# Find the Jordan Canonical Form from c(x) and m(x)

Given the matrix B: $$\begin{pmatrix} 2 & 1 & -2 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 1 & 0 & 0 & 0 \\ \end{pmatrix}$$

I have found that the characteristic polynomial is: $c(x)=(x-1)^{3}(x+1)$

and then found that the minimal polynomial is: $m(x)=c(x)$

so then can I conclude that is similar to the Jordan Canonical matrix: $$\begin{pmatrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 1 & -1\\ \end{pmatrix}$$

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You have too many 1's outside the diagonal, there should be none corresponding to $-1$. –  N. S. Aug 9 '12 at 19:28

From the comment of N.S., and CW. $$\left(\begin{matrix} 1 & 0 & 0 & 0\\ 1 & 1 & 0 & 0\\ 0 & 1 & 1 & 0\\ 0 & 0 & 0 & -1\\ \end{matrix}\right)$$

Note that, despite the multiple eigenvalue, any matrix that commutes with $B$ can be written as a polynomial $p(B).$

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If I have the matrix A = $$\begin{matrix} 0 & 0 & 0\\ 1 & 0 & 0 \\ 0 & 1 & 1\\ \end{matrix}$$ I am finding J = A. Is this correct? –  sarah jamal Aug 9 '12 at 20:04
$A$ is not a Jordan canonical form, I'm afraid. So $J\neq A$. –  a.r. Aug 9 '12 at 20:14
@AgustíRoig I found $c(x)=m(x)=x^{2}(x-1)$ what would $J=$? –  sarah jamal Aug 10 '12 at 9:59
@AgustíRoig would J= $$\begin{pmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 0 & 1\\ \end{pmatrix}$$ –  sarah jamal Aug 10 '12 at 10:06
@sarahjamal. Yes, you're right. The reason is because the $0$ eigenvalue has $\mathrm{dim}\ \mathrm{Ker}(A) = 3 - \mathrm{rank}(A) = 3 - 2 = 1$. That is, just one Jordan block. And the $1$ eigenvalue diagonalises trivially. –  a.r. Aug 10 '12 at 18:13

I'm not sure how you computed the minimal polynomial, but it's not a necessary step -nor its knowledge will give you, in general, the Jordan canonical form.

Instead, once I had the characteristic polynomial, I would procede as follows. Assume it looks like

$$Q(t) = (t-\lambda)^m \dots$$

That is: the eigenvalue $\lambda$ has algebraic multiplicity equal to $m$. That is, the Jordan block of $\lambda$ has size $m\times m$.

Then, I would compute the following ranks:

$$\mathrm{rank} (A - \lambda I) > \mathrm{rank} (A - \lambda I)^2 > \mathrm{rank} (A - \lambda I)^3 > \dots$$

Or, what is the same, the dimensions of the following subspaces:

$$\mathrm{Ker} (A - \lambda I) \subset \mathrm{Ker} (A - \lambda I)^2 \subset \mathrm{Ker} (A - \lambda I)^3 \subset \dots$$

When I got

$$d_\alpha = \mathrm{dim}\ \mathrm{Ker} (A - \lambda )^\alpha = m$$

I would stop. All the information you need about the Jordan canonical form is contained in the sequence of numbers

$$d_i = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I )^i \ , \qquad i = 1, \dots, \alpha \ .$$

For instance,

$$d_1 = \mathrm{dim}\ \mathrm{Ker} (A - \lambda I ) = n - \mathrm{rank} (A - \lambda I)$$

is the number of "boxes" inside the Jordan block of the eigenvalue $\lambda$. (Here, $n$ is the size of your matrix.)

$$d_1 = \mathrm{dim}\ \mathrm{Ker} (A - I ) = 4 - \mathrm{rank} (A - I) = 4- 3 = 1$$

means there is just one box in the Jordan block of $1$. So, it must be $3 \times 3$. That is, it must be

$$\begin{pmatrix} 1 & 0 & 0 \\ 1 & 1 & 0 \\ 0 & 1 & 1 \end{pmatrix}$$

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The knowledge of the minimal polynomial can be useful; see my answer. –  M Turgeon Aug 9 '12 at 20:53
@Agustí Roig: Note that the highest power of $t-\lambda$ that divides the minimal polynomial gives you the size of the largest Jordan block associated to $\lambda.$ –  Ehsan M. Kermani Aug 9 '12 at 21:05
@MTurgeon & ehsanmo: Yes, I know that. But, on the other hand, just the knowledge of the characteristic polynomial, together with the minimal one, does not determine the Jordan canonical form. For instance: $$\begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 \end{pmatrix}$$ and $$\begin{pmatrix} 0 & 0 & 0 & 0 \\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{pmatrix}$$ are different Jordan canonical forms with equal characteristic and minimal polynomials. –  a.r. Aug 10 '12 at 3:47
What can you do with this? Suppose the (distinct) eigenvalues of a matrix $A$ are $\lambda_1,\ldots,\lambda_r$, and suppose that each eigenspace is equal to 1. That is, the geometric multiplicity of each eigenvalue is 1. This means that there are exactly $r$ Jordan blocks in the Jordan canonical form, one for each eigenvalue. Therefore, the Jordan canonical form is $$\left(\begin{matrix}J_1& &\\ &\ddots& \\ & &J_r\end{matrix}\right),$$ where $J_i$ is a Jordan block of size $m_i\times m_i$ ($m_i$ is the algebraic multiplicity of $\lambda_i$) for the eigenvalue $\lambda_i$. In particular, the Jordan canonical form for your matrix $B$ is the one in Will Jagy's answer.