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Math Help - how to use part A in order to solve pard B

  1. #1
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401

    how to use part A in order to solve pard B

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

    solution:

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

    so i get
    <br />
\bigl(\begin{smallmatrix}<br />
 x_{11}\lambda_1& x_{12}\lambda_2 & x_{13}\lambda_3\\ <br />
 x_{21}\lambda_1&x_{22}\lambda_2  &x_{23}\lambda_3 \\ <br />
 x_{31}\lambda_1&x_{32}\lambda_2  &x_{33}\lambda_3 <br />
\end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix}<br />
 x_{11}\lambda_1& x_{12}\lambda_1 & x_{13}\lambda_1\\ <br />
 x_{21}\lambda_2&x_{22}\lambda_2  &x_{23}\lambda_2 \\ <br />
 x_{31}\lambda_3&x_{32}\lambda_3  &x_{33}\lambda_3 <br />
\end{smallmatrix}\bigr)
    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):
    <br />
A=\bigl(\begin{smallmatrix}<br />
 13& -42 & 0\\ <br />
 7&-22 &0\\ <br />
 0&0&3 <br />
\end{smallmatrix}\bigr)
    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
    \lambda_1=-1 and \lambda_2=-8
    and \lambda_3=3

    P^-1AP =\bigl(\begin{smallmatrix}<br />
 -1& 0 & 0\\ <br />
 0&-8 &0\\ <br />
 0&0&3 <br />
\end{smallmatrix}\bigr)
    i can substitute A by X but whats to do next??
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  2. #2
    MHF Contributor
    Joined
    Nov 2008
    From
    France
    Posts
    1,458
    Hi

    Your idea is good
    P^-1XPD=DP^-1XP

    Let Y be P^-1XP
    YD=DY

    From part A you know that Y must be of diagonal structure

    <br />
\bigl(\begin{smallmatrix}<br />
 y_{11}& 0 & 0\\ <br />
 0&y_{22}  &0 \\ <br />
 0&0  &y_{33}<br />
\end{smallmatrix}\bigr)

    And X=PYP^-1

    Find P and you will get the general form of X
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  3. #3
    MHF Contributor
    Joined
    Nov 2008
    Posts
    1,401
    but i found the diagonolized form
    <br />
\bigl(\begin{smallmatrix}<br />
 -1& 0 & 0\\ <br />
 0&-8 &0\\ <br />
 0&0&3 <br />
\end{smallmatrix}\bigr)<br />

    and i got the original as a given
    <br />
\bigl(\begin{smallmatrix}<br />
 13& -42 & 0\\ <br />
 7&-22 &0\\ <br />
 0&0&3 <br />
\end{smallmatrix}\bigr)<br />
    i cant really see how am i supposed to use part A
    ??
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