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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.

  • 0
    Note that the i$n$form$a$tion J$a$vier notes is enough in this situ$a$tion ($b$ecause you only have two possi$b$ilities, 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.2011-04-09

1 Answers 1

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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.