Prove that $HAH^{-1} = \left( \begin{smallmatrix} \lambda_{1} & b^{t} \\ 0 & B \\ \end{smallmatrix} \right) $ Consider $A \in \Bbb{R}^{nxn}$ with distinct eigenvalues $\lambda_{1}, ..., \lambda_{n}$ and respective eigenvectors $v_{1}, ..., v_{n}$.
$H \in \Bbb{R}^{nxn}$ an orthogonal matrix so that $Hv_{1} = \alpha e_{1}$. Show that $HAH^{-1} = \left( \begin{matrix} \lambda_{1} & b^{t} \\ 0 & B \\   \end{matrix} \right) $ with $b \in \Bbb{R}^{n-1}$ and $B \in \Bbb{R}^{(n-1)x(n-1)}$ and that $\lambda_{2}, ..., \lambda_{n}$ are eigenvalues for $B$.
Here is what I know:
As $A$ has n distinct eigenvalues I know it can be diagonalized into  $A = SDS^{-1}$ with the eigenvalues $\lambda_{1}, ..., \lambda_{n} $ in the diagonal $D$ and H is the householder reflection.
 A: HINT
Here's something to get it going and hopefully it helps you. First think about what happens if you multiply $HAH^T$ by the vector $e_1$. Second think about the eigenvalues of the matrix $HAH^T$ using properties of the similarity of matrices. You know that $A \sim SDS^{-1}$, where $D$ is a diagonal matrix with the eigevalues of $A$ on its diagonal. What can you conclude about $HAH^T$?
EDIT: 
You must find that 
$HAH^{-1} = \left( \begin{matrix} \lambda_{1} & b^{t} \\ 0 & B \\   \end{matrix} \right) $ with $b \in \Bbb{R}^{n-1}$.
Now I say this is the same as saying $e_1$ is an eigenvector of $HAH^{T}$ with $\lambda_{1}$ as it's eigenvalue and that the remaining matrix $B$ has as it's eigenvalues $\lambda_{2}, ... \lambda_{n}$ as $B \in \mathbb{R}^{(n-1)(n-1)}$. 
For the first part use that $Hv_1 = \alpha_{1}e_1$ which means that $v_1 =\alpha_{1} H^{T}e_1$ which implies that $H^{T}e_1 = v_{1} {\alpha_{1}}^{-1}$. Now a simple calculation shows that $HAH^{T}e_1 = $$\lambda_{1}e_1$.
For the second part use that as A is diagonalizable which means there exist matrices $S, S^{-1},  D$ where $D$ is a diagonal matrix with the eigenvalues of $A$ on it's diagonal such that $A=SDS^{-1}$. These implies that $HAH^{-1} = HSDS^{-1}H^{-1}$ which implies that it has the same eigenvalues as the matrix $A$, in other words they are similar matrices. Hope this clear's it up.
