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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 $-x^3+5 x^2-7 x+3 = 0$. – Pratik Deoghare Aug 5 '10 at 13:18
Please input {{2, -1, 1}, {-2, 3, -2}, {-1, 1, 0}} at and verify information you supported. – Pratik Deoghare Aug 5 '10 at 13:20
@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? – Charles Siegel Aug 5 '10 at 13:30
Characteristic equation incorrect, matrix not symmetric. Back to the drawing board. =) Thanks guys! – Gentry Aug 5 '10 at 14:00
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. – Charles Siegel Aug 5 '10 at 14:17

Well, first off, a vector isn't orthogonal, a set of vectors is. Remember that for two vectors to be orthogonal, it means that they are at right angles to each other, and orthonormal means that plus they're unit vectors.

Now, if I had to say where I think your first error is, you took a 3 x 3 matrix and got a quadratic equation somehow, but you should have a cubic. And also, as for your eigenvectors, where did $\lambda_4$ come from? What is it? The symbol just appears from nowhere. Perhaps you have typos and it's $\lambda_2,\lambda_3$ rather than 3 and 4, and each of them must have a nonzero eigenvector, because they are eigenvalues of multiplicity 1 (though with the equation error, they might not be eigenvalues), and so you would have to be incorrect about having a two dimensional eigenspace in the first place.

However, both of the eigenvectors for 1 check out, which means that you've incorrectly calculated the eigenvalues.

(instead of just an answer, I put in all the thinking I did to get there, because I thought it might help clarify how to check your work in the future)

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+1 for "a vector isn't orthogonal, a set of vectors is" :) – Pratik Deoghare Aug 5 '10 at 13:23
Thanks for your expedience, Charles! 1) Yes, and <b>x<sub>1</sub><b> and <b>x<sub>2</sub><b> are now orthonormal to each other. Sorry for not being clear about that, I haven't slept and it's 0630 my time. 2) I got a quadratic after factoring out (&lambda; - 1) from the cubic via rational root theorem, and the only solutions to the quadratic were the ones I posted. I checked it with a factoring applet to be sure. – Gentry Aug 5 '10 at 13:39
<html> 3) There is no &lambda;<sub>4</sub>, you're right, that was a typo, but that's the source of my confusion. &lambda;<sub>1</sub> gave me two vectors, <b>x<sub>1</sub></b> and <b>x<sub>2</sub><b>. &lambda;<sub>2</sub> and &lambda<sub>3</sub> each gave me a vector <0, 0, 0> -- do I just treat these two solutions as one vector/column when forming my P matrix? Why? </html> – Gentry Aug 5 '10 at 13:40
Wow, that turned out very poorly. First time using the site, my apologies for that being so unclear. (1) x1 and x2, from the solution space of (lambda)_1, are orthonormal. (2) The quadratic is from factoring (lambda - 1) via rational root theorem, and the quadratic had no "nice" solution, I had to brute force it with the quadratic formula. (3) There is no lambda_4, that was a typo, sorry about that. But why do I have four vectors, two from lambda_1, and one <0,0,0> from each lambda_2? Also, the <0,0,0> vector makes P singular. I've checked and quadruple-checked, all my arithmetic is right. – Gentry Aug 5 '10 at 13:45
Wow, I looked but did not see. The matrix, obviously, is not symmetric. I was assuming it was because that's all we covered in class today, haha. Also, my characteristic equation WAS incorrect - silly errors when calculating the determinant! Thank you very much for clearing that up for me gentlemen. – Gentry Aug 5 '10 at 13:59

Your characteristic polynomial is incorrect. Go to and enter: {{2, -1, 1},{-2, 3, 2}, {-1, 1, 0}}{{2,+-1,+1},{-2,+3,+2},+{-1,+1,+0}}

You will see that the characteristic polynomial is of 3rd degree. It must be of 3rd degree because your matrix is 3x3. Here is what you'll get:

p(x)=-x^3+5 x^2-3 x-1
lambda_1 = 4.23607
lambda_2 = 1
lambda_3 = -0.236068

And you will also get 3 eigenvectors. (Never 4 for a 3x3 matrix).

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You input the matrix incorrectly, there's no +2 in the second row, and all three eigenvalues are integral, and all three eigenvectors are defined over the integers. – Charles Siegel Aug 5 '10 at 14:19

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