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.

  • $\begingroup$ By any chance are you studying "Metodos numericos" at the UBA? $\endgroup$ – Leo Lerena May 27 '18 at 7:17
  • $\begingroup$ Yo have discovered me. $\endgroup$ – estebanquito May 27 '18 at 13:48


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.

  • $\begingroup$ I know that if I right multiply any vector of the standard basis $e_{i}$, I get the $i-th$ column of the matrix. In this case the first column of $HAH^{-1} $ which as you said because of orthogonality is equal to $HAH^{t}$. I am still thinking about the second thing. $\endgroup$ – estebanquito May 27 '18 at 13:54
  • $\begingroup$ But you can actually compute $(HAH^T)e_1$ and get the first column! And my second observation is to conclude that $HAH^T$ has the same eigenvalues as $A$ (although not the same eigenvectors). $\endgroup$ – Leo Lerena May 27 '18 at 15:31
  • $\begingroup$ If this still doesn't help you, I can write up the solution. $\endgroup$ – Leo Lerena May 27 '18 at 15:32
  • 1
    $\begingroup$ I would appreciate greatly $\endgroup$ – estebanquito May 30 '18 at 3:02

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