1
$\begingroup$

Let $V$ be an inner product space finitely-generated over $\mathbb{C}$ and let $\alpha \in End(V )$ satisfy the condition that every eigenvector of $β = \alpha + \alpha^*$ is also an eigenvector of $γ = \alpha − \alpha^*$. Prove that $\alpha$ is normal.

I'm stuck on this problem. have to show $\alpha^*\alpha=\alpha\alpha^*$. Since $\alpha^*\alpha-\alpha\alpha^* $ is selfadjoint, I guess it's enough to prove $\langle \alpha^*\alpha-\alpha\alpha^*(v),v\rangle=0$ for all $v\in V,$ butI'm sure this will work or it might be a wrong approach. Anyway, I would appreciate any hint.

1 Answers 1

1

Hint: $ \alpha + \alpha^* $ is Hermitian, and every eigenvector of $ \alpha + \alpha^* $ is also an eigenvector of $ \alpha $.

  • 1
    So if I'm correct you mean since $\alpha^*+\alpha$ is self-adjoint, so is normal and then it's orthogonally diagonalizable, then there exists a basis $B$ of $V$ composed of eigenvectors of $\alpha+\alpha^*$. As, eigenvectors of $\alpha^*+\alpha$ are also eigenvectors of $\alpha$ then $\alpha$ is orthogonally diagonalizable so it's normal. – 2017-01-05