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

#### TMO

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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#### fresh_42

Mentor
2018 Award
You could solve for $SJ_M=MS$ or transform it from step to step.

#### TMO

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?

#### fresh_42

Mentor
2018 Award
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 $\mathbb{E}_{-1}^{(2)}=\{\,v\in \mathbb{R}^3\,|\,(M+I)^2.v=0\,\}$

#### TMO

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 $\mathbb{E}_{-1}^2$?

#### fresh_42

Mentor
2018 Award
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 $J_M=\operatorname{diag}(-1,-1,1)$.

#### vela

Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
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 $\lambda = -1$.

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.

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