# prove that every eigenvalue $\lambda$ of $A$ must satisfy $\lambda=-i\bar \lambda$

let $A^*=iA$ where $A \in M_{2\times2}(\Bbb{C})$. How would you prove that every eigenvalue $\lambda$ of $A$ must satisfy $\lambda=-i\bar \lambda$

My Proof: I have already shown that $A$ is normal and thus unitarily diagonalizable therefore $U^*AU=D$ where $U$ is a unitary matrix, and $D$ is the diagonal matrix. Then $(U^*AU)^*=U^*A^*U=U^*(iA)U=i(U^*AU)=D^*$, and then from there we can conclude that $U^*AU=-iD^*=-i\bar{D}$ and thus all the eigenvalues of A satisfy $\lambda=-i\bar \lambda$.

Is this proof correct?

• Your proof is fine if you have shown that $A$ is unitarily diagonalizable. However, there is a shorter proof (which does not require diagonalizablity). Suppose $A^*=\gamma A$ with $\gamma\in \mathbb{C}$ (actually for this to be possible, you also need $|\gamma|=1$). Let $\lambda$ be an eigenvalue of $A$ ($n\times n$) with eigenvector $v$. Note that$$\overline{\lambda}v^*v =v^* A^* v = \gamma v^* A v =\gamma \lambda v^*v\Longrightarrow \overline{\lambda} = \gamma \lambda$$ Jul 10, 2018 at 16:57

Given $A^{*} = iA$.

To Prove: $\lambda = -i\overline{\lambda}$.
Let $\lambda$ be an eigenvalue of $A$, i.e., there exists a non-zero vector $x$, such that,

$$Ax = \lambda x$$ Therefore, $$A^{*}x = iAx = i\lambda x$$ Therefore, $i\lambda$ is an eigenvalue of $A^{*}$.

It is not difficult to prove that if $A$ has a non-zero eigenvalue $\lambda$ that $\overline{\lambda}$ is an eigenvalue of $A^*$. (Do mention in the comments if you aren't able to prove this part.)

Using these pieces of information we have,

$$\overline{\lambda} = i \lambda \implies \lambda = -i \overline{\lambda}$$

Hence,proved.

• how did you get $A^{*}x = iAx$? Jul 11, 2018 at 18:23
• Its given in the question : $A^* = iA$ Jul 11, 2018 at 18:24
• my apologies I completely glossed over that part. Thank you! Jul 11, 2018 at 18:26
• Not a problem!! Jul 11, 2018 at 18:26