If $\phi$ and its adjoint share a eigenvector with conjugate eigenvalue then normal

Question

Suppose for any eigenvector $v$ of an operator $\phi:\mathbb{C}^n\rightarrow \mathbb{C}^n$ with corresponding eigenvalue $\lambda$, $v$ is also the eigenvector of $\phi^*$(the adjoint of $\phi$) with corresponding eigenvalue $\bar{\lambda}$. Prove that $\phi$ is normal.

I know the converse is true, but I have no idea how to prove this. Easy to see $\phi \phi^*(v)=\phi^* \phi(v)$ for $v$ in eigenspaces, but how to prove the others? Thanks for any help.

Answer 1

Let $v_1$ be an eigenvector of $\phi$. Show that $V=\operatorname{span}\{v_1\}$ and $V^\perp$ are invariant subspaces of both $\phi$ and $\phi^\ast$. Hence prove that the stated property of $\phi$ in the problem statement is preserved on $V^\perp$. Proceed recursively, we obtain an orthonormal eigenbasis $\{v_1,\ldots,v_n\}$ of $\phi$. Hence $\phi$ is normal.

Answer 2

This is clearly true for $n=1$, since any $\phi:\mathbb{C}\rightarrow\mathbb{C}$ is normal. Suppose for induction that it is true for all $n<m$.

Let $\phi:\mathbb{C}^m\rightarrow\mathbb{C}^m$ be a linear map such that any eigenvector of $\phi$ is also eigenvector of $\phi^*$, with the same eigenvalue. By Schur triangulation, pick an orthonormal basis $(x_1,\ldots,x_m)$ such that the $m\times m$ matrix $M$ defined by $\phi x_i=\sum_{j=1}^m M_{ji} x_j$ is lower triangular. Note that the matrix $M$ has the same property as $\phi$, since $\phi x=\lambda x \Leftrightarrow M\tilde{x}=\lambda\tilde{x}$ and $\phi^* x=\lambda x \Leftrightarrow M^*\tilde{x}=\lambda\tilde{x}$, where $\tilde{x}$ denotes the coordinate vector of $x$ with respect to the basis $(x_1,\ldots,x_m)$. Let $e_j$ denote the $j$'th canonical basis vector in $\mathbb{C}^m$. Since $M$ is lower triangular, we have $M e_m=M_{mm} e_m$, i.e. $e_m$ is an eigenvector of $M$ with eigenvalue $M_{mm}$. But then $$ \bar{M}_{mm} e_m=M^* e_m = \sum_{j=1}^m \bar{M}_{mj}e_j, $$ which implies $M_{m1}=\ldots=M_{m,m-1}=0$. It follows that $M$ and $M^*$ leave the subspace $V=\mathrm{span}(e_1,\ldots,e_{m-1})$ invariant, so that we may naturally consider the restriction of $M$ to $V$, say $M_V$, which is a lower triangular matrix. Note that $(M_V)^*=(M^*)_V$, so that $M_V$ is an operator in $\mathbb{C}^{m-1}$ with the property that any eigenvector of $M_V$ is also an eigenvector of $M_V^*$. By the induction hypothesis, this implies that $M_V$ is normal. Since any normal lower triangular matrix is diagonal, this implies that $M$ is a diagonal matrix, which implies that $\phi$ has an orthonormal basis of eigenvectors, namely $(x_1,\ldots,x_m)$. Then $\phi$ is normal.