If one is asked to find the eigenvector(s) for a Householder transformation matrix, but one is not given the values of or dimensions of the unit vector $u$.

So if $H = I_n - 2uu^T$ where $I_n$ is the n x n identity matrix and has length/norm $||u||^2 = 1$.

It can easily be shown that H is symmetric so that $H = H^T$ and that the eigenvalues are either (and only) of values $\lambda = 1$ and $\lambda = -1$.

Am I correct when assuming that the eigenvectors of $H$ would be the same as those for the identity matrix $I_n$. But is it possible to be more specific/detailed than that when describing the eigenvectors/eigenspaces of these eigenvalues?

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    $\begingroup$ Every vector is an eigenvector of $I$, but not so for $H$. $\endgroup$ – kennytm May 19 '16 at 13:32

The geometric transformation encoded by $$ R=I_n - 2 uu^T $$ is easy to understand: for any $v\perp u$ we have $Rv=v$, while $Ru=-u$, so $R$ encodes a reflection with respect to the orthogonal subspace of $u$. In your case $$ H = I_n + 2uu^T$$ maps any $v\perp u$ into itself and $u$ into $3u$. In both cases, a base of eigenvectors is given by $u$ and a base of the orthogonal subspace of $u$.

  • $\begingroup$ what about eigenspaces? could you help with that? $\endgroup$ – Ernesto Iglesias Feb 9 '17 at 14:01
  • $\begingroup$ @ErnestoIglesias: just think to the geometry - what do you expect the eigenspaces of a reflection to be? $\endgroup$ – Jack D'Aurizio Feb 9 '17 at 16:53

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