If $A$ is an antisymmetric matrix then $A+I$ is invertible Let $A$ be an antisymmetric matrix with real entries. How can I show that $A+I$ is invertible?
 A: Another possible approach:
Since $A$ is normal, it follows that $A$ is diagonalizable (over $\mathbb{C}$, of course), say $P^{-1}AP = D$ is diagonal. Then $P^{-1}(A+I)P = D+I$, and so it suffices to show that $0$ does not appear as a diagonal entry of the complex matrix $D+I$. In other words, it suffices to show that $-1$ is not an eigenvalue of $A$.
This is true because if $Av=-v$, then $$-|v|^2 = \langle -v, v\rangle = \langle Av, v\rangle = \langle v, A^tv\rangle = \langle v,-Av \rangle=\langle v,v \rangle = |v|^2$$ Thus $|v|=0$, so $v=0$ and hence $v$ cannot be an eigenvector.
Actually, this argument (as well as the other ones) may easily be generalized to show that $A+dI$ is invertible, for any real $d \neq 0$.
A: \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 $A$.
A: Since $A^T = -A$,
$(I + A)(I - A) = I- A^2 = I + A^TA; \tag{1}$
for any vector $x \ne 0$,
$\langle x, (I + A^TA)x \rangle = \langle x, x \rangle + \langle x, A^TAx \rangle = \langle x, x \rangle + \langle Ax, Ax \rangle$
$ = \Vert x \Vert^2 + \Vert Ax \Vert^2 > 0, \tag{2}$
which implies
$(I + A^TA)x \ne 0; \tag{3}$
(3) in turn implies $I + A^TA$ is nonsingular;  thus the factor matrices $I \pm A$ in (1) are nonsingular as well.
And that's how it may be shown!
