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?