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

• Clandry
In summary, the homework statement asks for all 2x2 matrices X such that AX=XA for all 2x2 matrices. An algebraic solution is found if x=y=0.f

## 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?

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?

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. "

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.

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?

No, it just means the solution is not unique...