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How to use part A in order to solve pard B

  1. Mar 5, 2009 #1
    part A(i managed to solve it):
    X is a variable of [tex]M_{3X3}[/tex]
    [tex]
    D=\bigl(\begin{smallmatrix}
    \lambda_1 &0 &0 \\
    0 & \lambda_2& 0\\
    0&0 &\lambda_3
    \end{smallmatrix}\bigr)
    [/tex]
    where
    [tex]\lambda_1,\lambda_2,\lambda_3[/tex] are different rational numbers.
    solve XD=DX for X.

    solution:

    [tex]
    \bigl(\begin{smallmatrix}
    x_{11}& x_{12} & x_{13}\\
    x_{21}&x_{22} &x_{23} \\
    x_{31}&x_{32} &x_{33}
    \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix}
    \lambda_1 &0 &0 \\
    0 & \lambda_2& 0\\
    0&0 &\lambda_3
    \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}
    \lambda_1 &0 &0 \\
    0 & \lambda_2& 0\\
    0&0 &\lambda_3
    \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix}
    x_{11}& x_{12} & x_{13}\\
    x_{21}&x_{22} &x_{23} \\
    x_{31}&x_{32} &x_{33}
    \end{smallmatrix}\bigr)
    [/tex]

    so i get
    [tex]
    \bigl(\begin{smallmatrix}
    x_{11}\lambda_1& x_{12}\lambda_2 & x_{13}\lambda_3\\
    x_{21}\lambda_1&x_{22}\lambda_2 &x_{23}\lambda_3 \\
    x_{31}\lambda_1&x_{32}\lambda_2 &x_{33}\lambda_3
    \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}
    x_{11}\lambda_1& x_{12}\lambda_1 & x_{13}\lambda_1\\
    x_{21}\lambda_2&x_{22}\lambda_2 &x_{23}\lambda_2 \\
    x_{31}\lambda_3&x_{32}\lambda_3 &x_{33}\lambda_3
    \end{smallmatrix}\bigr)[/tex]
    so for both side to be equal X must be of diagonal structure
    every member must be zero except the diagonal
    because the lambda values are given as different.

    part B(the one that i dont understand):
    [tex]
    A=\bigl(\begin{smallmatrix}
    13& -42 & 0\\
    7&-22 &0\\
    0&0&3
    \end{smallmatrix}\bigr)[/tex]
    what is the solution space of XA=AX (use part A)??

    i tried
    XA=AX
    XPDP^-1=PDP^-1X (multiplying by p from the right)
    XPDP^-1P=PDP^-1XP
    XPD=PDP^-1XP (multiplying by p^-1 from the left)
    P^-1XPD=P^-1PDP^-1XP
    P^-1XPD=DP^-1XP

    another thing i could find is the eigen values of the matrix
    i got
    [tex]\lambda_1=-1[/tex] and [tex]\lambda_2=-8[/tex]
    and [tex]\lambda_3=3[/tex]

    P^-1AP[tex]=\bigl(\begin{smallmatrix}
    -1& 0 & 0\\
    0&-8 &0\\
    0&0&3
    \end{smallmatrix}\bigr)[/tex]
    i can substitute A by X but whats to do next??
     
    Last edited: Mar 5, 2009
  2. jcsd
  3. Mar 5, 2009 #2

    Mark44

    Staff: Mentor

    For the a part, you don't show what you actually got for matrix X, which is a 3 x 3 diagonal matrix with x11, x22, and x33 on the diagonal.

    For the b part, you're given a matrix A. The most obvious start would be to carry out the products AX and XA and see what you get.
     
  4. Mar 5, 2009 #3
    X matrix are all the matrices which are diagonalized

    i was told to use part A not do a straight multiplication
     
  5. Mar 5, 2009 #4

    Mark44

    Staff: Mentor

    Apparently you were also told to use part a, but you're not using it if you don't show what you get for matrix X. Regardless of what you were told, if I had to do this problem, I would multiply AX and XA and see what I got.
     
  6. Mar 5, 2009 #5
    in part A in order for the equality to be true
    x must look as
    [tex]
    \bigl(\begin{smallmatrix}
    x_{11}\lambda_1&0 & 0\\
    0&x_{22}\lambda_2 &0 \\
    0&0 &x_{33}\lambda_3
    \end{smallmatrix}\bigr)
    [/tex]
    does it help in using part B

    after i got this
    P^-1XPD=DP^-1XP
    ??
     
  7. Mar 5, 2009 #6

    Mark44

    Staff: Mentor

    No, your matrix X has to look like this:
    [tex]\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)[/tex]

    What you just showed was XD or DX, not X.

    You're given matrix A (you showed it in your first post). Use what you've found about matrix X to find the solution space of AX = XA.
     
    Last edited: Mar 5, 2009
  8. Mar 5, 2009 #7
    i get this expression
    P^-1XPD=DP^-1XP

    x is diagonal
    P^-1XP is a formula for diagonalizing a matrix
    what solution i need to find??
     
  9. Mar 5, 2009 #8

    Mark44

    Staff: Mentor

    Why are you still messing around with this? Do you know what problem you're trying to solve? It's not evident to me that you do.
    Do you want me to hold your hand? Look at my post #6.
     
  10. Mar 5, 2009 #9
    you are right
    i dont understand what i am asked to find
    i need to find X
    ??
     
  11. Mar 5, 2009 #10

    Mark44

    Staff: Mentor

    OK, here's X
    [tex]\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)[/tex]

    And here's A
    [tex]\bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)[/tex]

    Put them in the equation AX = XA. What do you get? The equations in x11, x22, and x33 will determine the solution space for the equation AX = XA.
     
  12. Mar 5, 2009 #11

    Mark44

    Staff: Mentor

    Regarding my previous post, I haven't worked this problem through, but I'm just trying to show you my thinking process. It's possible that what I've proposed is a deadend, but you can't know that until you've tried it.

    Now, why is it that I know what you're asked to find? I looked at the problem you showed in your first post. You have to know what the problem is, and what are the relevant definitions for your problem.
     
  13. Mar 6, 2009 #12
    [tex]
    \bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)
    \bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix} 13x_{11}&-42x_{11} & 0\\ 7x_{22}&-22x_{22} &0 \\ 0&0 &3x_{33} \end{smallmatrix}\bigr)
    [/tex]
    [tex]
    \bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)
    =\bigl(\begin{smallmatrix} 13x_{11}&-42x_{11} & 0\\ 7x_{22}&-22x_{22} &0 \\ 0&0 &3x_{33} \end{smallmatrix}\bigr)
    [/tex]

    from both multiplication i get the same matrix
    what to do now
    ?
     
  14. Mar 6, 2009 #13

    Mark44

    Staff: Mentor

    AX and XA don't come out the same. Try again. Since there are only three nonzero entries in X, you're less prone to making errors if you write it like this:
    [tex]X = \bigl[\begin x_{1}&0 & 0\\ 0&x_{2} &0 \\ 0&0 &x_{3} \end \bigr][/tex]
     
  15. Mar 6, 2009 #14
    i cant see my mistake
    [tex]
    \begin{bmatrix}
    x_1 &0 &0 \\
    0&x_2 &0 \\
    0& 0& x_3
    \end{bmatrix}

    \begin{bmatrix}
    13 &-42 &0 \\
    7&-22 &0 \\
    0& 0& 3
    \end{bmatrix}
    =
    \begin{bmatrix}
    13x_1 &-42x_1 &0 \\
    7x_2 &-22x_2 &0 \\
    0& 0& 3x_3
    \end{bmatrix}\\
    [/tex]
    same here
    [tex]
    \begin{bmatrix}
    13 &-42 &0 \\
    7&-22 &0 \\
    0& 0& 3
    \end{bmatrix}

    \begin{bmatrix}
    x_1 &0 &0 \\
    0&x_2 &0 \\
    0& 0& x_3
    \end{bmatrix}
    =
    \begin{bmatrix}
    13x_1 &-42x_1 &0 \\
    7x_2 &-22x_2 &0 \\
    0& 0& 3x_3
    \end{bmatrix}\\
    [/tex]
     
  16. Mar 6, 2009 #15

    Mark44

    Staff: Mentor

    The errors (two of them) are in the multiplication below.
     
  17. Mar 6, 2009 #16

    Mark44

    Staff: Mentor

    Here we are at post 16, still futzing around with something that I suggested you should try all the way back in post 2. I have to say it's pretty frustrating for me. I can only hope that you start to pick up on some of the guidance that I and others on this forum are trying to provide you so that you can do some of this on your own.
     
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