# Thread: how to use part A in order to solve pard B

1. ## how to use part A in order to solve pard B

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

solution:

$\displaystyle \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)=$$\displaystyle \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)$

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

P^-1AP$\displaystyle =\bigl(\begin{smallmatrix} -1& 0 & 0\\ 0&-8 &0\\ 0&0&3 \end{smallmatrix}\bigr)$
i can substitute A by X but whats to do next??

2. Hi

P^-1XPD=DP^-1XP

Let Y be P^-1XP
YD=DY

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

$\displaystyle \bigl(\begin{smallmatrix} y_{11}& 0 & 0\\ 0&y_{22} &0 \\ 0&0 &y_{33} \end{smallmatrix}\bigr)$

And X=PYP^-1

Find P and you will get the general form of X

3. but i found the diagonolized form
$\displaystyle \bigl(\begin{smallmatrix} -1& 0 & 0\\ 0&-8 &0\\ 0&0&3 \end{smallmatrix}\bigr)$

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