Spectral decomposition of a diagonal matrix

In summary, the task is to find the sine and cosine of J, which is a Jordan matrix with 1's below the diagonal. The solution is \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}, however, this is incorrect as J is not a Jordan matrix in the traditional sense. In order to find the correct solution, we need to consider the form of
  • #1
syj
55
0

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.
 
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  • #2
syj said:

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
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
syj said:
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
 

Related to Spectral decomposition of a diagonal matrix

What is spectral decomposition of a diagonal matrix?

Spectral decomposition, also known as diagonalization, is a process in linear algebra that involves breaking down a diagonal matrix into its constituent parts, including its eigenvalues and eigenvectors.

Why is spectral decomposition important?

Spectral decomposition is important because it allows us to simplify complex matrices into a form that is easier to work with. It also reveals important information about the matrix, such as its eigenvalues and eigenvectors, which can be used for further analysis.

How is spectral decomposition calculated?

The process of spectral decomposition involves finding the eigenvalues and eigenvectors of a diagonal matrix. This can be done using various methods, such as the power iteration method or the Jacobi method.

What are the applications of spectral decomposition?

Spectral decomposition has many practical applications, including in signal processing, image compression, and data analysis. It is also used in quantum mechanics to describe the behavior of quantum systems.

Is spectral decomposition the same as matrix diagonalization?

Yes, spectral decomposition and matrix diagonalization refer to the same process of breaking down a diagonal matrix into its eigenvalues and eigenvectors. However, spectral decomposition is often used in the context of linear algebra, while diagonalization is used more broadly in mathematics.

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