Spectral decomposition of a diagonal matrix

1. Oct 1, 2012

syj

1. The problem statement, all variables and given/known data
I have
$$J=\begin{bmatrix} \frac{\pi}{2}&0&0\\ 1&\frac{\pi}{2}&0\\ 0&1&\frac{\pi}{2}\\ \end{bmatrix}$$

I need to find $$\sin(J) \text{ and } \cos(J) \text{ and show that } \sin^{2}(J)+\cos^{2}(J)=I$$

2. Relevant equations

3. The attempt at a solution
I have the following:

$$\sin(J)= \begin{bmatrix} 1&0&0\\ 0&1&0\\ 0&0&1\\ \end{bmatrix}$$

and
$$\cos(J)= \begin{bmatrix} 0 & 0 & 0\\ 0 & 0 & 0\\ 0 & 0 & 0\\ \end{bmatrix}$$

I don't know if this is correct. All the questions I have examples of have more than one eigenvalue, this one only has one eigenvalue.

2. Oct 1, 2012

Ray Vickson

J is basically a Jordan form, except its 1's are below the diagonal instead of above; that is, it is a transpose of a Jordan matrix. So, you need to know what is the form of
$$\begin{pmatrix} \lambda & 0 & 0 \\ 1 & \lambda & 0 \\ 0 & 1 & \lambda \end{pmatrix}^n$$

See, eg., http://en.wikipedia.org/wiki/Jordan_normal_form to see why your final answers are incorrect.

RGV

3. Oct 1, 2012

syj

The textbook I have defines a Jordan matrix to be one where the 1s are below the diagonal. So the matrix given is already a Jordan matrix.

4. Oct 1, 2012

Ray Vickson

OK, fine. But did you read the rest of the message?

RGV