# Schur decomposition for $3 \times 3$ matrix

Suppose $A=\begin{bmatrix} 1 &-2 &2\\-1 &1 &1\\-2 &0 &3 \end{bmatrix}$, what is the Schur decomposition?

The eigenvalues of $A$ are $\lambda_1 = 1,\lambda_2 = 2+2i$ and $\lambda_3 = 2-2i$. For $2 \times 2$ matrix, if the eigenvalues are real, then I just pick one normalized eigenvector to construct an orthonormal basis. If the eigenvalues are complex numbers, then do the similar process (the other vector should be orthogonal to the conjugate of the normalized eigenvector); however, for $3 \times 3$, I'm not sure how to find $U$ such that $U^{H}AU=T$, where $U$ is unitary and $T$ is upper triangular. Thanks.

If $\lambda_1$ an eigenvalue of $A$ with associated eigenvector $u_1$ (unitary), we can get an orthonormal basis of $\mathbb{C}^3:$ $B=\left\{u_1,u_2,u_3\right\}$ so, $U_1=\begin{bmatrix}u_1,u_2,u_3\end{bmatrix}$ is an unitary matrix. Besides $Au_1=\lambda_1u_1$, hence $$AU_1=A\begin{bmatrix}u_1,u_2,u_3\end{bmatrix}=\begin{bmatrix}\lambda_1u_1,Au_2,Au_3\end{bmatrix}$$ $$=\begin{bmatrix}u_1,u_2,u_3\end{bmatrix}\begin{bmatrix} \lambda_1 & * & * \\ 0 & * & *\\ 0 & * & * \end{bmatrix}=U_1\left[\begin{array}{c|c} \lambda_1 & b_1 \\ \hline 0 & A_1 \end{array} \right].$$ That is, $$U_1^{-1}AU_1=\left[\begin{array}{c|c} \lambda_1 & b_1 \\ \hline 0 & A_1 \end{array} \right],\text{ with } U_1 \text{ unitary and } A_1\in\mathbb{C}^{2\times 2}.$$ If you find $M_2\in \mathbb{C}^{2\times 2}$ unitary such that $M_2^{-1}A_1M_2=\begin{bmatrix} \lambda_2 & t_{12} \\ 0 & \lambda_3 \\\end{bmatrix},$ define $$U_2=\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2 \end{array} \right].$$ $U_2$ is unitary and $$\left(U_1U_2\right)^{-1}A\left(U_1U_2\right)=U_2^{-1}U_1^{-1}AU_1U_2$$ $$=\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2^{-1} \end{array} \right]U_1^{-1}AU_1\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2 \end{array} \right]$$ $$=\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2^{-1} \end{array} \right]\left[\begin{array}{c|c} \lambda_1 & b_1 \\ \hline 0 & A_1 \end{array} \right]\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2 \end{array} \right]$$ $$=\left[\begin{array}{c|c} \lambda_1 & b_1 \\ \hline 0 & M_2^{-1}A_1 \end{array} \right]\left[\begin{array}{c|c} 1 & 0 \\ \hline 0 & M_2 \end{array} \right]=\left[\begin{array}{c|c} \lambda_1 & b_1M_2 \\ \hline 0 & M_2^{-1}A_1M_2 \end{array} \right]$$ $$=\begin{bmatrix} \lambda_1 & s_{12} & s_{1n}\\ 0 & \lambda_2 & s_{2n} \\ 0 & 0 &\lambda_3\end{bmatrix}=T.$$ The matrix $U=U_1U_2$ is unitary (product of unitary matrices), as a consequence $U^{-1}AU$ $=$ $U^*AU=T$ with $T$ triangular.