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I was reading "Linear Algebra" of G.Strang(4ed) and I encountered Schur's lemma. The statement and the proof go like this

Suppose $A$ is a complex square matrix. Then there exists a unitary matrix $U$ such that $$U^{-1}AU$$ is triangular.

Proof Suppose $A$ is a $4$ by $4$ matrix. $A$ has at least one unit eigenvector $x_1$, which we place in the first column of $U$. By the Gram-Schmidt process, there exists a unitary $U_1$ such that $${U_1}^{-1}AU_1= \begin{bmatrix} \lambda_1 &*&*&*\\ 0&*&*&*\\ 0&*&*&*\\ 0&*&*&* \end{bmatrix}$$ Now consider the $3$ by $3$ submatrix in the lower right-hand corner. It has a unit eigenvector $x_2$, which becomes the first column of a unitary matrix $M_2$. $$\text{Set } U_2=\begin{bmatrix} 1&0&0&0\\ 0\\ 0&&M_2\\ 0 \end{bmatrix} \qquad \text{then } \qquad {U_2}^{-1}{U_1}^{-1}AU_1U_2= \begin{bmatrix} \lambda_1 &*&*&*\\ 0&\lambda_2&*&*\\ 0&0&*&*\\ 0&0&*&* \end{bmatrix}$$

In a similar fashion, $${U_3}^{-1}{U_2}^{-1}{U_1}^{-1}AU_1U_2U_3= \begin{bmatrix} \lambda_1 &*&*&*\\ 0&\lambda_2&*&*\\ 0&0&\lambda_3&*\\ 0&0&0&* \end{bmatrix}$$

Here is the question: Can anyone give me a nontrivial(e.g. non-diagonal) matrix which demonstrate this lemma, with the corresponding $U_1$, $U_2$, $U_3$ and $T$?

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  • $\begingroup$ Haha I feel like every single author has a different statement they call "Schur's lemma", this is my first time seeing this one! :) $\endgroup$ Commented Apr 9, 2020 at 11:51

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The $4\times 4$ matrix is too complicated for us to demonstrate the lemma as an example. Instead, consider the following $3\times 3$ matrix; $$A=\begin{bmatrix} 2&1&-2\\ 1&0&0\\ 0&1&0 \end{bmatrix}$$ It has $\lambda=1,-1,2$ as eigenvalues. Choose $\lambda_1=1$. The corresponding eigenvector of length $1$ is $$x_1=\frac1{\sqrt3}\begin{bmatrix}1\\1\\1\end{bmatrix}\tag{1}$$ Since we have $$Ax_1=x_1,\tag{2}$$ construct a matrix $U_1$ which have $x_1$ as the first column. And impose $U_1$ to be unitary. We may set, for example, $$U_1=\begin{bmatrix} \frac1{\sqrt3} &\frac1{\sqrt2} &\frac1{\sqrt6} \\ \frac1{\sqrt3} &-\frac1{\sqrt2} &\frac1{\sqrt6} \\ \frac1{\sqrt3} &0 &-\frac2{\sqrt6} \end{bmatrix}$$ Note that $U_1$ is not unique. From (2), we have $$AU_1=U_1 \begin{bmatrix} 1&*&*\\ 0&*&*\\ 0&*&*\\ \end{bmatrix}\tag{3}$$ Let the unkown matrix on the right hand side be $B$. We get $$B={U_1}^{-1}AU_1=\begin{bmatrix} 1& \frac1{\sqrt6} &\frac3{\sqrt2} \\ 0& 0 &\sqrt3 \\ 0& \frac2{\sqrt3}&1 \end{bmatrix}.$$ Let $$\bar B=\begin{bmatrix} 0 &\sqrt3\\ \frac2{\sqrt3} &1 \end{bmatrix}.$$ It has $\lambda=-1,2$. (Note that $A$ and $B$[defined as in (7)] have exactly the same eigenvalues, which is trivial since $A$ and $B$ are similar). Choose $\lambda_2=-1$. The corresponding unit eigenvector is $$\overline{x_2}=\frac12\begin{bmatrix} \sqrt3\\-1\tag{4} \end{bmatrix}$$ We have $$\bar B\overline{x_2}=-\overline{x_2}.\tag{5}$$ Let $\overline{U_2}$ have $\overline{x_2}$ as the first column. Again, $U_2$ is unitary: $$\overline{U_2}=\begin{bmatrix} \frac{\sqrt3}2 &\frac12\\ -\frac12 &\frac{\sqrt3}2 \end{bmatrix}.$$ (Here, $\overline{U_2}$ is not unique, but we only have two possibilities.) From (5), we have $$\bar B\overline{U_2}=\overline{U_2} \begin{bmatrix} -1 &*\\ 0 &* \end{bmatrix}.\tag{6}$$ Let the unknown matrix on the right hand side be $\bar C$. We get $$\bar C=\overline{U_2}^{-1}\bar B\overline{U_2}=\begin{bmatrix} -1 &\frac1{\sqrt3}\\ 0 &2 \end{bmatrix}.$$ Let $B$, $U_2$ and $C$ be $3\times 3$ matrices and let $x_2$ be a column vector in $\mathbb R^3$ such that $$B=\begin{bmatrix} 1&0\\0&\bar B \end{bmatrix} ,\quad U_2=\begin{bmatrix} 1&0\\0&\overline{U_2} \end{bmatrix} ,\quad C=\begin{bmatrix} 1&0\\0&\bar C \end{bmatrix} ,\quad x_2=\begin{bmatrix} 0\\\overline{x_2} \end{bmatrix}. \tag{7}$$ Then (6) reduces to $$BU_2=U_2C.\tag{8}$$ By (3) and (8), $$C={U_2}^{-1}BU_2={U_2}^{-1}{U_1}^{-1}AU_1U_2.$$ Note that $C$ is a triangular matrix.

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