Let $A \in \mathbb{C}^{n \times n}$ be upper triangular. Let $u$ be a unit-norm eigenvector of $A$ whose eigenvalue is $a_{11}$ (the top-left entry of $A$). Let $U \in Unitary(n)$ be a lower-Hessenberg unitary matrix whose first column is $u$.
Question: Show that $U^* A U$ is an upper triangular matrix. (Or find a counterexample.)
Comments:
(0) Write $U = [u, U_\perp]$, where $U_\perp\in \mathbb{C}^{n \times n-1}$. Note that $U_\perp^* A U_\perp$ is upper triangular if and only if $U^* A U$ is upper triangular. (Since $U_\perp^* A u = 0$.) It is not hard to check that the (nonsquare) matrix $U_\perp^* A$ is upper triangular.
(1) I tried to find counterexamples when $n=3, 4$, but have had no success - seems like the statement might be true. It is not hard to come up with examples showing that the assumption that $U$ is lower Hessenberg is necessary.
(2) As $A$ is upper triangular, $e_1$ is an eigenvector of $A$. If $a_{11}$ is a distinct eigenvalue, then $e_1$ is the unique eigenvector associated to $a_{11}$ (up to scaling). In this case, it seems like it might be easier to prove the statement, but I'm still stuck. Anyway, I am particularly interested in the case where the eigenvalues of $A$ are not distinct.
Ok, I think I figured it out. The key is the following lemma, which for example can be found in Proposition 3.1 here.
Lemma: Let $U \in \mathbb{C}^{n \times n}$ be unitary and lower-Hessenberg, and let $u_j$ denote the $j$-th column of $U$. Then $u_j \in \mathrm{span}(u_1, e_1, \ldots, e_{j-1})$ for $j \in[2, n]$.
Now onto the proof. Since $A$ is upper triangular and $u_1$ is an eigenvector of $A$ (by assumption), $A u_j \in \mathrm{span}(u_1, e_1, \ldots, e_{j-1})$ (applying the Lemma). Let $i > j$. We know $u_i^* u_1 = 0$ (by unitary-ness) and $u_i^* e_{k} = 0$ for $k < i-1$ (by Hessenberg-ness). So $u_i^* A u_j = 0$.