Proving that for when AB = 2BA, then B is not invertible if A is invertible. Suppose that A and B are 2x2 matrices, A is invertible, and AB = 2BA. Prove that B is not invertible.
My first thought is that I need to find what B is in terms of the terms of A, and proving B's invertibility from there, but I don't know how to start that..
Im really stumped. Any help/tips/advice would be much appreciated!
 A: All answers so far suggest the result is false in characteristic$~3$. Indeed (and I see this is already in a comment by user1551) over a field of characteristic$~3$ one has
$$
  A=\pmatrix{0&1\\1&0}, B=\pmatrix{1&0\\0&-1}
  \quad\text{then }
  AB=\pmatrix{0&-1\\1&0\\}=-\pmatrix{0&1\\-1&0\\}=-BA=2BA.
$$
Probably not intended, but the question does not specify where the matrix entries live.
I may add another example to show that while (outside characteristic $3$) "$B$ singular" may be strengthened to "$B$ nilpotent" (as the answer by 'math' indicates), one cannot strengthen it to "$B=0$":
$$
  A=\pmatrix{2&0\\0&1}, B=\pmatrix{0&1\\0&0}
  \quad\text{then }
  AB=\pmatrix{0&2\\0&0\\}=2\pmatrix{0&1\\0&0\\}=2BA.
$$
Finally (a pet subject of mine), while certain other matrix sizes will allow counterexamples in other odd characteristics, there is one size where counterexamples exist independently of the characteristic: for $0\times0$ matrices, $AB=2BA$ holds "always" (there is just one case), but $A,B$ are both invertible.
A: Taking determinant, we have
$$
\det(A)\det(B)=2^2\det(B)\det(A)\implies \det(A)\det(B)=0\implies \det(B)=0.
$$
The last implication above uses $\det(A)\neq 0$.
A: Hint : Take $\det$ on both sides...
