Find $\sin(A)$ and $p(x)$ such that $p(A)=\sin(A)$

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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How did you find these $E_1,E_2,E_3$? and how could you split up $\sin A$ using them? –  Berci Sep 30 '12 at 17:22
You don't need to write \operatorname{sin}; just \sin is enough. It's a standard operator name in $\TeX$. –  Michael Hardy Sep 30 '12 at 17:23
Do you intend $p$ to be a polynomial? If there are no constraints on what $p$ can be, why not just say $p=\sin$ and you're done? Also $p$ could be the usual power-series expansion of the sine function. –  Michael Hardy Sep 30 '12 at 17:24
@MichaelHardy p needs to be a polynomial. –  sarah jamal Sep 30 '12 at 17:28
@Berci the E's are found using the lagrange interpolation polynomials, $$E_{1}=\lambda_{1}h_{1}(A)$$. There is an example in my text book (Matrices and Linear Transfromations, Cullen) that found sin(A) by that method, so I followed that. However, I don't know how to get the p(x). I know the answer for p(x) because it is in the back of the book, but I don't know how to get it. However, the book does not give me sin(A). –  sarah jamal Sep 30 '12 at 17:32

1 Answer

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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And $\sin(A')=\begin{bmatrix} 1 &0&0 \\ 0&0&0 \\ 0&0&0 \end{bmatrix}$. –  vesszabo Sep 30 '12 at 17:52
@Berci the textbook gives me $$p(x)=\frac{4}{3}-(\frac{4}{3}\pi{^2})x^{2}$$ –  sarah jamal Sep 30 '12 at 17:53
See my comment above. –  copper.hat Sep 30 '12 at 18:11
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. –  Berci Sep 30 '12 at 18:54