Demonstration Schur's lemma(Linear algebra)

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 $$T$$?

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

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.