Special property of matrix $A\in\mathbb{R}^{n,n}$ and it's determinant

Question

How to prove for matrix $A\in\mathbb{R}^{n,n}$ that if in places where $x$ columns crosses with $y$ rows are placed $0$, then we can be sure, that $$ \det A = 0\qquad \qquad \text{if }\quad x+y > n $$ .

Answer 1

Let $A=[a_{ij}]_n=\begin{bmatrix}a_{11}&a_{12}&a_{13}&...&a_{1n}\\a_{21}&a_{22}&a_{23}&...&a_{2n}\\\vdots&\vdots&\vdots&&\vdots\\a_{n1}&a_{n2}&a_{n3}&...&a_{nn}\end{bmatrix}_n$

Now, $a_{ij}=0$ where the $x$ columns cross the $y$ rows. In totality, we have $xy$ number of $0$s. We know that the determinant of a matrix gets multiplied by $-1$ on interchange of any two rows/columns, but its magnitude is unchanged. We can use elementary row and column interchanges to gather/cluster all the above zeroes together. The transformed matrix $A'$ looks like this:

$A'=\begin{bmatrix}0_{y\times x}&B_{y\times(n-x)}\\C_{(n-y)\times x}&D_{(n-y)\times(n-x)}\end{bmatrix}_n$

where $0_{y\times x}$ is the zero matrix of indicated order.

Also note that $\det A'=(-1)^k\cdot\det A$ or $|\det A'|=|\det A|$

Observe that the rows of $B$ are linearly dependent, since $y$ rows/vectors in $n-x(<y)$ dimesional vector space cannot be linearly independent. This means the rows/vectors of the submatrix $\begin{bmatrix}0&B\end{bmatrix}_{y\times n}$ are linearly dependent. This means $\det A'=0\implies\det A=0$.

Answer 2

If I rightly understand your question, I have a solution:

Consider $$\det A = \det \begin{pmatrix} O& B\\ C& D \end{pmatrix} $$ where $O$ is a $x\times y$ zero matrix, $B$ is a $(n-x)\times y$ matrix, $C$ is a $x\times(n-y)$ matrix and $D$ is a $(n-x)\times(n-y)$ matrix. Notice that $y>n-x$, so the rank of $B$ is no more than $n-x$ and hence the first $y$ lines of $A$ must be linearly dependent, which implies that $\det{A}=0$.

In general, by alternating rows and columns, every determine of $A$ satisfies the given condition can be transfer to the form above (or the opposite number of it).