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Eigenvalue/vector and similar matrices

  1. Mar 15, 2009 #1
    1. The problem statement, all variables and given/known data
    Find a matrix B such that B^2 = A

    A = 3x3 =

    9 -5 3
    0 4 3
    0 0 1

    2. Relevant equations

    B^2 = A

    A = XDX^(-1) (similar matrices rule)

    also used to find eigenvectors: A - λI

    3. The attempt at a solution

    Thoughts: If A = XDX^(-1), then B^2 = XDX(-1), and B = X * D^(1/2) * X^-1
    So, if I could find D and X for A, I could find B. D = diagonal matrix where the diagonal elements are the eigenvalues of A, and A is lower triangular, so it's eigenvalues are:

    λ1 = 9, λ2 = 4, λ3 = 1,

    so D = 3x3 =

    9 0 0
    0 4 0
    0 0 1

    To find X, I need the eigenvectors:

    A-9I =
    0 -5 3
    0 -5 3
    0 0 -8
    so (-5)v2 + (3)v3 = 0, or v2 = (3/5)v3
    and the eigenvector (representative) = [0 1 3/5]^T

    I suspect there is something wrong here because my bottom row in A-9I isn't all 0 and I think it is supposed to be... I don't know what I did wrong though.

    For the other eigenvalues, I got eigenvectors that were:

    λ2 = 4,
    [1 1 0]^T

    λ3 = 1
    [1 1 -1]

    X = 3x3 =

    0 1 1
    1 1 1
    (3/5) 0 -1

    X^-1 =
    -1 1 0
    (8/5) (-3/5) 1
    (-3/5) (3/5) -1

    To get D^(1/2) I square-rooted the diagonal elements:
    3 0 0
    0 2 0
    0 0 1

    Then find X * D^(1/2) * X^-1, which = B, but when I square B, it doesn't = A...
    Any input would be great! Thanks
     
    Last edited: Mar 16, 2009
  2. jcsd
  3. Mar 16, 2009 #2

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi A_lilah! :smile:

    Hint: is the square of a triangular matrix triangular?

    if so, what happens to the diagonal elements? :wink:
     
  4. Mar 16, 2009 #3

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    You've completely ignored the third row: -8v3= 0 so v3= 0. Yes, the second row then gives v2= (3/5)v3 and since v3= 0, v2= 0. Now the important point: the fact that the first column is 0 does NOT mean that v1= 0! It says there is NO CONDITION on v1. With v2= v3= 0, 0v1-5v2+ 3v3= 0, 0v1- 5v2+ 3v3= 0, and 0v1+ 0v2- 8v3= 0 are true for ALL VALUES of v3: [1, 0, 0]^T is an eigenvector.

    yes, this is correct.

    Did you check it?
    [tex]\left[\begin{array}{ccc}9 & -5 & 3 \\ 0 & 4 & 3 \\ 0 & 0 & 1\end{array}\right]\left[\begin{array}{c}1 \\ 1 \\ 1\end{array}\right]= \left[\begin{array}{c} 7 \\ 7 \\ 1\end{array}\right][/tex]
    No, that is not [1 1 -1]^T again.

    For eigenvalue 1, you get the three equations 8v1- 5v2+ 3v3= 0, 3v2+ 3v3= 0, and 0= 0. The last equations says that v3 can be anything. From the second equation, 3v2= -3v3 so v2= -v3 and, putting that into the first equation, 8v1+ 5v3+ 3v3= 0, v1= -v3. You can write down the eigenvector corresponding to eigenvalue 1 from that.

     
    Last edited: Mar 16, 2009
  5. Mar 16, 2009 #4
    Thank you both so much! Of course I should have looked at the other rows ~ silly of me. Thanks again :)
     
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