# Jordan normal form

1. Mar 10, 2013

### Mihulik

Hi

1. The problem statement, all variables and given/known data
Let $f:\:\mathbb{Z}_{5}^{3}\rightarrow \mathbb{Z}_{5}^{3}$ be a linear operator and let $[f]_{e_{3}}^{e_{3}}=A=\begin{pmatrix}3 & 1 & 4\\ 3 & 0 & 2\\ 4 & 4 & 3 \end{pmatrix}$ over $\mathbb{Z}_{5}$.
Find a basis B of $\mathbb{Z}_{5}^{3}$ such that $[f]_{B}^{B}=\begin{pmatrix}2 & 1 & 0\\ 0 & 2 & 0\\ 0 & 0 & 2 \end{pmatrix}$.

3. 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 $v_{1}=\begin{pmatrix}1\\ 0\\ 1 \end{pmatrix}$ and $v_{3}=\begin{pmatrix}4\\ 1\\ 0 \end{pmatrix}$.

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!

2. Mar 10, 2013

### HallsofIvy

Staff Emeritus
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.

3. Mar 10, 2013

### 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.:-/

4. Mar 11, 2013

### 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.