1. ## Commuting Matrices

[$\displaystyle \star \star$] Let $\displaystyle A,B$ be any $\displaystyle 2 \times 2$ matrices with entries from $\displaystyle \mathbb{C}$ and $\displaystyle AB=BA.$ Let $\displaystyle t_A$ and $\displaystyle I$ denote the trace of $\displaystyle A$ and the $\displaystyle 2 \times 2$ identity matrix respectively.

Find some $\displaystyle \alpha, \beta \in \mathbb{C}$ such that $\displaystyle (2A - t_AI)B=\alpha A + \beta I.$

2. i don't have to mention that i'm not interested in a direct solution. anyway, here is a hint for those who like the problem:

Spoiler:
a clever use of Cayley-Hamilton theorem will solve the problem very quickly!

by the way, the result can be extended to $\displaystyle n \times n$ commuting matrices but the solution to this general case is much harder.

3. ## Commuting(?) matrices

This is a particular consequence of a very well-known result. For any $\displaystyle 2\times2$ matrices $\displaystyle A$ and $\displaystyle B$

$\displaystyle AB+BA=t_B A+t_A B+(t_{AB}-t_At_B)I$

In particular, if $\displaystyle AB=BA$ then

Spoiler:

$\displaystyle 2AB-t_AB=\alpha A+\beta I$ where $\displaystyle \alpha=t_B$ and $\displaystyle \beta=t_{AB}-t_At_B$.

Incidentally, $\displaystyle (AB-BA)^2=\gamma I$ holds for all $\displaystyle 2\times2$ matrices $\displaystyle A$ and $\displaystyle B$. What is $\displaystyle \gamma$?

4. Originally Posted by halbard
This is a particular consequence of a very well-known result. For any $\displaystyle 2\times2$ matrices $\displaystyle A$ and $\displaystyle B$

$\displaystyle AB+BA=t_B A+t_A B+(t_{AB}-t_At_B)I$
well, assuming that the above holds, your solution would obviously be correct. my solution is different from yours.

Incidentally, $\displaystyle (AB-BA)^2=\gamma I$ holds for all $\displaystyle 2\times2$ matrices $\displaystyle A$ and $\displaystyle B$. What is $\displaystyle \gamma$?
$\displaystyle \gamma = -\det(AB-BA),$ by Cayley-Hamilton and the fact that $\displaystyle t_{AB-BA}=0.$