Null Eigenvector

Question

I calculated the three eigenvalues from a matrix and now I want to calculate the associated eigenvectors.

Initial matrix is: $ \begin{bmatrix} 1+\sigma^2 & a & b\\ a & 1+\sigma^2 & a \\ b & a & 1+\sigma^2 \end{bmatrix} $

the Eigenvalues I found are $1+\sigma^2-b, 1+\sigma^2-\dfrac{1}{2}(b + \sqrt{b^2 + 8a^2}), 1+\sigma^2-\dfrac{1}{2}(b - \sqrt{b^2 + 8a^2})$

Let's say $K = \dfrac{1}{2}(b + \sqrt{b^2 - 8a^2})$; the second eigenvalue gives a matrix like this:

$ \begin{bmatrix} K & a & b\\ a & K & a \\ b & a & K \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $

but the solution is $\begin{bmatrix}x, y, z\end{bmatrix}^T$ = $\begin{bmatrix}0, 0, 0\end{bmatrix}^T$

Wasn't an Eigenvector not null by definition, was it?

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EDIT

if $K = \dfrac{1}{2}(b + \sqrt{8a^2 + b^2})$

$ \begin{bmatrix} -K & a & b\\ a & -K & a \\ b & a & -K \end{bmatrix} $

gives as solutions: $x = z(\dfrac{K^2-a^2}{K + a^2})$

$y = z(\dfrac{K^2(K^2-a^2)-2a^2(K^2-a^2)}{K(K + a^2)})$

$z (b - \dfrac{K^4 - 3K^2a^2 + 2a^4}{K(K + a^2)}) = 0$

Answer 1

the Eigenvalues I found are $1+\sigma^2-b, 1+\sigma^2-\dfrac{1}{2}(b + \sqrt{b^2 + 8a^2}), 1+\sigma^2-\dfrac{1}{2}(b - \sqrt{b^2 + 8a^2})$

Careful here: the first eigenvalue is correct but the other two are: $$1+\sigma^2 \color{red}+\frac{1}{2}\left(b \color{blue}{\pm} \sqrt{b^2 + 8a^2}\right)$$

And this also means that in:

Let's say $K = \dfrac{1}{2}(b + \sqrt{b^2 - 8a^2})$; the second eigenvalue gives a matrix like this:

$ \begin{bmatrix} K & a & b\\ a & K & a \\ b & a & K \end{bmatrix} \begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix} $

you need $K = \frac{1}{2}\left(\color{red}{-}b + \sqrt{b^2 \color{red}{+} 8a^2}\right)$, the other one being $K = \frac{1}{2}\left(\color{red}{-}b \color{red}{-} \sqrt{b^2 \color{red}{+} 8a^2}\right)$.

To verify, that should lead to these and these results respectively.

Answer 2

The eigenvector $[x, y, z]^T$ is non-null, by definition. However, the eigenvalue (and thus the result of applying the linear transformation to the eigenvector) is very much allowed to be $0$.

If you mean that solving that equation gives you $[x, y, z]^T = [0,0,0]^T$ as the only solution, then that means that there is no eigenvector with eigenvalue $0$.

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Edit: Now that we have the actual matrix in front of us: Inserting $a = b = 1, \sigma = 0$ should give us eigenvalues of $0, 0$ and $3$ in some order. That's not what I get from your formulas (they give $0, -1$ and $2$ respectively). For next time (and every time after) I'm strongly advising you to do simple checks like that every time you encounter such difficulties. Checking and re-checking eigenvalues becomes much easier once you have a concrete (and comparatively simple) matrix to work with, rather than what almost amounts to a general symmetric matrix.

I am guessing that it should be $+$ rather than $-$ between $1+\sigma^2$ and $\frac12(b\pm \sqrt{b^2+8a^2})$.

Answer 3

Many times it can be easier to look for eigenvectors directly instead of following the rote process of computing roots of the characteristic polynomial and then computing null spaces for each eigenvalue. This is one of those times.

The matrix $A$ is real symmetric, so we know that it is orthogonally diagonalizable. Subtracting the last column from the first produces $(1+\sigma^2-b,0,b-(1+\sigma^2))^T$, so we know that $\mathbf v_1 = (1,0,-1)^T$ is an eigenvector with eigenvalue $1+\sigma^2-b$. We therefore look for other eigenvectors orthogonal to this one. The vectors $(1,0,1)^T$ and $(0,1,0)^T$ are an obvious basis for the orthogonal complement of $\mathbf v_1$, i.e., we want vectors of the form $\mathbf v = (c,d,c)^T$. If this is an eigenvector of $A$, then $A\mathbf v$ is a scalar multiple of $\mathbf v$, which holds iff $\mathbf v\times A\mathbf v=0$. This leads to the equation $$ad^2-2ac^2+bcd = 0.$$ (The cross product actually generates two nontrivial equations, but they aren’t independent.) If $a\ne0$, the quadratic formula gives us $c$ in terms of $d$ or vice-versa. It turns out that the two solutions correspond to orthogonal eigenvectors, so we don’t need to do anything else to find a third. If, on the other hand, $a=0$, the equation reduces to $bcd=0$. If $b=0$, then $A$ is a multiple of the identity matrix and every vector is an eigenvector, otherwise, either $c=0$ or $d=0$, so our two basis vectors $(1,0,1)^T$ and $(0,1,0)^T$ are the eigenvectors we seek. This is obvious from inspection of the matrix $A$ with $a=0$. In any of these cases, the corresponding eigenvalues are easily recovered by computing $A\mathbf v$.