Computing sinus for squared not diagonilized matrix $A$

Let $A = 2\begin{pmatrix}4 & -1 & -3 \\ -2 & 1 & 1 \\-2 & 3 & -4\end{pmatrix}$

I want to compute $\sin\left(\dfrac{\pi A}{2}\right)$ for this matrix. I know that sinus of a matrix can be expressed as sum of infinite series, but for this I have to know $A^k$. Am I on the right direction or there're other approaches to this problem? Any help would be greatly appreciated.

• I get three distinct eigenvalues, why wouldnt this be diagonalizable? – Olba12 Sep 20 '18 at 0:34
• @Olba12 I don't claim it's not diagonalizable in a different basis. It's not diagnolized in its original form. Do you think diagonalization may be of help? Could you please describe your approach? – Swistack Sep 20 '18 at 0:36
• If $A=PDP^{-1}$ is diagonalizable then $$f(A) = P \begin{bmatrix} f(d_1) & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & f(d_n) \end{bmatrix}P^{-1}$$ – Olba12 Sep 20 '18 at 0:38
• The matrix built by eigenvectors to corrensponding eigenvalues – Olba12 Sep 20 '18 at 0:41
• Swistack, check the numbers in the preparation paper against what you typed above. You cannot expect to do this problem unless the eigenvalues are very easy, indeed integers. Then $\sin W = (e^{iW} - e^{-iW})/(2i)$ and you can find the exponential parts if you can explicitly diagonalize $A$ – Will Jagy Sep 20 '18 at 0:56

Let $$A\in M_3(\mathbb{Q})$$ that admits $$3$$ distinct real eigenvalues $$(\lambda_i)_i$$ and let $$f:\mathbb{R}\rightarrow \mathbb{R}$$ be a $$C^{\infty}$$ function. Note that $$A$$ is diagonalizable over $$\mathbb{R}$$ and $$f(A)$$ commute with $$A$$, and consequently, is a polynomial in $$A$$.
Let $$P\in \mathbb{R}[x]$$ be the Lagrange interpolating polynomial of degree $$2$$ that sends the $$\lambda_i's$$ on the $$f(\lambda_i)'s$$ ($$P(x)=a+bx+cx^2$$). Then $$f(A)=P(A)=aI_3+bA+cA^2$$.
Unfortunately, here, the $$(\lambda_i)$$ are in an algebraic extension of $$\mathbb{Q}$$ of degree $$6$$; we can explicitly calculate the $$(\lambda_i)'s$$ but the calculations of $$a,b,c$$ is quasi unfeasible (except for Maple for example).