Could anyone tell me how to solve this one?
$A=\begin{bmatrix}5&-6&-6\\ -1&4&2\\3&-6&-4 \end{bmatrix}$, and given that $\sin A=B\times C\times E$, then find $B,C,E$.
I never solved such type of problem in past thanks for helping.
I see charpoly of $A$ is $\lambda^3-5\lambda^2+8\lambda-4=0$.
I also checked $P^{-1}AP=\text{diagonal}(2,2,1), P=\begin{bmatrix}2&2&1\\1&0&\frac{-1}{3}\\ 0&1&1\end{bmatrix}$.
I suppose the matrix is diagonalizable, i.e. $A = P^{-1} \Lambda P$, where $\Lambda = diag(\lambda_1, \ldots, \lambda_n)$ - all the eigenvalues of A on the main diagonal and all other entries are $0$'s. And $P$ is the eigenvectors of $A$ stacked in columns.
It is straight-forward to show $$A^n = P^{-1}\Lambda^n P.$$
Then just define functions of matrix as the Taylor expansion, e.g. $$\sin A = \sum\limits_{k=0}^{\infty} \frac{(-1)^k}{(2k+1)!}A^{2k+1}$$
and substitute $A^k$ in.