How to use part A in order to solve pard B

[transgalactic]

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 don't 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 what's to do next??

 

[Mark44]

For the a part, you don't show what you actually got for matrix X, which is a 3 x 3 diagonal matrix with x11, x22, and x33 on the diagonal.

For the b part, you're given a matrix A. The most obvious start would be to carry out the products AX and XA and see what you get.

 

[transgalactic]

X matrix are all the matrices which are diagonalized

i was told to use part A not do a straight multiplication

 

[Mark44]

Apparently you were also told to use part a, but you're not using it if you don't show what you get for matrix X. Regardless of what you were told, if I had to do this problem, I would multiply AX and XA and see what I got.

 

[transgalactic]

in part A in order for the equality to be true
x must look as
$$\bigl(\begin{smallmatrix} x_{11}\lambda_1&0 & 0\\ 0&x_{22}\lambda_2 &0 \\ 0&0 &x_{33}\lambda_3 \end{smallmatrix}\bigr)$$
does it help in using part B

after i got this
P^-1XPD=DP^-1XP
??

 

[Mark44]

No, your matrix X has to look like this:
$$\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)$$

What you just showed was XD or DX, not X.

You're given matrix A (you showed it in your first post). Use what you've found about matrix X to find the solution space of AX = XA.

 

[transgalactic]

i get this expression
P^-1XPD=DP^-1XP

x is diagonal
P^-1XP is a formula for diagonalizing a matrix
what solution i need to find??

 

[Mark44]

i get this expression
P^-1XPD=DP^-1XP

Why are you still messing around with this? Do you know what problem you're trying to solve? It's not evident to me that you do.

x is diagonal
P^-1XP is a formula for diagonalizing a matrix
what solution i need to find??

Do you want me to hold your hand? Look at my post #6.

 

[transgalactic]

you are right
i don't understand what i am asked to find
i need to find X
??

 

[Mark44]

OK, here's X
$$\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr)$$

And here's A
$$\bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)$$

Put them in the equation AX = XA. What do you get? The equations in x11, x22, and x33 will determine the solution space for the equation AX = XA.

 

[Mark44]

Regarding my previous post, I haven't worked this problem through, but I'm just trying to show you my thinking process. It's possible that what I've proposed is a deadend, but you can't know that until you've tried it.

Now, why is it that I know what you're asked to find? I looked at the problem you showed in your first post. You have to know what the problem is, and what are the relevant definitions for your problem.

 

[transgalactic]

$$\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr) \bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)=\bigl(\begin{smallmatrix} 13x_{11}&-42x_{11} & 0\\ 7x_{22}&-22x_{22} &0 \\ 0&0 &3x_{33} \end{smallmatrix}\bigr)$$
$$\bigl(\begin{smallmatrix} 13& -42 & 0\\ 7&-22 &0\\ 0&0&3 \end{smallmatrix}\bigr)\bigl(\begin{smallmatrix} x_{11}&0 & 0\\ 0&x_{22} &0 \\ 0&0 &x_{33} \end{smallmatrix}\bigr) =\bigl(\begin{smallmatrix} 13x_{11}&-42x_{11} & 0\\ 7x_{22}&-22x_{22} &0 \\ 0&0 &3x_{33} \end{smallmatrix}\bigr)$$

from both multiplication i get the same matrix
what to do now
?

 

[Mark44]

AX and XA don't come out the same. Try again. Since there are only three nonzero entries in X, you're less prone to making errors if you write it like this:
$$X = \bigl[\begin x_{1}&0 & 0\\ 0&x_{2} &0 \\ 0&0 &x_{3} \end \bigr]$$

 

[transgalactic]

i can't see my mistake
$$\begin{bmatrix} x_1 &0 &0 \\ 0&x_2 &0 \\ 0& 0& x_3 \end{bmatrix} \begin{bmatrix} 13 &-42 &0 \\ 7&-22 &0 \\ 0& 0& 3 \end{bmatrix} = \begin{bmatrix} 13x_1 &-42x_1 &0 \\ 7x_2 &-22x_2 &0 \\ 0& 0& 3x_3 \end{bmatrix}\\$$
same here
$$\begin{bmatrix} 13 &-42 &0 \\ 7&-22 &0 \\ 0& 0& 3 \end{bmatrix} \begin{bmatrix} x_1 &0 &0 \\ 0&x_2 &0 \\ 0& 0& x_3 \end{bmatrix} = \begin{bmatrix} 13x_1 &-42x_1 &0 \\ 7x_2 &-22x_2 &0 \\ 0& 0& 3x_3 \end{bmatrix}\\$$

 

[Mark44]

i can't see my mistake
$$\begin{bmatrix} x_1 &0 &0 \\ 0&x_2 &0 \\ 0& 0& x_3 \end{bmatrix} \begin{bmatrix} 13 &-42 &0 \\ 7&-22 &0 \\ 0& 0& 3 \end{bmatrix} = \begin{bmatrix} 13x_1 &-42x_1 &0 \\ 7x_2 &-22x_2 &0 \\ 0& 0& 3x_3 \end{bmatrix}\\$$
same here

The errors (two of them) are in the multiplication below.

$$\begin{bmatrix} 13 &-42 &0 \\ 7&-22 &0 \\ 0& 0& 3 \end{bmatrix} \begin{bmatrix} x_1 &0 &0 \\ 0&x_2 &0 \\ 0& 0& x_3 \end{bmatrix} = \begin{bmatrix} 13x_1 &-42x_1 &0 \\ 7x_2 &-22x_2 &0 \\ 0& 0& 3x_3 \end{bmatrix}\\$$

 

[Mark44]

Here we are at post 16, still futzing around with something that I suggested you should try all the way back in post 2. I have to say it's pretty frustrating for me. I can only hope that you start to pick up on some of the guidance that I and others on this forum are trying to provide you so that you can do some of this on your own.