Find the matrix $f(A)$ for $f(z)=cos(3\pi z)$ with $A =\begin{bmatrix} 0 & 3& -5& 2\\ 1& 3& 1& -1\\ 1 &-1 &4 &-1\\-2& 2& -4& 4 \end{bmatrix}$
I'm taking a linear algebra course and this was one of the homework questions (studying for the final). Although there was no reference to which chapter this related to. I think I have to utilize generalized eigenvectors and the Jordan Canonical form, but I'm not exactly sure how.
If I were to use J.C. Form, my characteristic polynomial would be $f(\lambda)=(\lambda -2)(\lambda - 3)^3$, so my eigenvalues are $\lambda=2,3$. Then the $Nul(A-2I)=Span\left\{\begin{bmatrix} 1\\0\\0\\1 \end{bmatrix} \right\}$. But then I believe $Nul(A-3I)=Span\left\{\begin{bmatrix} 3\\0\\-1\\2 \end{bmatrix} \right\}$? (I think)... Is this way I should start this?
One has $A=QFQ^{-1}$ where $F=diag(2,3I_3+J)$, where $J$ is the nilpotent Jordan block of dimension $3$. Then $f(A)=Qf(F)Q^{-1}$.
The conclusion comes from the calculation of $Q$ and
$f(F)=diag(f(2),f(3I+J))$ with $f(3I+J)=f(3I)+f'(3)J+(1/2)f''(3)J^2$.