Is there $2 \times 2$ symmetric matrix with eigenvectors $u, v$ where the angle between $u$ and $v$ is not $π/2$? Is there $2 \times 2$ symmetric matrix with eigenvectors $u, v $ where the angle between $u$ and $v$ is not $π/2$?
In other words, where the dot product between $u$ and $v$ is not $0$?
 A: A matrix being symmetric implies that $v\cdot (Aw) = (Av)\cdot w$ for any two vectors $v, w$. If $v$ and $w$ are eigenvectors of $A$ with different eigenvalues $\lambda_1, \lambda_2$, we get
$$
\lambda_1(v\cdot w) = (\lambda_1 v)\cdot w = (Av)\cdot w = v\cdot (Aw) = v\cdot(\lambda_2w) = \lambda_2(v\cdot w)
$$
and since $\lambda_1\neq\lambda_2$, we must have $v\cdot w = 0$.
A: If the eigenvalues $\lambda_u$ and $\lambda_v$ of $u$ and $v$ are different, it's impossible. Because,
$\left< u,Av \right>= \lambda_v\left<u,v\right>$
and $\left<Au,v\right>= \lambda_u\left< u ,v \right>$.
But,$\left< u,Av \right>=\left< Au,v \right>$.
So $(\lambda_u- \lambda_v)\left<u,v\right>=0$.
And $\left<u,v\right>=0$
A: The only matrices fulfilling your condition are multiples of $$\begin{pmatrix} 1&0\\0&1\end{pmatrix}.$$
In this case $(1,0)$ and $(1,1)$ are not orthogonal linear independent eigenvectors.
Edit: calculating the characteristic polynomial of $$\begin{pmatrix} a&b\\b&c\end{pmatrix}$$
its discriminant is $$(a-c)^2+4b^2.$$
Hence the matrix possesses only one eigenvalue iff $b=0$ and $a=c$.
