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Jordan Normal/Canonical basis

  1. Aug 25, 2011 #1
    1. The problem statement, all variables and given/known data

    Right, I know how to do questions on Jordan Normal Form and find a basis, but there is one part I don't understand.

    Let's take for example this matrix, call it A.

    \begin{bmatrix}
    -3 & 1 & 0 \\
    -1 & -1 & 0 \\
    -1 & -2 & 1
    \end{bmatrix}

    We find the characteristic polynomial of A using
    det(A-tI) = 0, where I is the identity matrix.

    The roots of this equation are our eigenvalues, let's call them [itex]\lambda_{1}[/itex] and [itex]\lambda_{2}[/itex]

    For each one we find A-[itex]\lambda[/itex]I.
    Call this matrix B.

    We find rk(B), and using the formula rk(B) + dimKer(B) = n, where n is the dimension of our vector space, find dimKer(B).

    We find B2, B3, ..., Bk-1, Bk
    until rk(Bk-1) = rk(Bk)

    We then have to find the vector(s) that spans the kernel of Bk-1 and the first part of our basis is given by a vector which isn't in the kernel. This tends to change depending on various cases which we're working on.
    Example, if rk(B) =/= rk(B2) = 0 then we find the vectors that span the kernel of
    B2, bring it to reduced column echelon form and the vector we choose for part of our basis is the vector which makes up for the "missing leading one".

    The last paragraph is basically what I don't understand.
    Could someone please explain why this is done? (in general, not for any particular case) :smile:
     
  2. jcsd
  3. Aug 25, 2011 #2

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You are talking very generally. For this particular matrix, the eigenvalues are 1 and -2, with -2 being a "double" eigenvalue- the characteristic equation is [itex](\lambda- 1)(\lambda+ 2)^2= 0[/itex]. The space of all eigenvectors corresponding to the eigenvalue 1 is, of course, of dimension 1 and it is fairly easy to find a basis vector for that space.

    However, it turns out that the dimension of the eigenspace for eigenvalue -2 is also one, leaving us one eigenvector short. If there were two independent eigenvectors, then, using those three eigenvectors as basis, we could diagonalize the matrix. What we can do is look for a "generalized eigenvector". Every matrix satisfies its own characteristic equation so we must have (A- I)(A+ 2I)^2v= 0. Of course, if Av= v, that is, if v is an eigenvector corresponding to eigenvalue 1, that is satisfied. Further, the exist a space of eigenvectors corresponding to eigenvalue -2, so that (A+ 2I)v= 0 for those vectors. But there exist other vectors for which neither of those is true. For those vectors we must have (A+2I)[(A+2I)v]= 0. That means that (A+ 2I)v is an eigenvector corresponding to eigenvalue -2. That is, to find a third vector, forming a basis in which the matrix is not diagonal but in "Jordan Normal Form", find a vector, x, satisfying (A+ 2I)x= v with v an eigenvector corresponding to eigenvalue -2.
     
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