$A^T = - A$
Prove that if $\lambda \neq 0$ is a constant, then $A+ \lambda I$ is always invertible.
(where $I$ is the identity matrix)
My idea is to show that $0$ is the only eigenvalue so that $|A+\lambda I|$
I have shown that if $\lambda$ is an eigenvalue, then so must $-\lambda$. But I'm not sure how to show $0$ is the only eigenvalue from there.
Is my approach correct? Or is there a simpler way to approach this?
Thank you!
On
The eigenvalues of a skew-symmetric matrix are all purely imaginary. One proof can be found here, by looking at how the matrix behaves with respect to the complex scalar product using a complex eigenvector of $A$.
$A$ is skew -symmetric. Try to show that each eigenvalue of $A$ is of the form $it$ with $t \in \mathbb R$, (where $i^2=-1$).