Prove that $A = G^{4}$ for some self-adjoint matrix $G$

Question

Let $A\in M_{n}(\mathbb{C})$, $n\geq 2$.

Let $P_A(x) = (x-\lambda_1)....(x-\lambda_n)$ be characteristic polynomial of $A$ and all $\lambda_i$ are positive reals.

Show that if $A$ is normal, then $A = G^{4}$ for some self-adjoint matrix $G$

I just started on the topic on adjoint, normal matrices, and self-adjoint. Thus I feel that I'm missing some concepts behind this topic.

From what the question has given, I know the following:

$A^*A = AA^*$

and $\exists \ \beta$, such that $\beta$ is an orthonormal basis consisting of eigenvector of $A$

Both of which are not very helpful in this case.

Hence are there some concept I neglected, or is there some trick to solving these sort of questions?

Any help or insights is deeply appreciated.

Answer 1

You have an orthonormal basis $e_1, \dotsc, e_n$, such that w.r.t. this basis $A$ has the form $\operatorname{diag}(\lambda_1, \dotsc, \lambda_n)$.

All $\lambda_i$ are positive reals, so why don't you just take $G=\operatorname{diag}(\sqrt[4]{\lambda_1}, \dotsc, \sqrt[4]{\lambda_n})$?

This is clearly a self-adjoint operator, since it is given by a symmetric matrix w.r.t an orthonormal basis. Of course $A=G^4$ holds.