I am studying for my exam and I am stuck in the following problem:
Let $A$ be an $r × r$ real valued matrix, and $I$ be the r × r identity matrix.
Prove that if the entries of A satisfy $$ \sum_{j=1}^r |A_{ij} | < 1$$ for each $i = 1, 2, \ldots, r$, then $I + A$ is invertible. I have no idea how to do this one.
Prove that if $A$ satisfies $A^T = −A$, then $det(aI + A) \neq 0$ for all $a ∈ R - \{0\}$.
For this I have tried contradiction; suppose that $det(aI + A) = 0$ which means that it is singular. Therefore we have a non zero $x$ such that $(aI + A)x = 0$.
$\implies ax+Ax=0 \implies ax^T+x^T A^T =0 \implies ax^T-x^T A =0 \implies ax^T x -x^T A x =0 $.
This means we have $ax^T x -x^T A x =0$ and $ax^T x +x^T A x =0$. Simplying gives $2ax^T x=0 \implies x=0$. Which is a contradiction.