# Commuting Matrices

• Aug 15th 2010, 10:44 AM
NonCommAlg
Commuting Matrices
[ $\star \star$] Let $A,B$ be any $2 \times 2$ matrices with entries from $\mathbb{C}$ and $AB=BA.$ Let $t_A$ and $I$ denote the trace of $A$ and the $2 \times 2$ identity matrix respectively.

Find some $\alpha, \beta \in \mathbb{C}$ such that $(2A - t_AI)B=\alpha A + \beta I.$
• Aug 16th 2010, 06:29 AM
NonCommAlg
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 $n \times n$ commuting matrices but the solution to this general case is much harder.
• Aug 20th 2010, 12:49 AM
halbard
Commuting(?) matrices
This is a particular consequence of a very well-known result. For any $2\times2$ matrices $A$ and $B$

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

In particular, if $AB=BA$ then

Spoiler:

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

Incidentally, $(AB-BA)^2=\gamma I$ holds for all $2\times2$ matrices $A$ and $B$. What is $\gamma$?
• Aug 20th 2010, 07:15 AM
NonCommAlg
Quote:

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

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

well, assuming that the above holds, your solution would obviously be correct. my solution is different from yours.

Quote:

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