> [!tldr] Schur Decompositions > For an arbitrary square matrix $A \in \mathbb{C}^{n\times n}$, it always has the **Schur decomposition** $A=UTU^{\ast},$where $U\in \mathbb{C}^{n\times n}$ is unitary, and $T \in \mathbb{C}^n\times n$ is upper-triangular. > [!proof] Proof of existence > Induction on the size of $A$. > Let $\lambda_{1},v_{1}$ be an eigenpair of $A$. Then with orthogonal matrix $U_{1}:= [v_{1},V_{1}^\perp]$ (with $V_{1}^{\perp}$ being an orthogonal complement of $v_{1}$), $AU_{1}=(\lambda_{1}v_{1},\dots)= \begin{pmatrix} \times & \times &\dots&\times \\ & \times & \dots & \times \\ & \times & \dots & \times \\ & \times & \dots & \times \end{pmatrix}U_{1} \Rightarrow U_{1}^{\ast}AU_{1}=\begin{pmatrix} \times & \times &\dots&\times \\ & \times & \dots & \times \\ & \times & \dots & \times \\ & \times & \dots & \times \end{pmatrix}.$ > > Now using induction and padding gives $U_{2},\dots$ so that $U_{n}^{\ast}\dots U_{1}^{\ast}AU_{1}\dots U_{n}=T$. Defining $U:= U_{1}\dots U_{n}$ gives the Schur decomposition. Since $A,T$ are similar, we can find the eigenvalues of $A$ on the diagonal of $T$. Also, because of the nature of [[Eigenvalue Problems]], the Schur decomposition can only be found iteratively.