I try to get the eigenvalue m and the vectors a & h from this eigenvector problem:
$$A = \begin{bmatrix}0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0\end{bmatrix}$$
$$A⋅A^T⋅h=m⋅h$$ $$A^T⋅A⋅a=m⋅a$$
I started with this:
$$A^T = \begin{bmatrix}0 & 1 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0\end{bmatrix}$$
$$A⋅A^T = \begin{bmatrix}1 & 0 & 0 \\ 0 & 2 & 1 \\ 0 & 1 & 1\end{bmatrix}$$
$$A^T⋅A = \begin{bmatrix}2 & 0 & 1 \\ 0 & 1 & 0 \\ 1 & 0 & 1\end{bmatrix}$$
and then i got the highest m with (Characteristic Polynomial):
$$-x^3 + 4x^2 - 4x + 1 = 0$$
$$m≈\frac{3+√5}{2}$$
Now i don't really know what to do. I think i have to do this to get the core of the matrix?:
$$(A⋅AT−mE)⋅h⃗ =0$$
to get this:
$$ \begin{bmatrix} 1-(\frac{3+√5}{2}) & 0 & 0 \\ 0 & 2-(\frac{3+√5}{2}) & 1 \\ 0 & 1 & 1-(\frac{3+√5}{2})\end{bmatrix} * \begin{bmatrix} h_{1} \\ h_{2} \\ h_{3} \end{bmatrix} = 0 $$
But when i try to solve it i only get:
$$ h1 = 0 $$
$$ \frac{√5+1}{2} * h_{2} + h_{3} = 0 $$
$$ \frac{√5-1}{2} * h_{3} + h_{2} = 0 $$
=> so this??:
$$h⃗ = \begin{bmatrix} 0 \\ 0 \\ 0 \end{bmatrix}$$
Am i solving it wrong or am i doing a mistake somewhere else? Anything could help!
EDIT:
After the fix i get this:
$$(A⋅AT−mE)⋅h⃗ =0$$
=>
$$
\begin{bmatrix} 1-(\frac{3+√5}{2}) & 0 & 0 \\ 0 & 2-(\frac{3+√5}{2}) & 1 \\
0 & 1 & 1-(\frac{3+√5}{2})\end{bmatrix}
*
\begin{bmatrix} h_{1} \\ h_{2} \\ h_{3} \end{bmatrix} = 0
$$
=>
$$ h_{1} = 0 $$
$$ \frac{1-√5}{2} * h_{2} + h_{3} = 0 $$
$$ -\frac{√5+1}{2} * h_{3} + h_{2} = 0 $$
=>
$$h_{3} * 0 = h_{3}$$
$$h_{2} = \frac{√5+1}{2} * h_{3}$$
=>
$$h⃗ = \begin{bmatrix} 0 \\ 1,62*x \\ x \end{bmatrix}$$
Is this the right solution for vector h?