The question is to find the eigenvalues and vectors of the matrix $$\begin{bmatrix}9 & 8 \\ 8 & 9 \end{bmatrix}$$

Which by getting to the form:

$$\begin{bmatrix}9 - \lambda & 8 \\ 8 & 9 - \lambda \end{bmatrix}v = 0$$

and using the determinant of the matrix = 0, to get a quadratic equation to solve, you get the eigenvalues 1 and 17

If you substitute 17 back into the equation you get:

$$\begin{bmatrix}-8 & 8 \\ 8 & -8 \end{bmatrix}v = 0$$

and end up with the simultaneous equations

$-8v_1 + 8v_2 = 0 $ and $ 8v_1 - 8v_2 = 0 $

which ends up with $ v_1 = v_2 $

So the eigenvector could be any vector as long as the two components are equal?

(Also doesn't this apply to any matrix and it's eigenvectors, in that if you map an eigenvector to new vector and then map that again, you'll get a vector that still hasn't changed direction, which is the definition of an eigenvector. i.e.: $A(Av) = v' $, where v is an eigenvector. Won't the direction of v' be the same as v and Av and therefore v, Av, and v' are all eigenvectors? )

  • $\begingroup$ yes. (as long as the resulting vector is nonzero). $\endgroup$ – Michael Sep 8 '15 at 0:18
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    $\begingroup$ Right. What you really want, then, are bases of the eigenspaces. $\endgroup$ – Omnomnomnom Sep 8 '15 at 0:20
  • $\begingroup$ If $\mathbf v$ is an eigenvector, then $a\mathbf v$ is an eigenvector, for any non-zero real $a$. These are usually treated as equivalent eigenvectors, for most purposes. $\endgroup$ – Thomas Andrews Sep 8 '15 at 0:42

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