# Thread: Matrices went in 2x2

1. ## Matrices went in 2x2

Another challenge problem, inspired by an earlier question.

Given arbitrary 2x2 matrices $\displaystyle A$ and $\displaystyle B$, prove that

$\displaystyle AB+BA=\beta A+\alpha B+(\gamma-\alpha\beta)I$

where $\displaystyle \alpha$ is the trace of $\displaystyle A$, $\displaystyle \beta$ is the trace of $\displaystyle B$ and $\displaystyle \gamma$ is the trace of $\displaystyle AB$ or of $\displaystyle BA$. $\displaystyle I$ is of course the 2x2 identity matrix.

Although a brute force approach would work, a more elegant solution would be welcome.

Enjoy.

Moderator approved. CB

2. We use the Cayley-Hamilton identity for $\displaystyle A, \ B, \ A+B$

$\displaystyle A^2-\alpha A+\det A\cdot I_2=O_2$

$\displaystyle B^2-\beta B+\det B\cdot I_2=O_2$

$\displaystyle (A+B)^2-Trace(A+B)+\det (A+B)\cdot I_2=O_2$

We have $\displaystyle Trace(A+B)=Trace(A)+Trace(B)=\alpha+\beta$

and $\displaystyle \det(A+B)=\det A+\det B+\gamma-\alpha\beta$

Then

$\displaystyle A^2+B^2+AB+BA-\alpha A-\alpha B-\beta A-\beta B+\det A\cdot I_2+\det B\cdot I_2+(\gamma-\alpha\beta)I_2=O_2$

or $\displaystyle AB+BA=\beta A+\alpha B+(\gamma-\alpha\beta)I_2$

3. Originally Posted by red_dog
$\displaystyle (A+B)^2-Trace(A+B)+\det (A+B)\cdot I_2=O_2$
Should be

$\displaystyle (A+B)^2-Trace(A+B)\cdot(A+B)+\det (A+B)\cdot I_2=O_2$

but otherwise spot on! Thanks.