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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.

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  • $\begingroup$ I don't think there's a shortcut that avoids using Schur's lemma $\endgroup$ Feb 15, 2016 at 11:02

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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 $A$.

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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 $UAU^{*} = UIU^{*} = I = B$.

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  • $\begingroup$ So is Schur decomposition unique when our matrix $A$ is upper triangular to begin with? Could we find a $T$ such that $UAU^*=T$ with $A \neq T$? Or do we have $A=T$ for all unitary matrices $U$ such that $UAU^*=T$? $\endgroup$ Feb 15, 2016 at 23:21
  • $\begingroup$ Found a counter example! Not true in general. $\endgroup$ Feb 16, 2016 at 0:13

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