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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?

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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.

  • 0
    Once again: the new pair of vectors $\,x_1,x_2\,$ you found *are not* orthogonal.2012-12-11
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    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? ($x_1=\frac{1}{\sqrt{2}}(-1,1,0), x_2=\frac{1}{\sqrt{6}}(1,1,-2)$)2012-12-11
  • 0
    I calculated it and they are orthogonal. Their inner product produces zero.2012-12-11
  • 0
    *Now* they are...2012-12-11

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