- #1
transgalactic
- 1,395
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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 don't 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 what's to do next??
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 don't 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 what's to do next??
Last edited: