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A $3\times 3$ matrix $$A = \begin{pmatrix} 2 & -1 & 1\\ -2 & 3 & -2\\-1 & 1 & 0\end{pmatrix}$$

produces this characteristic equation: $\lambda^2 - 4\lambda + 3 = 0$, these eigenvalues: $\lambda_1 = 1$, $\lambda_2 = \frac{-1 + \sqrt{13}}{2}$, $\lambda_3 = \frac{-1 - \sqrt{13}}{2}$, and these eigenvectors:

  1. For $\lambda_1$, $\vec{x} = s\left<1,1,0\right> + t\left<-1,0,1\right>$ (or $\operatorname{span}\{\left<1,1,0\right>, \left<-1,0,1\right>\}$), $\vec{x}_1 = \left<1,1,0\right>$, $\vec{x}_2 = \left<-1,0,1\right>$
  2. For $\lambda_2$, $\vec{x}_3 = \left<0,0,0\right>$
  3. For $\lambda_3$, $\vec{x}_4 = \left<0,0,0\right>$

After normalizing the already orthogonal $\vec{x}_1$, I get $\operatorname{span}\{\left<\sqrt{2}/2, \sqrt{2}/2, 0\right>, \left< -\sqrt{2}/2, 0, \sqrt{2}/2\right>\}$.

Setting up my $P$ matrix for diagonalization, it appears to be a $3\times 4$ matrix that is singular: $\begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 & 0 & 0\\ \sqrt{2}/2 & 0 & 0 & 0\\0 & \sqrt{2}/2 & 0 & 0 \end{pmatrix}$

Questions: Why do I have four vectors, and why are two of them zero vectors? Symmetric matrices are always diagonalizable, where there's a $P$ such that $P^{-1} = P^T$.

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    Are you sure your characteristic equation is correct? I think its $-x^3+5 x^2-7 x+3 = 0$.2010-08-05
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    Please input {{2, -1, 1}, {-2, 3, -2}, {-1, 1, 0}} at http://www.wolframalpha.com and verify information you supported.2010-08-05
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    @TheMachineCharmer: isn't wolfram alpha a bit much, considering that the problem is that the polynomial has the wrong degree? Also, to Gentry...symmetric: I do not think it means what you think it means. That matrix is not symmetric, so why did you mention symmetric matrices at the end?2010-08-05
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    Characteristic equation incorrect, matrix not symmetric. Back to the drawing board. =) Thanks guys!2010-08-05
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    @Charles Siegel : Yes it is :-) Well, do we have to have non-zero co-efficient for $\lambda^{3}$ for any 3 x 3 matrix?2010-08-05
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    Yes. The characteristic polynomial (if you define it by $\det(\lambda I-A)$, which is just $(-1)^n$ times the other definition) is always monic. By the usual def, the lead coefficient is $(-1)^n$ where $n$ is the size of the matrix.2010-08-05

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