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Find th Jordan canonical form of a matrix

  1. Aug 10, 2008 #1
    I've followed and understood this small example of calculating jordan forms all the way to the last line where they say "Therefore, the jordan form is...". When they say "therefore", it's NEVER obvious :smile:

    Anyway, I get why the diagonal entries are -1. And that a minimal polynomial (t+1)^2 means I have two Jordan blocks with dimensions 1 and 2. But why is (3,2) in the final answer one? Or perhaps I should ask: how do you know what "non-diagonal" numbers to put where, when you get to this point of the calculation. Thanks in advance!

    http://www.math.umn.edu/~jpeters/Prelims/F96/Problem5.pdf
     
    Last edited: Aug 10, 2008
  2. jcsd
  3. Aug 10, 2008 #2

    HallsofIvy

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    When you mean by "the diagonal entries are -1", you mean for this particular problem, which you hadn't shown us yet. The diagonal entries are the eigenvalues of the matrix which may or may not be "-1". By the way, some texts define the Jordan Normal Form to have "1"s above the diagonal, not below. In the url you give, the Jordan normal form for a matrix consists of "blocks" in which the diagonal numbers are the same number- an eigenvalue, and the numbers must below the diagonal are 1. Such a single block might be
    [tex]\left[\begin{matrix} \lambda & 0 & 0 & 0 \\ 1 & \lambda & 0 & 0 \\ 0 & 1 & \lambda & 0 \\0 & 0 & 1 & \lambda\end{array}\right][/tex]

    A general "Jordan form" might consist of a number of those: if a 4 by 4 matrix has eigenvalues 2 and 3, each of multiplicity 2, then its Jordan form might be
    [tex]\left[\begin{matrix} 2 & 0 & 0 & 0 \\ 1 & 2 & 0 & 0 \\ 0 & 0 & 3 & 0\\0 & 0 & 1 & 3 \end{array}\right][/tex]

    A complicating factor is that the same eigenvalue may give rise to more than one "Jordan block" depending on how many independent eigenvecotrs it has.

    The particular matrix given here is 3 by 3 but its characteristic polynomial is [itex](1+ \lambda)^3[/itex] so it has only -1 as an eigenvalue, of multiplicity 3. In order to determine whether this matrix is diagonalizable, or how many "Jordan blocks" it has, we must determine the corresponding eigenvectors. It turns out that any eigenvector <x, y, z> must satisfy x+ 2y-5z so any eigenvector is a linear combination of <-2, 1, 0> and <-5, 0, 1>. That means that the eigenspace is of dimension 2 so there will be a 2 by 2 "Jordan block" and a single separate eigenvalue:
    [tex]\left[\begin{matrix} -1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 1 & -1 & 0 \\0 & 0 & 1 & -1 \end{array}\right][/tex]
    Where the separate eigenvalue -1 is in the first row and the rest is the "Jordan block".
     
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