Schurs Lemma for Upper Triangular matrix Schurs Lemma Says : For any $A\in M_n(\mathbb{C})$ with eigen values $\lambda_1,\dots,\lambda_n$ There exists an unitary matrix $U$ such that $$UAU^*=T$$ where $T$ is an upper triangular matrix with diagonal $\lambda_1,\dots,\lambda_n$
If In particular $A$ is itself an upper triangular matrix with diagonals $(x_1,\dots,x_n)$ then $T$ and $A$ will be same matrix?
what if $B$ is another upper triangular matrix with diagonals $(y_1,\dots,y_n)$ which is a permutation of   $(x_1,\dots,x_n)$ , then how can I conclude $A,B$ are equivalent?
Thanks for helping.
 A: They are not necessarily unitarily equivalent. For example, if 
$$ A = \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix}, B = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $$
then $A$ and $B$ are upper triangular and have the same diagonal elements, but you can't find $U$ such that $UAU^{*} = UIU^{*} = I = B$.
A: To answer your first question: no, $A$ and $T$ are not equal in general (since Schur decomposition is not unique--even in the case where each $\lambda_1,\dots,\lambda_n$ are distinct). Take
$$A = \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix},$$
we can Schur factor $A$ as follows:
$$\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}^{-1}=\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix}$$
$$=\begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix}$$
$$=A$$
And the matrix $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{pmatrix} $ is definitely unitary. So we found a Schur decomposition that was our original matrix $A$.
But we can also Schur factor $A$ as follows:
$$\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}^{-1}=\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix} \begin{pmatrix} 5 & 5 &5 \\ 0 & 6 & 5 \\ 0 & 0 & 7 \end{pmatrix} \begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$$
$$=\begin{pmatrix} 5 & 5 &-5 \\ 0 & 6 & -5 \\ 0 & 0 & 7 \end{pmatrix}$$
$$ \neq A$$
and it is easily checked that $\begin{pmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & -1 \end{pmatrix}$ is unitary, and so we found another Schur decomposition for $A$ that was not equal $A$.
