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Matrix Diagonalizable Question

  1. Nov 19, 2008 #1
    1. The problem statement, all variables and given/known data

    1. Let A =
    [-8 k
    0 -8]
    Then A is diagonalizable exactly for the following values of k

    2. Let B =
    [-8 k
    0 1]
    Then B is diagonalizable exactly for the following values of k

    2. Relevant equations

    -Equations for eigenvalues, eigenvectors... and D=PA(P^-1)
    -A matrix is invertible if and only if its determinant does not equal 0
    -If matrix P created from the eigenvectors [X_1, X_2] is invertible, then matrix A is diagonalizable.

    3. The attempt at a solution

    I seem to only get half of this answer correct, and am not sure why. I tried solving for the eigenvectors and then checking if the matrix P (from the eigenvectors) has a determinant = to zero.

    For the first part, I got the eigenvalue -8 with an algebraic multiplicity of 2; then when solving for the eigenvectors, I get:
    [0 -k
    0 0]
    and the eigenvector t[-k,0] and the matrix P =
    [-k -k
    0 0]

    Therefore, I say that the matrix A can never be diagonalizable for any value of k, as the number of parameters will always be less than the algebraic multiplicity, and the determinant will always be = 0, regardless of the k value.

    For the second part, I get eigenvalues -8 and 1.

    I then get the eigenvectors t[1,0] and t[(-k/9),1], and the matrix P =
    [1 (-k/9)
    0 1]

    Therefore, I said this matrix will be diagonalizable for all values of k, since the determinant will always equal 1, and is therefore always diagonalizable, regardless of the k value.

    Any ideas on what I am doing wrong; and if my logic is messed up?

  2. jcsd
  3. Nov 19, 2008 #2


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    Science Advisor
    Homework Helper

    You aren't that messed up. The second one is diagonalizable for any value of k. It has two distinct eigenvalues, hence two distinct eigenvectors. I don't think [-k/9,1] is an eigenvector though. A sign doesn't seem right. You are only a little bit wrong on the first one. If k=0 it's is diagonalizable. In fact, it's already diagonal.
  4. Nov 20, 2008 #3
    Hi Dick,

    Thanks a lot for the help!

    The question was only asking for which k values, so your explanation of k=0 for the first one solved my mistake. But, if you could just clear something up for me that I don't really understand:

    As you mention the first one is diagonalizable if k=0, to make it basically the identity matrix. But, isn't there a theorem that says "A matrix A is diagonalizable if and only if the multiplicity of every eigenvalue lambda of A equals the number of basic eigenvectors corresponding to lambda , which is the number of parameters in the solution of (lambda*I - A)X = 0." In other words, I got:
    [0 -k
    0 0]
    and the eigenvector t[k,0]
    for the first part; so there is only 1 parameter for the eigenvalue with a multiplicity of 2... what am I not understanding here?

    Also, that was a silly mistake that I had (-k/9); I just forgot to change the sign, as it as negative within my augmented matrix. Is that the mistake you were referring to?

    Thanks again!!
  5. Nov 20, 2008 #4


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    Science Advisor

    No, it makes it -8 times the identity matrix.

    How do you arrive at the idea that there is only one parameter? "k" is NOT the parameter meant if that is what you are thinking. With k= 0, "lambda*I- A" is the 0 matrix. The solutions to the equation 0X= 0 are all vectors in R2 and that has two parameters: any vector <x, y>. x and y are the two parameters.

    Last edited by a moderator: Nov 20, 2008
  6. Nov 20, 2008 #5


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    Science Advisor
    Homework Helper

    Yeah, that was the silly mistake I was referring to. And, sure, like Halls (and you) said, if k is not zero you only have the eigenvector [1,0]. If k=0 then [0,1] (and every other vector) also becomes an eigenvector. So you don't just have one anymore.
  7. Nov 20, 2008 #6
    Ahhh, Okay. I don't know why I didn't see that before... obviously then, there are two parameters for the 0 matrix, which would make it invertible... the parameters there would be t[1,0] and s[0,1] (or as you said any vector <x, y>.
    And thanks Dick, I see that when you have a k value other than 0, then you get:
    [0 1
    0 0]
    which would create only one parameter. And
    [0 0
    0 0]
    would create two.

    Thanks guys!
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