No, this is a so called Jordan block, and not much more can be done in its decomposition.
Let $A=\begin{bmatrix} 0&0&0\\1&0&0\\0&1&0 \end{bmatrix}$, and say, we have a matrix $J= A+\alpha\cdot I$. Denote the elements of the standard basis in $\mathbb R^3$ by $e_1,e_2,e_3$. Since $A$ takes $e_1\mapsto e_2$, $e_2\mapsto e_3$, $e_3\mapsto 0$, we have $A^2=\begin{bmatrix} 0&0&0\\0&0&0\\1&0&0 \end{bmatrix}$ and already $A^3=0$.
Hence, $J^2 = A^2+2\alpha A+\alpha^2 I$, $\ J^3=3\alpha A^2+3\alpha^2 A+\alpha^3 I$, and find similarly $J^n$ by the binomial theorem.
Then, substituting it in a power series we will practically get 3 distinct power series for the diagonal lanes (with coefficients of $A^2$, $A$ and $I$). Anyway, at the part 'Functions of matrices' of the wikipage, the result is described, and $f(J)$ will also contain elements related to $f'(\alpha)$ and $f''(\alpha)$. [Now $\alpha=\frac\pi2$ and $f=\sin$.]