$A+ cE$ invertible when $A^T = - A$

Question

$A^T = - A$

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!

This is clearly wrong. The matrix $A = \begin{pmatrix} 0 & 1 \ -1 & 0 \end{pmatrix}$ has eigenvalues $\pm i$ so $A \pm iI$ is noninvertible. — mechanodroid, Feb 27 '18 at 13:29
@mechanodroid Actually, the eigenvalues of your matrix are $\pm i$. — José Carlos Santos, Feb 27 '18 at 13:30
@JoséCarlosSantos Thanks. So $\lambda$ in the question is assumed to be real, I suppose. — mechanodroid, Feb 27 '18 at 13:31
@mechanodroid I have no doubts about that (althougn I think that the OP should have stated it). — José Carlos Santos, Feb 27 '18 at 13:32
@JoséCarlosSantos, Yeah. My bad. $\lambda$ is presumed to be real. — David, Feb 27 '18 at 13:48

score 2 · Answer 1 · answered Feb 27 '18 at 13:27

2

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 $A$.

answered Feb 27 '18 at 13:27

Arthur

199,419

score 2 · Accepted Answer · answered Feb 27 '18 at 13:28

2

$A$ is skew -symmetric. Try to show that each eigenvalue of $A$ is of the form $it$ with $t \in \mathbb R$, (where $i^2=-1$).

answered Feb 27 '18 at 13:28

Fred

77,394

Thank you! I think I have understood how to approach this question from this hint. – David Feb 27 '18 at 13:49

$A+ cE$ invertible when $A^T = - A$

2 Answers2