1
$\begingroup$

Let $A = [a_{ij}]_{n \times n}$ and suppose the minimal polynomial is $ m_A(x) = (x-2)^4 (x-1)^4 $ How many Jordan canonical forms are possible?

Since the exponent $4$ forces the largest Jordan block for each eigenvalue to be of size $4$, I counted the partitions of $4$ for eigenvalue $1$ and again for eigenvalue $2$, giving $$ P(4)\cdot P(4)=5\cdot 5 = 25. $$

However, the answer key states $20$, and I am confused about the discrepancy.

My doubt is whether the partition $1+1+1+1$ should be allowed or excluded, because the minimal polynomial requires at least one Jordan block of size exactly $4$.

Should the correct number of Jordan forms be $25$ or $20$?

New contributor
Prahallad is a new contributor to this site. Take care in asking for clarification, commenting, and answering. Check out our Code of Conduct.
$\endgroup$
3
  • 8
    $\begingroup$ My opinion is that with the given data the solution should be a function of $n$, and that you are therefore both wrong. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @Prahallad Did you forget to say what $n$ is equal to? $\endgroup$ Commented yesterday
  • $\begingroup$ Should "partitions of 4" be partions of $n$? $\endgroup$ Commented yesterday

1 Answer 1

Reset to default
1
$\begingroup$

If $d$ is the maximal exponent of a factor $(x-\lambda)^d$ of $m_A$, it means that there is at least one Jordan block associated with $\lambda$ of size $d$, and any Jordan block associated with $\lambda$ has size no greater than $d$.

We know that $n\ge 8$, or else $A$ can't have a degree $8$ minimal polynomial. The only freedom left for constructing the remaining Jordan blocks are:

  1. the number of basis vectors associated with 1 (this also determines the number of basis vectors associated with 2)
  2. the sizes of blocks associated with 1
  3. the sizes of blocks associated with 2

Consider the case where there are $s$ basis vectors associated with $\lambda$. Let $p(m)$ be the number of partitions of $m$ where each term does not exceed 4. Then $p(s-4)$ is the number of possible configurations for blocks associated with $\lambda$ (the first 4 slots must be occupied by a size 4 block, hence the $s-4$). Thus, the total number of configurations is $\sum_{s=4}^{n-4} p(s-4)p(n-s-4)$. Let's call this number $C(n)$.

Having done the calculations, I found that $C(12)=20$ and $C(13)=34$, so the $n$ in your original problem formulation must be $12$ and the answer must be $20$ instead of $25$.

$\endgroup$
2
  • $\begingroup$ But the original problem formulation says nothing about $n$ (unless OP has forgotten something in recounting the problem), so why can't $n$ be $13$, or $42$, or ....? $\endgroup$ Commented yesterday
  • 1
    $\begingroup$ @GerryMyerson OP stated that the answer key claims that the answer for OP's problem is 20, but OP thinks that it should be 25. I did the calculation for general $n$ and see that there is a value of $n$ for which $C(n)=20$ and there can't be a value of $n$ for $C(n)=25$ (because $C$ is strictly increasing). $\endgroup$ Commented yesterday

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.