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I want to prove that the columns of a Householder Reflector (i.e. $H = I_n - 2 uu^t$, with $||u|| = 1$) are its eigenvectors.

Well, $H^2 = I_n$ and $H=H^t$.

So, $Spec(H) \subseteq \{ -1,1 \} $.

Now, $Hu=-u$ so $-1 \in Spec(H)$.

Then, if v is a vector perpendicular on u, i.e. $<v,u>=u^tv=v^tu=0$, then $Hv=v$, so $1 \in Spec(H)$.

Moreover, if $U=Span(u)$, then $U^{\perp} \cap U = \emptyset$ and $U + U^{\perp} = \mathbb{R}^n$, so $U \oplus U^{\perp} = \mathbb{R}^n$. From here we get that $dim U + dim U^\perp = n$. But, $dim U = 1$ so $dim U^\perp = n-1$.

We deduce that 1 is an eigenvalue with multiplicity n-1 and -1 is an eigenvalue with multiplicity 1 and eigenvector u.

Unfortunately I don't know how to prove that the columns of H are eigenvectors of H.

I tried to prove this by writing $H=G^tDG$, where G is an orthogonal matrix with the columns the eigenvectors of H (because H is symmetric) and D is diag(-1,1,...,1), but, nothing.

Can anybody hekp, me please ? Thank you!

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1 Answer 1

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This is in general wrong.

Let $h_i = He_i$ denote the $i$ th column of $H$. If $h_i$ is an eigenvector, then we have $$ \pm h_i = Hh_i = e_i. $$ That is obviously in general wrong.

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  • $\begingroup$ I have one more question, if you don't mind: Is everything that I wrote above (especially the proof for the multiplicacities of the 2 eigenvalues) correct? Thank you! $\endgroup$
    – npatrat
    Commented Apr 6, 2016 at 21:08
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    $\begingroup$ @npatrat everything about the eigenvalues and their multiplicity are fine. $\endgroup$
    – user251257
    Commented Apr 6, 2016 at 21:09

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