2
$\begingroup$

Find the orthogonal $Q$ so that $Q^{-1}AQ=B$ if
$A=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$ and $B=\begin{pmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{pmatrix}$
then find a second pair of orthonormal eigenvectors $x_1,x_2$ for $\lambda=0$.
Show that $P=x_1x_1^{T}+x_2x_2^{T}$ is the same for both pairs.

My solution
The eigenvalues of A is 0,0,3. The eigenvectors corresponding that is $x_1=(-1, 1, 0) , x_2=(1,-1, 0) , x_3=(1, 1, 1)$
Using gram-schmidt, I found
$Q=\begin{pmatrix} \frac{-1}{\sqrt{2}} &\frac{1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ \frac{1}{\sqrt{2}} & \frac{-1}{\sqrt{2}} & \frac{1}{\sqrt{3}} \\ 0 & 0 & \frac{1}{\sqrt{3}} \end{pmatrix}$

Actually, I'm not sure what 'second pair' means but I picked another pair of $x_1=(-1/2, 1/2, 0) , x_2=(1/2, -1/2, 0)$
Then by the formula given, P is equal to both pairs.

I'm not sure that I did well. Is that correct?

  • 0
    @JulianK, done.2013-10-06

1 Answers 1

1

Expanding comment to answer:

First, one must choose linearly independent eigenvectors for the eigenvalue 0. Having chosen $x_1=(-1,1,0)$, one may, for example, choose $x_2=(1,0,-1)$. But since we want something orthogonal to $x_1$, a better choice is $x_2=(1,1,-2)$. This leads to the orthogonal matrix $Q=\pmatrix{-1/\sqrt2&1/\sqrt6&1/\sqrt3\cr1/\sqrt2&1/\sqrt6&1/\sqrt3\cr0&-2/\sqrt6&1/\sqrt3\cr}$ A different pair of (normalized) eigenvectors for the eigenvalue 0 is given by $x_1=(1/\sqrt{14})(1,-3,2)$, $x_2=(1/\sqrt{38})(5,-3,-2)$. We leave it to the reader to compute $x_1x_1^t+x_2x_2^t$ for both (normalized) pairs.