Find all 2x2 matrices X such that AX=XA for all 2x2 matrices

  • Thread starter Clandry
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  • #1
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Homework Statement


Find all 2x2 matrices X such that AX=XA for all 2x2 matrices


The Attempt at a Solution


Let A =
a b
c d

and X =
w x
y z

Then AX = XA ==>

aw+by=wa+xc .........(1)
ax+bz=wb+xd .........(2)
cw+dy=ya+zc .........(3)
cx+dz=yb+zd .........(4)

(1) ==> by = xc, which holds for all b and c only if x=y=0.
(2) ==> bz = wb, which hods for all b only if z=w.
(3) ==> w=z
(4) ==> x=y=0

So the answer is x=y=0, and w=z. That is, X = k*I with k being any real number and I is the identity matrix.



How does this look?
 

Answers and Replies

  • #2
Simon Bridge
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I see the algebra for (1) ... not the others. Perhaps you should be more explicit.
But your result works. Does k have to be a real number?
 
  • #3
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Hello. Thank you for the reply. The algebra for (2)-(4) were based on the results of (1).

For (2), we have ax+bz=wb+xd; in (1) we decided that x=y=0, thus this expression becomes bz=wb
For (3), if x=y=0, then w=z
For (4), we have cx+dz=yb+zd=> xc=yb, thus x=y=0.


I am not too confident with what I did for part 1. "(1) ==> by = xc, which holds for all b and c only if x=y=0. "
Alternatively, could I have also said "(1) ==> by = xc, which holds for all x and y only if b=c=0. "
 
  • #4
Simon Bridge
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I think I see what you mean. It kinda looks like you are saying that x=y=0 as a strong assertion - you got there by reason more than algebra - and that feels weak right?

I am not too confident with what I did for part 1. "(1) ==> by = xc, which holds for all b and c only if x=y=0. "
Alternatively, could I have also said "(1) ==> by = xc, which holds for all x and y only if b=c=0. "
In the second case, then you have just observed that for any A where b=c=0, the values of x and y in X don't matter.
But your problem was to find the values of x and y (and the rest) that work when b and c are other numbers too.

If you want a more rigorous approach, that depends less on direct reasoning, then try simplifying each line independently, building a matrix of coefficients, then using echelon reduction to arrive at the relationships.
 
  • #5
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Ah I see what you are saying. I had forgotten the essence of the problem statement.

Regarding the rigorous approach, it seems there are too many unknowns for that to come out nicely?
 
  • #6
Simon Bridge
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No, it just means the solution is not unique...
 

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