 2$ matrix, $[0\quad 1]$. Clearly this matrix has rank 1.
The above assertion says this is so if and only all $2\times 2$ minors vanish. There are none, so one might be tempted to say the criterion is satisfied "vacuously". However then it would also be true for rank $r=0$, which is inconsistent with the definition of rank of a matrix being the dimension of its row space (equiv. dimension of its column space).
A correct statement would be that an $m\times n$ matrix has rank $r$ if and only if some $r\times r$ minor does not vanish and every $(r+1)\times (r+1)$ minor does vanish, i.e. $r$ is the largest number such that some $r\times r$ minor does not vanish (is not zero).
For this to work we need the technical convention that a $0\times 0$ minor is $1$, i.e. that the $0\times 0$ matrix has determinant $1$. Such a convention is consistent with the notion of an empty product being $1$, though it may strike some as counterintuitive.