# Yet Another Question Regarding Jordan Form [duplicate]

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?

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

• Ddi you see already this question here? – Dietrich Burde Jan 11 at 20:38
• @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...) – thisisourconcerndude Jan 11 at 20:44
• @thisisourconcerndude The right thing to do is leave it up and mark as duplicate :) – 6005 Jan 11 at 21:05

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 $$0$$.