So here is my problem,
Let A be a symmetric Matrix (2×2) with EV=(-2, 3) and the Eigenvalue being -5.
Find the 2. Eigenvector to the second Eigenvalue.
The only info i can think of is that the EV of a symmetric matrix have to be orthogonal, so the dot product has to be 0.
So -2×1 +3×2 =0
I then can fix x1=a for example and then calculate, x2=2/3
$$ (1, 2/3)^t *a, $$
for all a Elements of R without 0
Is this the correct answer it seems a bit too simple….
Best Answer
You know that $\begin{pmatrix}-2\\3\end{pmatrix}$ is an eigenvector relative to $-5$. You correctly argue that the eigenvectors relative to the other eigenvalues are orthogonal to this one, so we find $-2x_1+3x_2=0$. Any such vector is good, so we can take $\begin{pmatrix}3\\2\end{pmatrix}$.
You are correct.
The spectral theorem says that $$ A=-5vv^T+\lambda ww^T $$ where $v$ and $w$ are the normalization of the two eigenvectors. There is no way to determine the second eigenvalue from the given data. You have $$ A=-\frac{5}{13}\begin{pmatrix} 4 & -6 \\ -6 & 9\end{pmatrix} +\frac{\lambda}{13}\begin{pmatrix} 9 & 6 \\ 6 & 4\end{pmatrix} $$