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Jordan normal form

  1. Mar 10, 2013 #1
    Hi

    1. The problem statement, all variables and given/known data
    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].

    3. 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!
     
  2. jcsd
  3. Mar 10, 2013 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    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.
     
  4. Mar 10, 2013 #3
    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.:-/
     
  5. Mar 11, 2013 #4
    I've double-checked my attempt at solution but I didn't find any mistakes.
    Now, I'm really desperate.:confused:

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