Define the operation $*$ on the set $M_2(\mathbb{Z})$ as: $A*B = AB+aBA$. Determine $a \in \mathbb{R}$ such that $*$ is binary.
My attempt:
\begin{bmatrix}
z_1w_1+z_2w_3+aw_1z_1+aw_2z_3 & z_1w_2+z_2w_4+aw_1z_2+aw_2z_4 \\
z_3w_1+z_4w_3+aw_3z_1+aw_3z_3 & aw_3z_2+aw_4z_4+z_3w_2+z_4w_4
\end{bmatrix}
This is simply the resulting matrix written by its components. Taking the trace one gets:
$$z_1w_1+z_2w_3+aw_1z_1+aw_2z_3+aw_3z_2+aw_4z_4+z_3w_2+z_4w_4 = k$$
For all integers $z_1,…,z_4$ and $w_1,…,w_4$ and some integer $k$. If the sum of terms multiplied by $a$ is not equal to $0$, it follows that $a=-k$. Plugging in $a=-k$ we get $\frac{-k}{k+1}=-k$. It follows that $k=0$ and thus $a = 0$. My questions:
1) What happens when the sum of terms multiplied by $a$ is not $=0$? Are there any additional solutions, is there a contradiction, etc.?
2) With this approach, are all $a$ accounted for? Why?
3) Is there a faster way to do this?
Any help would be greatly appreciated.
Best Answer
If $a\in\Bbb Z,$ then clearly this is a binary operation, as addition and multiplication of matrices are binary operations on $M_2(\Bbb Z).$ Explicitly, if $A,B\in M_2(\Bbb Z),$ then $AB,BA\in M_2(\Bbb Z).$ Moreover, if $a\in\Bbb Z,$ and $M\in M_2(\Bbb Z),$ then $$aM = \begin{pmatrix}a & 0\\0 & a\end{pmatrix} M\in M_2(\Bbb Z).$$ Take $M = BA$ to see that $aBA\in M_2(\Bbb Z)$ as well. Finally, adding any two matrices in $M_2(\Bbb Z)$ produces another matrix in $M_2(\Bbb Z).$
On the other hand, suppose that $a\in\Bbb R\setminus\Bbb Z,$ and take $A = B = I.$ Then we have $$I*_a I = I^2 + aI^2 = I + aI = \begin{pmatrix}1 + a & 0\\ 0 & 1 + a.\end{pmatrix},$$ and $1 + a\in\Bbb Z$ if and only if $a\in\Bbb Z.$
Thus, $A*_a B$ is a binary operation on $M_2(\Bbb Z)$ if and only if $a\in\Bbb Z.$ (Remember that a binary operation on a set $X$ is nothing more than a function $f : X\times X\to X,$ so as long $AB + aBA\in M_2(\Bbb Z),$ you get a binary operation.)