# Thread: Show that skew-Hermitian is nonsingular

1. ## Show that skew-Hermitian is nonsingular

Problem statement:
Let $S$ be skew-Hermitian, i.e, $S=-S^*$.
Show that $I-S$ is nonsingular.

Attempt:
Well, I know that an arbitrary square matrix can be written as the sum of a Hermitian matrix and a skew-Hermitian matrix.
This tells me that $I-S$ is Hermitian, and since the determinant of a Hermitian matrix is real it is $I-S$ is not singular.

Now, I would like to express this in an more elegant manner using actual math
Hope someone can give me a hint or two, thanks.

2. Originally Posted by Mollier
Problem statement:
Let $S$ be skew-Hermitian, i.e, $S=-S^*$.
Show that $I-S$ is nonsingular.

Attempt:
Well, I know that an arbitrary square matrix can be written as the sum of a Hermitian matrix and a skew-Hermitian matrix.
This tells me that $I-S$ is Hermitian, and since the determinant of a Hermitian matrix is real it is $I-S$ is not singular.

First, $I-S$ is not hermitian in general, as you can easily check by using $(A+B)^{*}=A^{*}+B^{*}$. Second, even if it were then its determinant is real so...so what? It could be zero as zero is a real number! In fact, the zero matrix is Hermitian, and thus your argument doesn't work.

Tonio

Now, I would like to express this in an more elegant manner using actual math
Hope someone can give me a hint or two, thanks.
.

3. Some bad thinking on my part, thanks for the heads up.
Here's another try:

If $(I-S)y=0$ for $y \neq 0$ it means that $I-S$ is singular.
Then $Sy = 0$ which says that $S$ has $1$ as an eigenvalue.
This contradicts the fact that skew-Hermitian matrices have pure imaginary eigenvalues.

4. Originally Posted by Mollier
Here's another try:

If $(I-S)y=0$ for $y \neq 0$ it means that $I-S$ is singular.
Then $Sy = 0$ which says that $S$ has $1$ as an eigenvalue.
This contradicts the fact that skew-Hermitian matrices have pure imaginary eigenvalues.

Now this looks correct, but I think you meant $(I-S)y=0\Longrightarrow Sy=y \Longrightarrow\,\,if\,\,\,y\ne 0\,\,\,then\,\,\,\lambda=1$ is an eigenvalue of S, which is impossible since S is skew-Hermitian...nice!

Tonio

5. Yes, I meant $Sy=y$. Was writing this up during work so I had to be quick
Thanks!