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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 $p(A)=\sin(A)$.

Michael Hardy
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sarah jamal
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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
  • Find the degree 2 polynomial that agrees with $\sin$ on the eigenvalues of $A$. – copper.hat Sep 30 '12 at 18:11
  • @copper.hat I have found a "lagrange interpolation formula" in my textbook. It says $$ p(x)=\sum^{n}{i=1}p(\lambda{i})h_{i}(x)$$ I used this with my working above and I got $$ \frac{4}{3}-\frac{4}{3\pi^{2}}x^{2}$$ could the textbook perhaps have a typo in it? – sarah jamal Sep 30 '12 at 18:22
  • @sarahjamal: Yes, if the answer you gave below is what the textbook gave, then it is wrong. Your answer is correct, to check, note that $p(-\pi) = p(\pi) = 0$, $p(\frac{\pi}{2}) = 1$. You could also check by just evaluating $p(A)$. – copper.hat Sep 30 '12 at 18:53
  • @copper.hat thanks so much!! now I have another matrix $$ J_{3}(\frac{\pi}{2}) $$ and I need to find $$\sin(J) \text{ and } \cos(J)$$ however, since I only have one eigenvalue, I don't know how to use the spectral decomposition method. :( – sarah jamal Sep 30 '12 at 19:14
  • 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 $J^k$. – copper.hat Sep 30 '12 at 20:31

1 Answers1

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

Berci
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