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Good evening,

I would love your help with this.

I want to know what's jordan normal form of matrix that it's $Characteristic$ $ $ $polynomial$ : $(t-3)^{4}\cdot (t-5)^{4}$

and it's $Minimal$ $ $ $ polynomial$ : $(t-3)^{2}\cdot (t-5)^{2}$.

I believe that these are the options:

$J_{A}=diag(J_{2}(3),J_{2}(3),J_{2}(5),J_{2}(5))$,

$J_{A}=diag(J_{2}(3),J_{2}(3),J_{2}(5),J_{1}(5),J_{1}(5))$,

$J_{A}=diag(J_{2}(3),J_{1}(3),J_{1}(3),J_{2}(5),J_{1}(5),J_{1}(5))$,

$J_{A}=diag(J_{2}(3),J_{1}(3),J_{1}(3),J_{2}(5),J_{2}(5))$.

My Question are:

1.Am I right?

2.Was there any difference if Jordan matrices of eigenvalue 5 were before eigenvalue 3? and why?

3.Is there any way to know more about what is the specific jordan normal form with this information?

Thank you guys.

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2) theres no difference, its just switching the basis around some more –  yoyo Apr 8 '11 at 21:42
    
@yoyo:ok, thanks. –  user6163 Apr 8 '11 at 21:45
3  
3) you need to know the rank of the matrices $(A-3)$ and $(A-5)$ or equivalently the dimension of $\ker(A-3)$ and $\ker(A-5)$ to know which specific form it is with the given information you have. $\dim \ker(A-\lambda)$ is the number of Jordan blocks for eigenvalue $\lambda$, and the exponents of the minimal polynomial are the length of the longest block for each eigenvalue. –  Javier Álvarez Apr 8 '11 at 22:14
    
Note that the information Javier notes is enough in this situation (because you only have two possibilities, which yield different dimensions of the eigenspaces, so knowing the dimension of the eigenspaces determines which possibility you are in), but it may not be enough in general. –  Arturo Magidin Apr 9 '11 at 4:53
    
+1 for thinking about the problem before posting –  Ross Millikan Apr 9 '11 at 5:06

1 Answer 1

up vote 2 down vote accepted

If the characteristic polynomial is $(t-3)^4(t-5)^4$, then the eigenvalues are $3$ (with multiplicity $4$) and $5$ (with multiplicity $4$ as well).

If the minimal polynomial is $(t-3)^2(t-5)^2$, this tells you that the largest Jordan block corresponding to $3$ is $2\times 2$ (and there is at least one block of size $2\times 2$), and likewise for $5$.

The possible "dot diagrams" for each of the eigenvalues are then $$\begin{array}{cc} \bullet & \bullet\\ \bullet &\bullet \end{array},\qquad\text{and}\qquad \begin{array}{ccc} \bullet & \bullet & \bullet\\ \bullet \end{array},$$ corresponding to two blocks of size $2$; or to a single block of size $2$ and two blocks of size $1$.

So you have two possibilities for each eigenvalue. This gives a total of $4$ possibilities, which are the ones you list. The information you have is insufficient to distinguish between the four possibilities.

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