If $A$ is real skew-symmetry. $I-A$ and $I+A$ are nonsingular

Question

I asked these questions to my TA during workshop but none of them figured it out.

Can anyone help me out?

Answer 1

From http://en.wikipedia.org/wiki/Skew-symmetric_matrix#Spectral_theory the eigenvalues of a real skew-symmetric matrix are imaginary.

If $I-A$ were singular, there exists a vector $x$ such that $(I-A)x=0 \Rightarrow Ax=x$, which implies 1 is an eigenvalue of $A$. Contradiction.

Answer 2

This is in fact true. The reason is as follows. We work over the complex numbers and write $A^* := \left( \overline A \right)^T$. Now, since $A$ is real and skew-symmetric, we have $A^* = A^T = -A$. This implies $AA^* = A^*A$, i.e. $A$ is normal. It is well known that a complex matrix which is normal can be diagonalized over $\mathbb C$ via a unitary transformation, i.e. there is a complex matrix $U$ with the properties $U^*U = I$ (identity matrix), or equivalently $U^{-1} = U^*$, and $U^*AU = D$ where $D$ is a complex diagonal matrix. Using all these facts, we get $$D^* = \left( U^*AU \right)^* = U^*A^*U = U^* A^T U = -U^*AU = -D.$$ This shows that the diagonal entries of $D$ (i.e. the eigenvalues of $A$) are purely imaginary. Moreover, we get $$U^*(I + A)U = U^*U + U^*AU = I + D.$$ Here, the r.h.s is a diagonal matrix with nonzero entries on the diagonal, since in the sum $I + D$ there's no cancellation between the diagonal entries of $I$ and $D$. So the r.h.s. is invertible, and thus also the l.h.s. This implies that $I + A$ is invertible. Since $-A$ is also real and skew-symmetric, we get the invertibility of $I-A$ by the same argument.