# Finding Jordan canonical form of these matrices

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1. Mar 30, 2016

1. The problem statement, all variables and given/known data
For each matrix A, I need to find a basis for each generalized eigenspace of $L_A$ consisting of a union of disjoint cycles of generalized eigenvectors. Then I need to find the Jordan canonical form of A.
The matrices are:
$a) \begin{pmatrix} 1 & 1\\ -1 & 3 \end{pmatrix} b) \begin{pmatrix} 1 & 2\\ 3 & 2 \end{pmatrix} c) \begin{pmatrix} 11 & -4 & -5\\ 21 & -8 & -11\\ 3 & -1 & 0 \end{pmatrix} d) \begin{pmatrix} 2 & 1 & & \\ & 2 & 1 & \\ & & 3 & \\ & 1 & -1 & 3 \end{pmatrix}$

2. Relevant equations
$P_A(t)=det(A-tI)$
$K_\lambda = \left \{ (A-\lambda I)^j \right \}, 1\leq j \leq p$ where p is the minimum value for which $(A-\lambda I)^j(x)=0$ for a generalized eigenvector x

3. The attempt at a solution
I hope you don't mind if I link to a picture of what I have so far, as I don't want to go through the trouble of typing it all up.
http://prntscr.com/am7si6
The squiggles next to some of the matrices are just messily written words indicating that there is only one generalized eigenvector for a given basis.

For (a), I determined that the only basis element was (1,1)
For (b), I found 2 single-element basis which contained (1,-1) and (1,1.5)
For (c), I once again found that each generalized eigenspace only had one generalized eigenvector. However, what I'm not sure of is how to determine what the Jordan blocks are in this case- i.e, I don't know how to determine if
$J=\begin{pmatrix} -1 & 1 & \\ & -1 & \\ & & 2 \end{pmatrix}$ or
$J=\begin{pmatrix} -1 & & \\ & 2 & 1\\ & & 2 \end{pmatrix}$
For (d), I have not yet started. I just need to confirm that my current approach of raising the power of the generalized matrices to find generalized eigenvectors is correct, and I need to know how to figure out what the Jordan blocks are.

2. Mar 31, 2016

### Ray Vickson

The eigenvalues for (c) are -1,2,2 so the eigenvalue 2 has multiplicity two but eigenspace dimensionality one. Therefore, the Jordan block is for 2.

3. Apr 1, 2016

### vela

Staff Emeritus
For (a), you have a 2x2 matrix, so the basis should comprise two vectors.

4. Apr 5, 2016

How do I find the other one? For the matrix U, I don't know how I would find another eigenvector for what it is, and U2 is the zero matrix. 2 was also the only eigenvalue I was able to find.

Not on any of my assignments, but how would I do it for a matrix with multiple blocks but only one eigenvalue?

5. Apr 5, 2016

### vela

Staff Emeritus
If $\vec{p}_1$ is the eigenvector, the generalized eigenvector satisfies $(A-\lambda I)\vec{p}_2 = \vec{p}_1$. That'll give you a more useful equation to solve.