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Let $A \in M_n(\mathbb{C})$ show that if every eigenvector of $A$ is an eigenvector of $A^*$ then $A$ is normal.

This question is equivalent to If every eigenvector of $T$ is also an eigenvector of $T^{*}$ then $T$ is a normal operator. The first answer uses unitary triangularization but I didn't understand why the fact that $e_1$ is an eigenvector of $T^*$ implies $a_{1j}=0$ for $j>1$. In the second answer there is a hint but how does it complete the proof?

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user91015
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3 Answers3

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First we show:

If $T$ is triangular and the eigenvectors of $T$ are the eigenvectors of $T^*$, then $T$ is diagonal.

Proof by induction: The statement is trivially satisfied for $n=1$. Assume it's true for $(n-1)\times(n-1)$ matrices and let the (upper) triangular $T$ have the form $$ T=\begin{bmatrix}\tau&t^*\\0&\tilde{T}\end{bmatrix}. $$ The matrix $T$ has an eigenvalue $\tau$ with the eigenvector $e_1=[1,0,\ldots,0]^T$, that is, $Te_1=\tau e_1$. We have $$ T^*e_1=\begin{bmatrix}\bar{\tau}&0\\t&\tilde{T}^*\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}=\begin{bmatrix}\bar{\tau}\\t\end{bmatrix}. $$ By the assumption, $e_1$ is the eigenvector $T^*$. This is possible only if $t=0$, that is, $T$ has the form $$ T=\begin{bmatrix}\tau&0\\0&\tilde{T}\end{bmatrix}. $$ By the induction assumption, the matrix $\tilde{T}$ is diagonal since it has the same eigenvectors as $\tilde{T}^*$.

Second, consider the Schur decomposition of $T$, $T=QRQ^*$, where $Q$ is unitary and $R$ upper triangular. Note that $Tx=\lambda x$ is equivalent to $(Q^*TQ)(Q^*x)=\lambda(Q^*x)$, that is, $R(Q^*x)=\lambda(Q^*x)$ and $T^*x=\mu x$ is equivalent to $(Q^*T^*Q)(Q^*x)=\mu(Q^*x)$, that is, $R^*(Q^*x)=\mu(Q^*x)$. Hence $x$ is a common eigenvector of $T$ and $T^*$ if and only if $Q^*x$ is a common eigenvector of the triangular factor $R$ of the Schur decomposition of $T^*$. Using the statement we proved, we see that $T$ and $T^*$ have common eigenvectors if and only if $R$ is diagonal and hence $T=QRQ^*$ (with diagonal $R$) is normal.

Algebraic Pavel
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Here is a start. Fi rst of all notice this

if $\lambda $ is an eigenvalue of $A$, then $\overline{\lambda}$ is an eigenvalue of $A^*$.

Now, we have

$$ Au = \lambda u $$

which implies

$$ A^* u= \overline{\lambda }u, $$

since they have the same eigenvector. Apply $A^*$ and $A$ to the last two equations respectively and try to finish the problem.

Mhenni Benghorbal
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let v is eigen vector of T for eigen value ɳ

and v is eigen vector of T* for eigen value µ then

Tv = ɳv

T*v = µv, operating T* on 1st and T on 2nd:

T*Tv = ɳT*v = ɳµv

TT*v = uTv = ɳµv; Thus TT*v = T*Tv for all L.I. eigen vectors, Hence TT* = T*T

Sry
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