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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
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3 Answers 3

up vote 4 down vote accepted

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
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$A$ 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
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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.)

In your example,

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

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
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@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
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@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
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I am not sure what you really are asking for, but here is a result you might find useful. As Arturo Magidin proved here:

The minimal polynomial is equal to the characteristic polynomial if and only if the dimension of each eigenspace is 1.

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.

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