Find a basis of $\mathbb{F_5^5}$ in which the matrix has the Jordan normal form

Question

I have the following task:

Let $A = \left( \begin{array}{ccc} 2 & 2 & 3 & 3 & 0 \\ 0 & -3 & 0 & 0 & 0 \\ 0 & 0 & 1 & 0 & 0 \\ 0 & 3 & 0 & 1 & 0\\ 0 & 0 & 1 & 1 & 2\\ \end{array} \right) \in M_5(\mathbb{F_5})$.

Find a basis of $\mathbb{F_5^5}$ in which the matrix A has the Jordan normal form.

My first question: Can I consider $\mathbb{F_5^5}$ as modulo 5? So as the set $\{0,1,2,3,4\}$?

My solution:

I computed the characteristic polynomial:

$(1- \lambda)^2(2- \lambda)^3$. So the eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = 2$.

Now, I computed the dimension of the eigenspaces.

$\dim(\ker(A-Id))=2$

$\dim(\ker(A-2Id))=2$

So since I know the eigenvalues and the dimension of the corresponding eigenspaces I could write the jordan matrix, right?

So:

$J= \left( \begin{array}{ccc} 1 & 0 & 0 & 0 & 0 \\ 0 & 1 & 0 & 0 & 0 \\ 0 & 0 & 2 & 0 & 0 \\ 0 & 0 & 0 & 2 & 1\\ 0 & 0 & 0 & 0 & 2\\ \end{array} \right) $

Is that right?

That what is still unclear to me is:

• What is meant by a basis of $\mathbb{F_5^5}$?