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Let $A$ be a $n \times n$ matrix and suppose $A$ has a zero submatrix of order $p \times q$ where $p + q \ge n+1$. Then $\det(A) = 0$.

I can see this happening when doing Laplace expansion. I can also prove it using induction but it fills more than a page and with a lot of if-then-else. Can somebody suggest a short and 'elegant' proof?

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up vote 2 down vote accepted

In the $q$ columns, $p$ rows are zero and at most $n-p$ rows aren't. Since $n-p\lt q$, the $q$ column vectors span a subspace of dimension less than $q$, so they are linearly dependent.

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Let $K$ be the scalar field, and $f$ be the linear map from $K^n$ to $K^n$ associated with $A$. The fact that $A$ has a $p \times q$ zero submatrix implies that $K^n$ has a $q$-dimensional subspace $Q$ and an $(n-p)$-dimensional subspace $P$ such that $f(q) \in P$ whenever $q \in Q$. Since $p+q \ge n+1$, $q>n-p$, i.e. $Q$ is of smaller dimension than $P$. Thus $f|_Q$ is not injective, hence $f$ is not injective, hence $\det(A)=\det(f)=0$.

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Consider the rank $r$ of the matrix $A$, computed, for example, by row. Then $r \leq n-p + \min\{p,n-q\}$, since there are $n-p$ rows that can be different from the null row, and the other $p$ rows form a submatrix of size $p \times (n-q)$. This is equal to $\min\{n,2n-(p+q)\}\leq \min\{n,n-1\} = n-1$, and then $A$ doesn't have full rank, so it's singular, with zero determinant.

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by interchanging row and columns assume that there is a $p \times q$ block of zeros in the upper right hand corner. Also, assume $p < q$, otherwise, turn it sideways. When computing the determinant by summing over all permutaions etc., in the first p rows you need to pick a elt in the first n-q columns, since the last q are all zeros. Since p > n-q you run out of distinct cols to choose from.

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I think the OP already mentions this proof. – Yuval Filmus May 9 '12 at 14:46

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