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Given $ A= \begin{bmatrix} \frac{\pi}{2} & 1 & 2 \\ 0 & \pi & 3 \\ 0 & 0 & -\pi \end{bmatrix} $ $ \text{find} \sin(A) $ $ \text{I have found the spectral decomposition of A to be} $ $ \frac{\pi}{2}E_1 + \pi E_2 - \pi E_3 $ with $ E_{1}= \begin{bmatrix} 1 & -2\pi & \frac{4\pi -12}{3 \pi^{2}}\\ 0 & 0 & 0 \\ 0 & 0 & 0 \end{bmatrix} $ $ E_{2}= \begin{bmatrix} 0 & \frac{2}{\pi} & \frac{3}{\pi^{2}}\\ 0 & 1 & \frac{3}{2\pi}\\ 0 & 0 & 0\\ \end{bmatrix} $ $ E_{3}= \begin{bmatrix} 0 & 0 & \frac{3-4\pi}{3\pi^{2}}\\ 0 & 0 & \frac{-3}{2\pi}\\ 0 & 0 & 1\\ \end{bmatrix} $ I then used this to say $ \sin(A)=\sin\left(\frac{\pi}{2}\right)E_{1} + \sin(\pi)E_{2} - \sin(\pi)E_{3} \therefore \sin(A)=E_1 $ I now need to find $p(x)$ such that $p(A)=\sin(A)$.

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    If $d$ is the geometric multiplicity of an eigenvalue $\lambda$ (ie, the size of the largest Jordan block), then you need $p^{(k)}(\lambda) = f^{(k)}(\lambda)$ for all $k=0,...,d-1$, where $f$ is the function in question. In your example above, $d=1$ for all eigenvalues, so there was no need to match derivatives. Alternatively, you could try computing the matrices directly (ie, using the power series expansion) once you have figured out a formula for $J^k$.2012-09-30

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I'm not sure that your 'spectral decomposition' is good.

However, you can reduct the problem to the case for $A':=\begin{bmatrix} \pi/2 &0&0 \\ 0&\pi&0 \\ 0&0&-\pi \end{bmatrix}$.

This is because $A$ has 3 different eigenvalues, and take one eigenvector for each: $v_1,v_2,v_3$, it is a basis, then the matrix $B:=\left[ v_1|v_2|v_3 \right]$ is invertible and it will conjugate $A$ to $A'$: $A=BA'B^{-1}$ With this, using $(BAB^{-1})^n = BA^nB^{-1}$, we get that both $p(A)=B\cdot p(A')\cdot B^{-1}\ \text{ and }\ \sin(A) = B\cdot \sin(A')\cdot B^{-1} $ (because $\sin$ is defined by its power series..)

So, $p(A)=\sin(A) \iff p(A')=\sin(A')$.

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    I haven't calculated the polynom, this might be good, we can verify if $p(A')=\sin(A')$. Ah, I see, a typo, yes, the $\pi^2$ should go in the *denominator*, and then it seems to work.2012-09-30