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Jordan Normal Form & Generalized Eigenvectors

  1. Apr 17, 2012 #1
    I've been having some trouble with conceptually understanding the idea of a generalized eigenvector. If we have a linear operator A and want to diagonalize we get it's eigenvalues and eigenvectors but if the algebraic multiplicity of one of the eigenvalues is greater than the geometric multiplicity we can't diagonalize. Now, it makes sense that we want to make A as diagonal as possible.

    Here's my problem, what is the motivation or idea of getting these generalized eigenvectors such that they come from computing the null spaces of (A-tI) where t is the eigenvalue with greater algebraic multiplicity than geometric and then taking the powers of the null spaces of the above matrix (a-tI) until its stabilizes at j such that dim(ker(A-tI))^j=dim(ker(A-tI))^x where x>j, that is until the dimension of the null space becomes constant after taking a finite number of powers.

    I've realised that if (1) (A-tI)v=0 then what we are doing is something like(2) (A-tI)w=v, where w will be the first generalized eignevector because when we sub back into (1) we get (A-tI)^2w=0 and thus if we take ker(A-tI)^2 we will see what w is in this space and w will be the eigenvector of (A-tI)^2 with eigenvalue t and at some point this will stabilize.

    My problem is that I don't know why we do this, why will generalized eigvectors work? Essentially, in most textbooks I've seen that they just put it down that we do this and when we have J=B^-1AB it means that B will consist of the generalized eigenvectors of A but thus far I cannot see why this works.

    I would be very grateful if anyone could help. Thanks.
     
  2. jcsd
  3. Apr 17, 2012 #2

    micromass

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    Good question. I hope to answer it as good as I can.

    So given a 4x4 matrix A. Let's say that the eigenvalues are 2 (with algebraic multiplicity 3) and 1 (I do this to ease the notation, but it works in general). We want to put it in Jordan canonical form. This will have the form (for example)

    [tex]J=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\
    0 & 2 & 1 & 0\\
    0 & 0 & 2 & 1\\
    0 & 0 & 0 & 2
    &\end{array}\right)[/tex]

    Of course, given the eigenvalues and multiplicties, other Jordan forms are possible. Indeed, it might even be possible that our matrix is diagonalizable. But let's assume that our matrix will actually end up in the form of J.

    We will want to find invertible matrices P such that

    [tex]J=P^{-1}AP[/tex]

    or in other words

    [tex]AP=JP[/tex]

    Let's suppose the P has columns [itex](p_1~ p_2 ~p_3 ~p_4)[/itex]. Then we have that

    [tex]A(p_1~ p_2 ~ p_3 ~ p_4)=(p_1~ p_2 ~ p_3 ~ p_4)J=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\
    0 & 2 & 1 & 0\\
    0 & 0 & 2 & 1\\
    0 & 0 & 0 & 2
    &\end{array}\right)
    =(p_1 ~ 2p2 ~ p_2 + 2p3 ~ p_3+2p_4)[/tex]

    Thus we see that

    [tex]Ap_1=p_1,~Ap_2=2p_2,~Ap_3= p_2+2p_3,~Ap_4=p3+2p_4[/tex]

    Thus

    [tex](A-I)p_1=0,~(A-I)p_2=0,~(A-I)p_3=p_2,~(A-I)p_4=p_3[/tex]

    Combining the last equations, we get that

    [tex](A-I)^2p_3=(A-I)p_2=0,~(A-I)^3p_4=(A-I)^2p_3=0[/tex]

    So we see that [itex]p_1,p_2,p_3[/itex] and [itex]p_4[/itex] must be generalized eigenvectors.

    So if we were able to write the Jordan form as above, then the columns of the transformation matrix would have to be generalized eigenvectors. This is where the concept of a generalized eigenvector comes from!! And this is why you have to look for the generalized eigenvectors!
     
  4. Apr 18, 2012 #3
    I never thought of taking the approach of letting AP=PJ & work from that. I can now see the results of using generalized eigenvectors & why they are useful. Thanks!
     
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