TITLE: If $A$ is an antisymmetric matrix then $A+I$ is invertible
QUESTION [9 upvotes]: Let $A$ be an antisymmetric matrix with real entries. How can I show that $A+I$ is invertible?

REPLY [12 votes]: \begin{align*}
(A + I)x = 0 &\implies x^T (A + I) x = 0 \\
&\implies \underbrace{x^T A x}_0 + \underbrace{x^T I x}_{\|x\|^2} = 0 \\
&\implies x = 0.
\end{align*}
This shows that the null space of $A + I$ is trivial, which means that $A + I$ is invertible.  I used the fact that $x^T A x = 0$ for all $x$, which follows from the anti-symmetry of $A$.