Spectral decomposition of a diagonal matrix

[syj]

Homework Statement

I have
<br /> J=\begin{bmatrix}<br /> \frac{\pi}{2}&amp;0&amp;0\\<br /> 1&amp;\frac{\pi}{2}&amp;0\\<br /> 0&amp;1&amp;\frac{\pi}{2}\\<br /> \end{bmatrix}<br />

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

Homework Equations

The Attempt at a Solution

I have the following:

<br /> \sin(J)=<br /> \begin{bmatrix}<br /> 1&amp;0&amp;0\\<br /> 0&amp;1&amp;0\\<br /> 0&amp;0&amp;1\\<br /> \end{bmatrix}<br />

and
<br /> \cos(J)=<br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix}<br />

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.

 

[Ray Vickson]

Homework Statement

I have
<br /> J=\begin{bmatrix}<br /> \frac{\pi}{2}&amp;0&amp;0\\<br /> 1&amp;\frac{\pi}{2}&amp;0\\<br /> 0&amp;1&amp;\frac{\pi}{2}\\<br /> \end{bmatrix}<br />

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

Homework Equations

The Attempt at a Solution

I have the following:

<br /> \sin(J)=<br /> \begin{bmatrix}<br /> 1&amp;0&amp;0\\<br /> 0&amp;1&amp;0\\<br /> 0&amp;0&amp;1\\<br /> \end{bmatrix}<br />

and
<br /> \cos(J)=<br /> \begin{bmatrix}<br /> 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0\\<br /> 0 &amp; 0 &amp; 0\\<br /> \end{bmatrix}<br />

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.

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 &amp; 0 &amp; 0 \\<br /> 1 &amp; \lambda &amp; 0 \\<br /> 0 &amp; 1 &amp; \lambda<br /> \end{pmatrix}^n

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

RGV

 

[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.

 

[Ray Vickson]

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.

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

RGV