Normal transformation based on eigenvectors Let $T:V\rightarrow V$ be a linear transformation over the finite vector space $V$. It is known that each eigenvector of $T$ is a eigenvector of $T^*$ as well. How can it be shown that $T$ is normal?
 A: Without loss of generality, take $T$ to be an $n \times n$ matrix.
Let $U$ be a unitary matrix such that the first column is an eigenvector of $T$, and let $A = U^*TU$.  Note that if $A$ is normal, then $T$ must be normal. We note that $A$ has the form
$$
A = \pmatrix{\lambda_1 & x^*\\ 0&\tilde T}
$$
Note that $A$ must satisfy the hypothesis of the question.  That is, every eigenvector of $A$ is an eigenvector of $A^*$.  Now, we note that $e = (1,0,\dots,0)^*$ is an eigenvector of $A$.  It follows that $A^*e$ must me a multiple of $e$.  Conclude that $x = 0$.
Thus, we have
$$
A = \pmatrix{\lambda_1 & 0\\0&\tilde T}
$$
We now see that $\tilde T$ has the property that every eigenvector of $\tilde T$ is also an eigenvector of $\tilde T^*$.
At this point, it suffices to show that $\tilde T$ is normal.
Using this analysis towards an inductive step, you may now prove by induction on the dimension of $V$ that if every eigenvector of $T$ is also an eigenvector of $T^*$, then $T$ must be normal.
Note: you may find it easier to use a Schur triangulariztion in the first step (so that $A$ is upper triangular).  Even with this change, I think a proof by induction works best.
