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Consider a matrix A (5x5) with all entries = 1. Here the entries are considered as elements of $F_5$ ,the finite field of order 5.

What is the Jordan canonical form?

I have found out that $A^2=0$ and thus the minimal polynomial is $x^2$. So I know there are (two 2x2 blocks and one 1x1 block) OR (one 2x2 block and three 1x1 blocks)

How do I tell which?

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    $\begingroup$ You haven't stated the size of the matrix explicitly, but from your work it sounds like $A$ is $5 \times 5$. Please add this to your question. $\endgroup$ Dec 12, 2019 at 15:30
  • $\begingroup$ added. thank you $\endgroup$ Dec 12, 2019 at 15:32

2 Answers 2

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Hint: If $A$ is an $n \times n$ matrix, then $n - \operatorname{rk}(A)$ is the total number of Jordan blocks that $A$ has associated with $\lambda = 0$.

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Hint. Consider the rank of $A$.

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  • $\begingroup$ rank =4 and is same for JCF. So it contains two 2x2 blocks and one 1x1 block. Am I right? and thank you Edit - bad mistake. rank = 1 $\endgroup$ Dec 12, 2019 at 15:28
  • $\begingroup$ also curious,if for some higher order matrix both wold have same ranks then how do I tell which is the JCF? $\endgroup$ Dec 12, 2019 at 15:28
  • $\begingroup$ @Angry_Math_Person The rank isn't 4. How did you get that? $\endgroup$
    – user1551
    Dec 12, 2019 at 15:28
  • $\begingroup$ sorry. major error. It's 1. hence the latter option. $\endgroup$ Dec 12, 2019 at 15:29
  • $\begingroup$ @Angry_Math_Person Yes, it's rank-1. In general, if $A$ is nilpotent, you need to consider the ranks of $A,A^2,A^3$ and so on. The sequence of ranks will determine the Jordan form uniquely. $\endgroup$
    – user1551
    Dec 12, 2019 at 15:30

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