# To find the Jordan Canonical Form

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?

• 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. – Ben Grossmann Dec 12 '19 at 15:30
• added. thank you – Angry_Math_Person Dec 12 '19 at 15:32

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 $$\lambda = 0$$.
Hint. Consider the rank of $$A$$.
• @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. – user1551 Dec 12 '19 at 15:30