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- Apr 14, 2013

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We have the matrix $A=\frac{1}{3}\begin{pmatrix}1 & 2 & 2 \\ 2 & 1 & -2 \\ 2 & -2 & 1\end{pmatrix}$. Show that there is an unit vector $v_1$, such that $A=I-2v_1v_1^T$.

We consider an orthogonal matrix $Q=\begin{pmatrix}v_1 & v_2 & v_3\end{pmatrix}$. Show that $Q^TAQ=\begin{pmatrix}-1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1\end{pmatrix}$.

Using the last relation show that $\det A=1$.

For the first part I have done the following:

Let $v_1=\begin{pmatrix}x_1 & x_2 & x_3\end{pmatrix}^T$ with $|v_1|=\sqrt{x_1^2+x_2^2+x_3^2}=1$.

Then we have that \begin{equation*}I-2v_1v_1^T=\begin{pmatrix}1-2x_1^2 & -2x_1x_2 & -2x_1x_3 \\ -2x_1x_2 & 1-2x_2^2 & -2x_2x_3 \\ -2x_1x_3 & -2x_2x_3 & 1-2x_3^2\end{pmatrix}\end{equation*} We set this equal to $A$ and we have to solve a system.

Is this correct or should we show that in an other way?