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Given $$ J=\begin{bmatrix} \frac{\pi}{2}&0&0\\ 1&\frac{\pi}{2}&0\\ 0&1&\frac{\pi}{2}\\ \end{bmatrix} $$

find $\sin(J) \text{ and } \cos(J)$

I know I need to find the spectral decomposition, but I am not sure what to do because the examples and exercises in the textbook I have all have more than one eigenvalue, here there is only one eigenvalue.

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$J=\frac{\pi}{2}I$? –  Patrick Li Oct 1 '12 at 16:20
    
You seem to be assuming you can diagonalize $J$. But $J$ isn't normal. –  Kevin Carlson Oct 1 '12 at 16:24
    
@PatrickLi omg i can't believe i wrote that. how do i go about answering this question? I know how to find lagrange polynomials and spectral decomposition when I have more than one eigenvalue. –  sarah jamal Oct 1 '12 at 16:25
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If $AB=BA$ for two matrices, then $\cos(A+B)$ and $\sin(A+B)$ satisfy the usual formulas. Also, if $\sin(\theta I) = \sin(\theta) I$ when $\theta$ is real, and same for $\cos(\theta I)$. Finally, you can hand-compute $\sin(A)$ and $\cos(A)$ when $A^n=0$ for some $n$ –  Thomas Andrews Oct 1 '12 at 16:27
    
@ThomasAndrews I actually need to prove that property you just mentioned. Can you help me with that? –  sarah jamal Oct 1 '12 at 16:29
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2 Answers 2

up vote 4 down vote accepted

As was discussed in the comments, let $K=e_{21}+e_{32}$. $J=\frac{\pi}{2}I+K$, and certainly $I$ and $K$ commute, so $$\sin J=\sin\left(\frac{\pi}{2}I+K\right)=\cos\left(\frac{\pi}{2}I\right)\sin K+ \sin\left(\frac{\pi}{2}I\right)\cos K$$ The first term vanishes since $\cos(\pi/2)$ does, and $\sin(\pi/2 I)=I$, so all we need is $\cos K$. As noted in another answer $K^n=0$ for $n\geq 3,$ so we can compute this directly from the power series: $$\sin J=\cos K=I-\frac{1}{2}K^2$$

This solution uses the angle-sum formula brought up in the comments, which again is implied by $\exp(A+B)=\exp A\exp B\ $ when $A$ and $B$ commute.

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Shouldn't $\cos K = I-\frac{1}{2}K^2$? –  Thomas Andrews Oct 1 '12 at 17:00
    
Indeed, edited! –  Kevin Carlson Oct 1 '12 at 17:01
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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 $f=\sin$.]

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