Finding the householder transformation given $P = P(w)$ such that $P(w) x = e_{1}$

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!

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 $$18$$.

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