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Let $A=\begin{pmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{pmatrix}$
Its eigenvalues are $\lambda=0,0,3$.
Find two pairs of orthonormal eigenvectors for $\lambda=0$ and
show that $P=x_1 x_1^{T}+x_2 x_2^{T}$ is the same for both pairs.
(These eigenvectors make orthogonal matrix $Q$, and $A=Q\lambda Q^{-1}$)

I found first pair $x_1=\frac{1}{\sqrt{2}}(-1,1,0), x_2=\frac{1}{\sqrt{2}}(-1,0,1)$
and second pair $x_1=\frac{1}{\sqrt{6}}(2,-1,-1), x_2=\frac{1}{\sqrt{2}}(0,1,-1)$
but then $P$ of these two pairs are not same.
Where am I wrong?


I solve it. Put $x_1=\frac{1}{\sqrt{2}}(-1,1,0), x_2=\frac{1}{\sqrt{6}}(1,1,-2)$ in the first pair.
I'm not sure there is any particular formula for this question,
but I just put all possible eigenvectors here.
Thanks for your answers and comments.

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Once again: the new pair of vectors $\,x_1,x_2\,$ you found are not orthogonal. –  DonAntonio Dec 11 '12 at 3:15
Actually, I'm not sure what "orthogonal" means. Before normalizing these two eigenvetors, they are orthogonal ($x_1=(-1,1,0), x_2=(1,1,-2)$ and then after normalizing them, they are not orthogonal? ($x_1=\frac{1}{\sqrt{2}}(-1,1,0), x_2=\frac{1}{\sqrt{6}}(1,1,-2)$) –  niagara Dec 11 '12 at 3:41
I calculated it and they are orthogonal. Their inner product produces zero. –  niagara Dec 11 '12 at 3:44
Now they are... –  DonAntonio Dec 11 '12 at 10:18

1 Answer 1

The first pair is not orthonormal. The inner product is $1/2$.

Since any nonzero constant multiple of an eigenvector is also an eigenvector, we take $x_1=(-1,1,0)$ and $x_2=(-1,0,1)$. Then $\langle x_1,x_2 \rangle = -1\cdot -1 + 1 \cdot 0 + 0 \cdot 1 = 1 \neq 0$.

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You're right! But then how can I find the first pair? I have to try all possible eigenvectors? –  niagara Dec 11 '12 at 3:00
I don't think this answer is very helpful. It should be a comment rather than an answer, but @niagara you should orthogonalize the given matrix which then will give you the desired vectors. –  diimension Dec 11 '12 at 6:34

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