Suppose we want to perform eigenvalue decomposition on this matrix:
$$P = \begin{bmatrix} 1.5 && -0.5\\ -0.5 && 1.5 \end{bmatrix}$$
I obtain the corresponding eigenvectors as $\lambda_1 = 1, \lambda_2 = 2$, and eigen-vector $v_1 = [1, 1]^T$, $v_2 = [1, -1]^T$.
So our $Q$ as in $P = Q\Lambda Q^T$ is given by
$$Q = \begin{bmatrix} 1 && 1\\ 1 && -1 \end{bmatrix}$$
Which is not a rotational matrix after we normalize it!
But we can alternatively pick our eigenvectors as $v_1 = [1, 1]^T$, $v_2 = [-1, 1]^T$. Notice $v_2$ has changed by a sign
Then obviously $$Q = \begin{bmatrix} 1 && -1\\ 1 && 1 \end{bmatrix}$$
Is a rotational matrix after normalization. In fact, it is trivial to see it rotates by $45^\circ$
Can someone resolve this paradox?