Rotated matrix and rotated eigenvectors.

Question

Let $R_\theta$ be a rotational matrix, my intuition (from experiences with quadratic forms) tells me that if $A$ has an eigenvector $v$, then $R_\theta^T A R_\theta$ has an eigenvectors $R_\theta v$.

Is this true? If so, how can I prove this?

Answer 1

Not quite. The matrix $R_\theta^TAR_\theta$ has eigenvector $R^T_\theta v$, because $$R_\theta^TAR_\theta(R_\theta^Tv)=R^T_\theta A(Iv)=R^T_\theta Av=R^T_\theta(\lambda v)=\lambda (R^T_\theta v)$$ where I've used the fact that $R_\theta^TR_\theta=I$, because $R_\theta$ is an orthogonal matrix.

Answer 2

Close.

If $v$ is an eigenvector of $A$, then $v' = R_\theta^T v$ is an eigenvector of $R' =R_\theta^T A R_\theta$, with the same eigenvalue.

Proof:

$$ A v = \lambda v $$ $$ R'v' = (R_\theta^T A R_\theta) (R_\theta^T v) = R_\theta^T A (R_\theta R_\theta^T) v $$ $$ = R_\theta^T A v = R_\theta^T \lambda v = \lambda (R_\theta^T v) = \lambda v' $$

This proof relies on the equation: $R_\theta R_\theta^T = I$, which is true for rotations.