Spectral decomposition of a diagonal matrix

  • Thread starter syj
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  • #1
syj
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Homework Statement


I have
[tex]
J=\begin{bmatrix}
\frac{\pi}{2}&0&0\\
1&\frac{\pi}{2}&0\\
0&1&\frac{\pi}{2}\\
\end{bmatrix}
[/tex]

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


Homework Equations





The Attempt at a Solution


I have the following:

[tex]
\sin(J)=
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{bmatrix}
[/tex]

and
[tex]
\cos(J)=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
[/tex]

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.
 

Answers and Replies

  • #2
Ray Vickson
Science Advisor
Homework Helper
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Homework Statement


I have
[tex]
J=\begin{bmatrix}
\frac{\pi}{2}&0&0\\
1&\frac{\pi}{2}&0\\
0&1&\frac{\pi}{2}\\
\end{bmatrix}
[/tex]

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


Homework Equations





The Attempt at a Solution


I have the following:

[tex]
\sin(J)=
\begin{bmatrix}
1&0&0\\
0&1&0\\
0&0&1\\
\end{bmatrix}
[/tex]

and
[tex]
\cos(J)=
\begin{bmatrix}
0 & 0 & 0\\
0 & 0 & 0\\
0 & 0 & 0\\
\end{bmatrix}
[/tex]

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
[tex] \begin{pmatrix} \lambda & 0 & 0 \\
1 & \lambda & 0 \\
0 & 1 & \lambda
\end{pmatrix}^n [/tex]

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

RGV
 
  • #3
syj
55
0
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
Ray Vickson
Science Advisor
Homework Helper
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1,722
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
 

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