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The matrix $$A =\begin{bmatrix} 2 & 10 & 2 \\ 10 & 5 & -8 \\ 2 & -8 & 11 \\ \end{bmatrix}$$

has an eigenvalue $\lambda = 9$ with the corresponding eigenvector $x = (2/3, 1/3, 2/3)^{T}$. Find the Householder transformation, $P = P(w)$ such that $P(w) x = e_{1}$ and compute the remaining two eigenvalues from the $2 \times 2$ minor of $PAP^{T}$.

I'm not quite sure how to start this problem. I understand how to do householder transformations but Im not understand what this question is asking

Any help is appreciated!

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Build the difference vector ;

$$U^T = (2/3, \ 1/3, \ 2/3)^T - (1, \ 0, \ 0)^T = (-1/3, \ 1/3, \ 2/3)^T$$

Normalize it : $V^T=U^T/\|U^T\|=\dfrac{1}{\sqrt{6}}(-1 \ 1 \ 2)^T$,

then build Householder matrix : $I-2VV^T =$

$$=\begin{pmatrix}1&0&0\\0&1&0\\0&0&1\end{pmatrix}-2\dfrac{1}{\sqrt{6}} \begin{pmatrix}-1\\ \ \ 1\\ \ \ 2\end{pmatrix}\dfrac{1}{\sqrt{6}}(-1 \ 1 \ 2)^T $$

$$P=\begin{pmatrix}2/3&1/3& \ \ 2/3\\1/3& \ \ 2/3&-2/3\\2/3&-2/3&-1/3\end{pmatrix}$$

A quick glance at the first column shows that indeed the image of $e_1$ is vector $(2/3,1/3,2/3)^T$.

Besides, as a Householder matrix is an isometry matrix, $P^T=P^{-1}$ ; therefore $PAP^T=PAP^{-1}$ ; this matrix, being similar to $A$, is known to share the same (eigen)values.

A quick computation shows that $$PAP^{-1}=diag(-9,9,18)$$

Thus the other eigenvalues are $-9$ and $18$.

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  • $\begingroup$ Ah I see. Thanks so much! $\endgroup$ – Michael Dec 10 '18 at 19:23

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