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

[Mihulik]

Hi

Homework Statement

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

The Attempt at a Solution

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

Now, let [itex]B=\left(v_{1},\: v_{2},\: v_{3}\right)[/itex].
We have [itex]f(v_{1})=2\cdot v_{1}[/itex] and [itex]f(v_{3})=2\cdot v_{3}[/itex] as desired.
We want [itex]v_{2}[/itex] to satisfy the equation [itex]Av_{2}=f(v_{2})=v_{1}+2v_{2}[/itex], which we can rewrite as [itex](A-2I_{3})v_{2}=v_{1}[/itex].
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 [itex](\lambda- 2)^3= 0[/itex] and every matrix satisfies its own characteristic equation- that is, [itex](A- 2I)^3v= 0[/itex] 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 [itex](A- 2I)^2v= (A- 2I)[(A- 2I)v]= 0[/itex]. 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 [itex]\mathbb Z_{5}[/itex]
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.