# Symmetric matrices and also orthogonal diagonalization.

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

generates this particular formula : $\lambda^2 - 4\lambda + 3 = 0$, these eigenvalues : $\lambda_1 = 1$, $\lambda_2 = \frac{-1 + \sqrt{13}}{2}$, $\lambda_3 = \frac{-1 - \sqrt{13}}{2}$, and also 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 stabilizing the currently orthogonal $\vec{x}_1$, I get $\operatorname{span}\{\left<\sqrt{2}/2, \sqrt{2}/2, 0\right>, \left< -\sqrt{2}/2, 0, \sqrt{2}/2\right>\}$.

Establishing my $P$ matrix for diagonalization, it seems a $3\times 4$ matrix that is single : $\begin{pmatrix} \sqrt{2}/2 & -\sqrt{2}/2 & 0 & 0\\ \sqrt{2}/2 & 0 & 0 & 0\\0 & \sqrt{2}/2 & 0 & 0 \end{pmatrix}$

Questions : Why do I have 4 vectors, and also why are 2 of them absolutely no vectors? Symmetrical matrices are constantly diagonalizable, where there's a $P$ such that $P^{-1} = P^T$.

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2019-05-07 11:23:56
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Answers: 2

Well, first of all, a vector isn't orthogonal, a set of vectors is. Bear in mind that for 2 vectors to be orthogonal, it suggests that they go to appropriate angles per various other, and also orthonormal methods that plus they're device vectors.

Currently, if I needed to claim where I assume your first mistake is, you took a 3 x 3 matrix and also obtained a square formula in some way, yet you need to have a cubic. As well as additionally, when it comes to your eigenvectors, where did $\lambda_4$ originated from? What is it? The icon simply shows up from no place. Probably you have typos and also it's $\lambda_2,\lambda_3$ as opposed to 3 and also 4, and also each of them have to have a nonzero eigenvector, due to the fact that they are eigenvalues of multiplicity 1 (though with the formula mistake, they could not be eigenvalues), therefore you would certainly need to be wrong concerning having a 2 dimensional eigenspace to begin with.

Nonetheless, both of the eigenvectors for 1 check out, which suggests that you've inaccurately computed the eigenvalues.

(as opposed to simply a solution, I place in all the reasoning I did to arrive, due to the fact that I assumed it could aid make clear just how to examine your operate in the future)

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2019-05-09 06:48:21
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Your particular polynomial is wrong. Most likely to wolframalpha.com and also enter : 2, - 1, 1 , - 2, 3, 2 , - 1, 1, 0

www.wolframalpha.com/input/?i = 2,+- 1,+1 , - 2,+3,+2 ,+ - 1,+1,+0

You will certainly see that the particular polynomial is of 3rd level. It has to be of 3rd level due to the fact that your matrix is 3x3. Below 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 certainly additionally get 3 eigenvectors. (Never 4 for a 3x3 matrix).

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2019-05-09 06:43:39
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