- 1,397

- 0

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??

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: