Jordan Normal Form: Find Basis B in $\mathbb{Z}_{5}^{3}$

[Mihulik]

Hi

Homework Statement

Let f:\:\mathbb{Z}_{5}^{3}\rightarrow \mathbb{Z}_{5}^{3} be a linear operator and let [f]_{e_{3}}^{e_{3}}=A=\begin{pmatrix}3 &amp; 1 &amp; 4\\<br /> 3 &amp; 0 &amp; 2\\<br /> 4 &amp; 4 &amp; 3<br /> \end{pmatrix} over \mathbb{Z}_{5}.
Find a basis B of \mathbb{Z}_{5}^{3} such that [f]_{B}^{B}=\begin{pmatrix}2 &amp; 1 &amp; 0\\<br /> 0 &amp; 2 &amp; 0\\<br /> 0 &amp; 0 &amp; 2<br /> \end{pmatrix}.

The Attempt at a Solution

I found out that the characteristic polynomial of f is -(\lambda-2)^{3} and that there are two linearly indenpendent vectors corresponding to \lambda =2.
These vectors are v_{1}=\begin{pmatrix}1\\<br /> 0\\<br /> 1<br /> \end{pmatrix} and v_{3}=\begin{pmatrix}4\\<br /> 1\\<br /> 0<br /> \end{pmatrix}.

Now, let B=\left(v_{1},\: v_{2},\: v_{3}\right).
We have f(v_{1})=2\cdot v_{1} and f(v_{3})=2\cdot v_{3} as desired.
We want v_{2} to satisfy the equation Av_{2}=f(v_{2})=v_{1}+2v_{2}, which we can rewrite as (A-2I_{3})v_{2}=v_{1}.
However, this is the point I got stuck on.
This equation doesn't have a solution... Why?
I was hopping somebody could tell me what I was doing wrong because I've spent a few hours trying to figure out what's wrong...

Thank you!

 

[HallsofIvy]

You need to find a third basis vector. What you need is a "generalized eigen vector". The characteristic equation for this matrix is (\lambda- 2)^3= 0 and every matrix satisfies its own characteristic equation- that is, (A- 2I)^3v= 0 for every vector v in the vector space. But you were not able to find three indepedent vectors such that (A- 2I)v= 0. That means there must exist a one dimensional space of vectors, v, for which (A- 2I)v is not equal to 0 but (A- 2I)^2v= (A- 2I)[(A- 2I)v]= 0. And that means that (A- 2I)v must be an eigenvector. A "generalized eigenvector" must satisfy either (A- 2I)v= <1, 0, 1> or (A- 2I)v= <4, 1, 0>. Try to solve those equations.

 

[Mihulik]

I tried to solve those equations before I asked here but neither of them has a solution over \mathbb Z_{5}
That's the reason why I'm so puzzled.:-/

 

[Mihulik]

I've double-checked my attempt at solution but I didn't find any mistakes.
Now, I'm really desperate.

I was hopping somebody could tell me what I was doing wrong.