How do I find the change of basis matrix for the JCF of M?

In summary: In this case, it would be## \begin{align}S=\begin{pmatrix} 1& 0& 1 \\ 1& 0& 0 \\ 0& 1& 0 \end{pmatrix} \end{align}. ##In summary, the JCF for the given matrix is found by first finding the characteristic polynomial and then using AMGM and integer partition equations to guess the JCF. If this method does not work, one can also find the JCF by using eigenvectors and generalized eigenvectors. The transition matrix can be found by forming a matrix with the columns being the generalized eigenvectors.
  • #1
TMO
45
1
Let

## \begin{align}M =\begin{pmatrix} 2& -3& 0 \\ 3& -4& 0 \\ -2& 2& 1 \end{pmatrix} \end{align}. ##

Here is how I think the JCF is found.

STEP 1: Find the characteristic polynomial

It's ## \chi(\lambda) = (\lambda + 1)^3 ##

STEP 2: Make an AMGM table and write an integer partition equation

The AM is given by looking at the power. The GM is found by finding the nullspace for each eigenvalue. For this matrix this is the table:

Code:
+-----+------+------+
|  λ  |  AM  |  GM  |
+-----+------+------+
| -1  |  3   |  2   |
+-----+------+------+

which gives the integer partition equation ## J_{1, \lambda_{-1}} + J_{2, \lambda_{-1}} = 3 ##. Because there's only one integer partition possible (up to permutation: remember that the JCF is unique only up to permutation not in general), we can guess the JCF is

## \begin{align}J_M =\begin{pmatrix} -1& 1& 0 \\ 0& -1& 0 \\ 0& 0& -1 \end{pmatrix} \end{align}. ##

But I don't know how to compute the transition matrix. I know it involves generalized eigenvectors. Can someone help me?
 
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  • #2
You could solve for ##SJ_M=MS## or transform it from step to step.
 
  • #3
fresh_42 said:
You could solve for ##SJ_M=MS## or transform it from step to step.

I need to find the change of basis matrix using eigenvectors and generalized eigenvectors. Why? Because for larger matrices I may not be able to get a nice number partition that allows me to guess the JCF. I know there's a way to do this. How do I do this?
 
  • #4
O.k., another method is to calculate ##M.v=-v## which should give a decomposition ##\mathbb{R}^3= \mathbb{E}_{-1}^{(1)} \oplus \mathbb{E}_{-1}^{(2)}## with a one dimensional eigenspace ##\mathbb{E}_{-1}^{(1)}## and a two dimensional generalized eigenspace ##\mathbb{E}_{-1}^{(2)}=\{\,v\in \mathbb{R}^3\,|\,(M+I)^2.v=0\,\}##
 
  • #5
Taking ## (M + I) = 0 ## and transforming it into RREF gives

##\begin{align}\begin{pmatrix} 1& -1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix} \end{align}##.

The non-pivot columns are two, so the eigenspace is given by

##\begin{align} v_1 - v_2 =& 0& \\ v_2 =& r \\ v_3 =& s \end{align}##

Rewriting this in terms of linear span gives

## \begin{align}\left\{\begin{pmatrix} 1 \\ 1 \\ 0\end{pmatrix}, \begin{pmatrix} 0 \\ 0 \\ 1\end{pmatrix}\right\}\end{align} ##.

Is this the eigenspace ## \mathbb{E}_{-1}^2 ##?
 
  • #6
I haven't done your homework, but it is easy to check that ##(1,1,0)^\tau## is an eigenvector to ##-1## and ##(0,0,1)^\tau## an eigenvector to ##1##, which contradicts your characteristic polynomial. I also think that ##J_M=\operatorname{diag}(-1,-1,1)##.
 
  • #7
TMO said:
Taking ## (M + I) = 0 ## and transforming it into RREF gives

##\begin{align}\begin{pmatrix} 1& -1& 0 \\ 0& 0& 0 \\ 0& 0& 0 \end{pmatrix} \end{align}##.
I got a 1 in the third column of the second row after row reduction.

I also got 1 and -1 as the eigenvalues, and you get only one eigenvector for ##\lambda = -1##.

TMO said:
But I don't know how to compute the transition matrix. I know it involves generalized eigenvectors. Can someone help me?
You just form a matrix where the columns are the generalized eigenvectors.
 

1. How do I determine the Jordan canonical form (JCF) of a matrix M?

To determine the JCF of a matrix M, you must first find the eigenvalues of M. Then, for each eigenvalue, find the corresponding eigenvectors and determine their algebraic and geometric multiplicities. Finally, use these values to construct the JCF of M.

2. What is the purpose of finding the JCF of a matrix?

The JCF of a matrix provides a simplified and canonical form of the matrix that makes it easier to perform calculations and understand the properties of the original matrix. It also reveals important information about the eigenvalues and eigenvectors of the matrix.

3. How do I find the change of basis matrix for the JCF of M?

To find the change of basis matrix for the JCF of M, you must first find the basis of eigenvectors for each eigenvalue of M. Then, arrange these eigenvectors in a matrix and invert it to get the change of basis matrix.

4. Can I find the change of basis matrix for the JCF of M without finding the JCF first?

No, the change of basis matrix is based on the JCF of M. Without knowing the JCF, it is not possible to determine the correct change of basis matrix.

5. Are there any shortcuts or tricks for finding the change of basis matrix for the JCF of M?

There are no shortcuts or tricks for finding the change of basis matrix for the JCF of M. It requires following the steps of finding the JCF and then constructing the change of basis matrix based on the JCF.

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