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Let $A\in \mathbb{M}^{n\times n}(\mathbb{C})$. Let $(\lambda,v)$ an eigenpair of $A$ and $\|v\|=1$. Let $V$ be a unitary matrix with $v$ as first column. Prove that there exists a $b \in \mathbb{C}^{n-1}$ and $B\in \mathbb{M}^{(n-1)\times (n-1)}$

$$ AV=V\left( \begin{array}{c|c} \lambda & b^* \\ \hline 0 & {{\mbox{$B$}}} \\ \vdots & \\ 0 & \end{array} \right) $$

It's quite some time ago that I followed my first linear algebra course. I forget how I can see such an equality. Any help would be really appreciated.

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I've edited the title. Feel free to change it to something that reflects the content better, but please don't put a huge matrix in the title. –  TMM Apr 23 '13 at 20:52
    
This simply says that there exists an orthonormal basis of $\mathbb{C}^n$ starting with $v$. This does not require anything but the existence of an orthonormal basis on $v^\perp$. –  1015 Apr 23 '13 at 20:52
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1 Answer 1

Let $(v,e_2,\cdots,e_n)$ a basis of $\mathbb{C}^n$ and by Gram–Schmidt process we can orthonormalise this basis $(v,e_2,\cdots,e_n)$ in a basis $\mathcal{B}=(v,v_2,\cdots,v_n)$ and then the matrix $A$ is similar to the matrix $$\left( \begin{array}{c|c} \lambda & b^* \\ \hline 0 & {{\mbox{$B$}}} \\ \vdots & \\ 0 & \end{array} \right) $$ written in the basis $\mathcal{B}$ and the change matrix from the canonical basis to the basis $\mathcal{B}$ is the unitary matrix $V=(v,v_2,\cdots,v_n)$.

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