# 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 $M_{3X3}$
$
D=\bigl(\begin{smallmatrix}
\lambda_1 &0 &0 \\
0 & \lambda_2& 0\\
0&0 &\lambda_3
\end{smallmatrix}\bigr)
$

where
$\lambda_1,\lambda_2,\lambda_3$ are different rational numbers.
solve XD=DX for X.

solution:

$
\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)
$

so i get
$
\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):
$
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
$\lambda_1=-1$ and $\lambda_2=-8$
and $\lambda_3=3$

P^-1AP $=\bigl(\begin{smallmatrix}
-1& 0 & 0\\
0&-8 &0\\
0&0&3
\end{smallmatrix}\bigr)$

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

$
\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
$
\bigl(\begin{smallmatrix}
-1& 0 & 0\\
0&-8 &0\\
0&0&3
\end{smallmatrix}\bigr)
$

and i got the original as a given
$
\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
??