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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?

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    Yes, you are very much correct.2012-11-25
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    You have $x_2=-x_1$ so these are not linearly independent eigenvectors. You want $x_2=(1,0,-1)$ (for example). When you write, $Q^1AQ$, do you mean $Q^tAQ$?2012-11-26
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    Sorry, that means inverse so I corrected it. Oh, I just noticed that two vectors are dependent. I have to change $x_2$ as you wrote. Now I have a question. The $x_1,x_2$ are the eigenvectors for $\lambda=0$ and I think it is possible that two eigenvectors are not independent. But to make orthogonal matrix Q, should I choose independent eigenvectors?2012-11-26
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    Yes. If you choose dependent vectors, $Q$ will have dependent columns, and won't be invertible.2012-11-26
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    I got it! Thanks!2012-11-26
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    @GerryMyerson Please consider converting your comment into an answer, so that this question gets removed from the [unanswered tab](http://meta.math.stackexchange.com/q/3138).2013-10-04
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    @JulianK, done.2013-10-06

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