Eigenvalues of an antihermitian matrix

Question

I have to prove that every eigenvalue of an antihermitian matrix is in the form of $bi $ for some $b \in R$.

I already know that if A is antihermitian , it is normal , thus we can diagonalise it with a unitary matrix.

I have tried doing this :

$ \exists$ U unitary and D diagonal such that

$ U^*AU=D$ $ \Rightarrow$ $A = UDU^*$

Multiplying by $A^*$:

$A^*A = -A^2 = (DU^*)^*(DU^*)$

From there , I don't know how to continue , or even if I'm going yhe right way.

Answer 1

Let $\lambda$ be an eigenvalue of $A$ and $v$ be the associated eigenvector. Then

$$Av= \lambda v$$ $$v^*Av = v^*\lambda v$$ Now take the hermitian conjugate on both sides:

$$v^*A^*v = \bar{\lambda}||v||$$ $$-v^*Av=\bar{\lambda}||v||$$ $$-v^*\lambda v = \bar{\lambda}||v||$$ Thus $$-||v||\lambda = \bar{\lambda}||v||$$

So since $||v|| \neq 0$

$$\lambda = -\bar{\lambda}$$

The only eigenvalues that satisfy this are eigenvalues of the form $bi$ for some $b \in \mathbb{R}$.