minors and rank of a matrix

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?

• The claim is at odds with the usual definition of $k\times k$ minor (see my Answer), and does not seem capable even of an alternative interpretation that would make it strictly true. Can you provide the name of the text where you found it? – hardmath Nov 4 '14 at 21:50

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.

• Also in the original question if you put $m=r=n$ you get that an $n \times n$ matrix has rank $n$ if and only if all its $1\times 1$ minors vanish - which is clearly nonsense. – Mark Bennet Nov 4 '14 at 16:31
• @MarkBennet: That is a good point! I've added a correct statement about ranks and vanishing minors. – hardmath Nov 4 '14 at 16:35
• 'r is the largest number such that some r×r minor does not vanish'. Can someone point me to a proof of this? – David Warren Katz Sep 11 '17 at 13:34
• @Auburn: Have a look at this previous Question, Determining the rank of a matrix based on its minors, to see if it is a sufficient proof for your purpose. – hardmath Sep 11 '17 at 13:42