Let $A = [a_{ij}]$ be an $n \times n$ matrix with entries in $\mathbb{R}$. Suppose there exists an $m$ with $a_{ij} = 0$ for $i \ge m$ and $j \le m$, and $a_{i,i} \ne 0$ for $1 \le i \lt m$. Show that $A$ has no inverse.
From my understanding no diagonal value is zero, but all the non-diagonal values more than some arbitrary $m$ is zero. And, I know I must show that the determinant is zero, but I'm not sure how to do this.
Best Answer
The determinant $\det(A)$ is a sum of products $$\pm\, a_{1\,\sigma(1)}\>a_{2\,\sigma(2)}\>\cdots\> a_{n\,\sigma(n)}\ ,\tag{1}$$ where $\sigma$ runs through the $n!$ permutations of $[n]$. For such a permutation $\sigma$ we cannot have $$m+1\leq \sigma(i)\leq n\quad(m\leq i\leq n)$$ (counting elements). It follows that in each product $(1)$ there is at least one factor $a_{i\,\sigma(i)}=0$. Therefore $\det(A)=0$; hence $A$ is singular.