If $A$ is an antisymmetric matrix then $A+I$ is invertible [closed]

Let $A$ be an antisymmetric matrix with real entries. How can I show that $A+I$ is invertible?

Solution 1:

\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 $A$.

Solution 2:

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!