Find the minimal polynomial of rotation $A$ in angle of $\frac {\pi}{4} $ of $\mathbb R^3$ around the vector $v=(1,2,1)$
I know the rotation matrix in $\mathbb R^3$ is $$\begin{bmatrix} \cos (t) & - \sin (t) & 0 \\ \sin (t) & \cos (t) & 0 \\ 0 & 0 & 1 \end{bmatrix},$$ yet I'm not sure how to rotate around $v$... Thanks in in advance for any help!
We can see geometricaly that $A^4=I_3$ so the polynomial $P(x)=x^4-1$ annihilates $A$ and then the eigenvalues of $A$ are in the set of the roots of $P$ but clearly $- 1$ isn't eigenvalue of $A$ otherwise we can find an eigenvector $x$ such that $$Ax=-x$$ which's geometricaly impossible, hence the polynomial $(x-1)(x^2+1)$ annihilates $A$ and since $A\ne I_3$ then $\mu_A=(x-1)(x^2+1)$ is the minimal polynomial. Notice that $A$ is only diagonalizable over the complex field.