Let A be an n x n mx with n distinct eigenvalues and let B be an n x n mx with AB = BA. If X is an eigenvector of A how can you show that BX is zero or is an eigenvector of A with the same eigenvalue?

Apparantly once you've done this you can conclude that X is also an eigenvector of B, but i'm not sure how exactly. If anyone could help me out that'd be great, thanks.

X is an eigenvector of A means:
AX = λΧ (where λ is a scalar)
ΒΑΧ = ΒλΧ
but (AB)X = (BA)X
so ABX = BAX = ΒλΧ = λΒΧ
or A(BX) = λ(BX)
which means BX is 0 or an eigenvector of A.

well that part seems easy enough, but how can you conclude that X is an eigenvector of B?

Off the top of my head I only remember how to prove the nondegenerate case, so here goes.

Your matrices have finite dimension, n. The set of eigenvectors that diagonalizes A is:

{x₁, x₂,...,x_n}

That means that for any i=1,2,...,n we have:

Ax_i=λ_ix_i

Now you've already been shown that A(Bx_i)=λ(Bx_i). That means that in the nondegenerate case
Bx_i must be proportional to the eigenvector x_i, or:

Bx_i=μ_ix_i.

I'll leave it to you to work it out when there are degeneracies.