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# Schur decomposition, proof of

The columns of the unitary matrix $Q$ in Schur’s decomposition theorem form an orthonormal basis of $\mathbb{C}^{n}$. The matrix $A$ takes the upper-triangular form $D+N$ on this basis. Conversely, if $v_{1},\ldots,v_{n}$ is an orthonormal basis for which $A$ is of this form then the matrix $Q$ with $v_{i}$ as its $i$-th column satisfies the theorem.

To find such a basis we proceed by induction on $n$. For $n=1$ we can simply take $Q=1$. If $n>1$ then let $v\in\mathbb{C}^{n}$ be an eigenvector of $A$ of unit length and let $V=v^{{\perp}}$ be its orthogonal complement. If $\pi$ denotes the orthogonal projection onto the line spanned by $v$ then $(1-\pi)A$ maps $V$ into $V$.

By induction there is an orthonormal basis $v_{2},\ldots,v_{n}$ of $V$ for which $(1-\pi)A$ takes the desired form on $V$. Now $A=\pi A+(1-\pi)A$ so $Av_{i}\equiv(1-\pi)Av_{i}(\mod v)$ for $i\in\{2,\ldots,n\}$. Then $v,v_{2},\ldots,v_{n}$ can be used as a basis for the Schur decomposition on $\mathbb{C}^{n}$.

## Mathematics Subject Classification

15-00*no label found*

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