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Diagonalizable Matrices

  1. Nov 30, 2008 #1
    1. The problem statement, all variables and given/known data

    find all real values of k for which A is diagonalizable.

    A = [ 1 1 ]
    [ 0 k ]

    3. The attempt at a solution

    let L = lamba = eigenvalue

    I did this:
    det(A - LI) = L2 - Lk - L + k

    so then it sorta looks like a quadratic so I did this:
    L2 - Lk - L + k = 0
    L2 - L(k+1) + k = 0
    and used the quadratic equation
    x = (-b +- sqrt(b2-4ac)) / 2a
    so then I got stuck at this point:

    ((k+1) +- (k-1)) / 2

    and this is where I got really really stuck. I'm not sure how or what to conclude about matrix A being diagonalizable for all real values of k.


    or can I skip the quadratic equation method and just do this instead?

    (1-L)(k-L) = 0
    then
    L = 1 and L = k
    so then matrix A is diagonalizable for any real number k except 0?
     
    Last edited: Nov 30, 2008
  2. jcsd
  3. Nov 30, 2008 #2
    What is the criteria for a matrix to be diagonalizable? If I remember correctly, one is that it's characteristic polynomial must split. If you are working in the reals then check for which real numbers k give an irreducible polynomial and exclude them from the final answer. Then check if the other criteria are satisfied for the remaining reals.
     
  4. Dec 1, 2008 #3

    HallsofIvy

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    k+1+ (k-1)= 2k. k+1- (k-1)= 2.

    Yes, you are allowed to do it the obvious way!

    No, under what conditions is a matrix diagonalizable?
     
  5. Dec 1, 2008 #4

    D H

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    Why did you do that? To get to that expression you had to expand (1-L)*(k-L)=0. There is no need to use the quadratic form when you already have a factorization. You can instead directly read the solutions L from (1-L)*(k-L)=0.

    How do the eigenvalues indicate whether a matrix is diagonalizable?
     
  6. Dec 1, 2008 #5

    Hurkyl

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    The most straightforward way is to actually perform the diagonalization procedure A. Make sure you divide into cases where appropriate: for example, if you would divide by k, then you need to split the problem into two cases: one where k is nonzero, and then handle the case where k is zero separately.

    In one (or more) cases, the diagonalization procedure will fail. Then you'll have your answer.


    This is standard fare -- the simplest response to "when can't you do X" is to actually do X, and see what obstacles impede your progress.
     
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