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This question already has an answer here:

The Problem:

Let $A$ be a $5 \times 5$ matrix with characteristic polynomial $(x-2)^3(x+1)^2$ and minimal polynomial $(x-2)^2(x+1)^2$. What are the possible Jordan forms for $A$.

My Approach:

There are many questions of this form here on StackExchange, but I seem to be encountering some contradictory interpretations when going through them all. So let me see if I understand what's going on...

Obviously (up to permuting the Jordan blocks), the Jordan form of $A$ is of the form \begin{pmatrix} 2 & a_1 & 0 & 0 & 0 \\ 0 & 2 & a_2 & 0 & 0 \\ 0 & 0 & 2 & a_3 & 0 \\ 0 & 0 & 0 & -1 & a_4 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}

where $a_i \in \{0,1\}$, for $i = 1,2,3,4$. That is, it's simply a matter of determining precisely which $a_i$ are $0$ and which are $1$. I believe that, since the multiplicity of the root $x = 2$ in the minimal polynomial of $A$ is $2$, this means that the largest possible Jordan block associated with the eigenvalue 2 is $2 \times 2$; and so at least one of $a_1, a_2$ must be $1$. But must there necessarily be such a Jordan block? (Note that I interpret a "Jordan Block" to necessarily be a matrix with $1$s along the superdiagonal--I've seen it defined differently elsewhere.)

Similarly, since the root $x = -1$ has multiplicity 2 in the minimal polynomial, I take this to mean that the largest Jordan block associated with the eigenvalue $1$ is $2 \times 2$. (Again, must there necessarily be such a Jordan block?)

This means that a possible Jordan form of $A$ is \begin{pmatrix} 2 & 0 & 0 & 0 & 0 \\ 0 & 2 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & -1 & 1 \\ 0 & 0 & 0 & 0 & -1 \end{pmatrix}

Now, I know that the Jordan form cannot be a diagonal matrix (as this is only true when the minimal polynomial is a product of 5 distinct linear factors); so we can't have $a_i = 0$, for each $i = 1, 2, 3, 4$. Moreover, we can't have $a_1 = 1$, for each $i = 1,2,3,4$, since there can be no Jordan blocks larger than $2 \times 2$. In fact, if we fix the diagonal entries as they appear above, it follows that $a_3 = 0$ and only one of $a_1, a_2$ can be 1.

Is there anything I can conclude?

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marked as duplicate by 6005, user416281, Alexander Gruber Jan 12 at 0:56

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

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    $\begingroup$ Ddi you see already this question here? $\endgroup$ – Dietrich Burde Jan 11 at 20:38
  • $\begingroup$ @DietrichBurde Somehow I missed that one. I believe that clears up my confusion. (I'm not sure if I should delete my post, mark it as duplicate, leave it up, or what...) $\endgroup$ – thisisourconcerndude Jan 11 at 20:44
  • $\begingroup$ @thisisourconcerndude The right thing to do is leave it up and mark as duplicate :) $\endgroup$ – 6005 Jan 11 at 21:05
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The knowledge of the characteristic and minimal polynomials completely determines the Jordan Form only for matrices of dimension $3\times 3$ ( or $2 \times 2$). In your case we can have, in principle, different Jordan forms for the given polynomials.

From the characteristic polynomial we know that the diagonal elements are three values $\lambda= 2$ and two values $\lambda=-1$ (this numbers are the algebraic multiplicities of the eigenvalues).

The minimal polynomial say us that for the eigenvalue $\lambda=2$, and also for the eigenvalue $\lambda=-1$, we have a Jordan bloc of dimension $2$. This means that we can have a jordan bloc with two eigenvalues $2$ on the diagonal and a value$1$ over these, and the same for the eigenvalue $-1$.

So, apart the position of the blocs, in this case we have one Jordan form, with yours $a_1=1$ (or $a_2=1$) and $a_4=1$. Note that $a_3$ is not an element of a Jordan bloch and must be $0$.

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