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The eigen values of following matrix are $4, -2,-2$ and corresponding eigen vectors are $(1,1,1), (-1,1,0), (-1,0,1)$. But as the matrix is symmetric the eigen vectors has to be orthogonal, where as last two or not. I know that $(1,-1,0)$ is a eigen vector for eigen value $-2$. But while I am solving I am not getting that, Can some one help me out.

\begin{pmatrix}0&2&2\\2&0&2\\2&2&0\end{pmatrix}.

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    $\begingroup$ The eigenvectors of DIFFERENT values of eigenvalues must be orthogonal. Eigenvectors of same eigenvalues (in your case, $-2$) need not to be orthogonal (also you can choose orthogonal ones). This is well-known in physics and it's called degeneracy. $\endgroup$ Commented Sep 21, 2014 at 5:11
  • $\begingroup$ Oh. But is there any procedure to choose orthogonal one. If possible can you provide some references to this type problems. $\endgroup$
    – Kumar
    Commented Sep 21, 2014 at 5:13
  • $\begingroup$ It's very easy. You see, the two eigenvectors of the value $-2$ generate a eigenspace of value $-2$, which is the subspace that contains all the possible eigenvectors of $-2$. You have a particular basis of this space, now you just need to orthogonalize it, by using Gram-Schimidt Orthogonalization maybe? $\endgroup$ Commented Sep 21, 2014 at 5:16

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There should be a way to get orthagonal eigenvectors (I believe)...but I need to go through how to solve it (your eigenvectors are obviously linear combinations of two orthagonal ones):

$$ \begin{pmatrix}2 & 2 & 2\\ 2 & 2 & 2 \\ 2& 2 & 2 \end{pmatrix} \times \begin{pmatrix} x \\ y \\ z \end{pmatrix} = \begin{pmatrix} 0\\ 0\\ 0 \end{pmatrix} $$

Now obviously this immediately degenerates to:

$2x + 2y + 2z = 0$

This gives:

$$ x = -y - z $$

So we come up with the following two eigenvectors:

$$ \left\langle-a - b, a, b\right\rangle\\ \left\langle-y - z, y, z \right\rangle $$

Find values of $a$, $b$, $y$, $z$ such that:

$$ \left\langle-a - b, a, b\right\rangle \circ \left\langle-y - z, y, z \right\rangle = ay + bz + az + by + ay + bz = 0 $$

Let's simplify:

$$ 2ay + 2bz + az + by = 0 $$

It's not surprising that there are many solutions to this, since you should be able to scale each of the two eigenvectors. Let's solve for one of the variables:

$$ a(2y + z) = -2bz - by \\ a = \frac{-2bz - by}{2y + z} $$

Now let's simply choose an arbitrary value of $b = 1$, this gives:

$$ a = \frac{-2z - y}{2y + z} $$

Furthermore, let's choose values of $y$ and $z$ such that the denominator above is not equal to zero...say $y = 1$ and $z = 1$:

$$ a = \frac{-2 - 1}{2 + 1} = -1 $$

This gives two orthogonal vectors of:

$$ \left\langle 0, -1, 1\right\rangle\\ \left\langle-2, 1, 1 \right\rangle $$

Finally, we can make them orthonormal:

$$ \left\langle 0, -\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right\rangle\\ \left\langle-\sqrt{\frac{2}{3}}, \frac{1}{\sqrt{6}}, \frac{1}{\sqrt{6}} \right\rangle $$

...I should say, that I do not believe that these particular, orthogonal eigenvectors are unique (i.e. I suspect there are multiple sets of two orthogonal eigenvectors).

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