$AB+BA=(trB)A + (trA)B +(trAB - trAtrB)I_2 .$ Let $A,B \in \mathbb{M}_2(\mathbb{F})$. I want to show that $$AB+BA=(trB)A + (trA)B +(trAB - trAtrB)I_2 .$$
I've tried to prove the assertion considering arbitrary two matrices $A,B$ in $\mathbb{M}_2(\mathbb{F})$. I have calculated $AB$ and $BA$. Then, I have obtained the equality above. However, it is too long and boring. Is there any short way? 
 A: We use Caley-Hamilton theorem, i.e. $X^2-{\rm tr}\ X \cdot X+{\rm
det}\ XI=0$, for $A+B$ :
For direct computation we have a claim $$ {\rm }
 {\rm tr }A \ {\rm tr}B- {\rm tr} (AB) =
 {\rm det} (A+B)- {\rm det}A- {\rm det} B $$
If $A=\left(
        \begin{array}{cc}
          a & b \\
          c & d \\
        \end{array}
      \right),\ B=\left(
                    \begin{array}{cc}
                      x & y \\
                      z & w \\
                    \end{array}
                  \right)$ then $$
{\rm det} (A+B)- {\rm det}A- {\rm det} B =aw +dx-bz-cy$$
Remaining thing is also followed from direct computation. 
A: Here is an answer that is light on computation but conceptually somewhat obscure.
If we perturb $A$ by some $\delta I_2$, both sides of the equality change by the same amount. The similar holds if we perturb $B$ by a scalar matrix. So, we may assume that both $A$ and $B$ are traceless. In this case, the equality reduces to
$$
AB+BA=\operatorname{tr}(AB)I_2\text{ when }\operatorname{tr}(A)=\operatorname{tr}(B)=0.\tag{1}
$$
As both sides are bilinear in $(A,B)$, it suffices to prove $(1)$ on any basis of the subspace of all traceless $2\times2$ matrices. 
We want to show that $(1)$ is a universal identity. So, it suffices to consider only the case $\mathbb F=\mathbb C$ (for the reason, see e.g. p.4 of the handout written by Keith Conrad). In particular, it suffices to prove it when $A$ and $B$ are some nonzero scalar multiples of Pauli matrices $\sigma_x,\sigma_y,\sigma_z$. Since the real linear span of $\{-i\sigma_x,-i\sigma_y,-i\sigma_z\}$ is isomorphic to the real algebra of quaternions under the isomorphism $-i\sigma_x\mapsto i,\ -i\sigma_y\mapsto j,\ -i\sigma_z\mapsto k$, equality $(1)$ further reduces to
$$
ab+ba=2\operatorname{Re}(ab),\quad a,b\in\{i,j,k\},\tag{2}
$$
which is completely trivial.
A: In general, a matrix $X\in \mathbb{M}_n(\mathbb{F})$ is determined by the linear map $\mathbb{M}_n(\mathbb{F})\to \mathbb{F}$ defined by $Y\mapsto \operatorname{tr}(XY)$.  This is because if $E_{ij}$ is the $n\times n$ elementary matrix that's all $0$ except for a $1$ in the $(i,j)$ entry, then $\operatorname{tr}(XE_{ij})$ is the $(i,j)$ entry of $X$.
Thus, it suffices to show that for all $A,B,C\in \mathbb{M}_2(\mathbb{F})$ that
$$
\operatorname{tr}(ABC+BAC)=\operatorname{tr}((\operatorname{tr}B)AC + (\operatorname{tr}A)BC +(\operatorname{tr}(AB) - \operatorname{tr}A\operatorname{tr}B)C ).
$$
Applying linearity of trace, this is equivalently
$$
\operatorname{tr}ABC+\operatorname{tr}BAC=\operatorname{tr}B \operatorname{tr}(AC) + \operatorname{tr}A \operatorname{tr}(BC) +\operatorname{tr}(AB)\operatorname{tr}C - \operatorname{tr}A\operatorname{tr}B\operatorname{tr}C
$$
and just for sake of putting things in order to show how symmetric the situation is, this is
$$
\operatorname{tr}ABC+\operatorname{tr}CBA= \operatorname{tr}A \operatorname{tr}(BC) + \operatorname{tr}B \operatorname{tr}(CA) + \operatorname{tr}C\operatorname{tr}(AB) - \operatorname{tr}A\operatorname{tr}B\operatorname{tr}C.
$$
(The theoretical context here is that we have these two trilinear forms on $\mathbb{M}_2(\mathbb{F})$ that we want to show are equal.  In fact, these are trilinear forms invariant under the action of $GL(2)$ on $\mathbb{M}_2(\mathbb{F})$ by conjugation. The vector space of such invariant trilinear forms for general $GL(n)$ and $\mathbb{M}_n(\mathbb{F})$ happens to be spanned by the six terms present in the above equation, and what we're wanting to show is that when $n=2$ that this spanning set is linearly dependent in this way. Linear dependence occurs exactly when $n=0,1,2$, and, fancifully, when $n=-1,-2$, otherwise the vector space of invariant trilinear forms is indeed 6-dimensional.)
Recall that the third exterior power of a two-dimensional vector space is zero-dimensional, i.e. $\bigwedge^3\mathbb{F}^2\approx 0$.  Regarding the exterior power as a subspace of $\mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2$, then the projection $p$ onto the third exterior power is given by
$$v_1\otimes v_2\otimes v_3 \mapsto \frac{1}{6}\sum_{\sigma\in S_3}(-1)^\sigma v_{\sigma(1)}\otimes v_{\sigma(2)}\otimes v_{\sigma(3)}$$
where $(-1)^\sigma\in\{-1,1\}$ is the sign of the permutation $\sigma$ in the symmetric group $S_3$ on three elements.  (If you care about finite-characteristic fields, you can get rid of the $\frac{1}{6}$, and it's ok using a scale multiple of a projection for the following.) Since the subspace is zero-dimensional, this projection function is identially zero, hence we get the identity
$$
v_1\otimes v_2 \otimes v_3 - v_1\otimes v_3 \otimes v_2 + v_2\otimes v_3 \otimes v_1 - v_2\otimes v_1 \otimes v_3 + v_3\otimes v_1 \otimes v_2 - v_3\otimes v_2 \otimes v_1 = 0
$$
for all $v_1,v_2,v_3\in\mathbb{F}^2$.
This next part is tricky to explain.  I'm staring at some Penrose notation that convinced me of it, but I'll try to do it symbolically.  The idea is that we can take the three matrices as linear operators $\mathbb{F}^2\to\mathbb{F}^2$, tensor them together to get an operator $A\otimes B\otimes C:\mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2\to \mathbb{F}^2\otimes \mathbb{F}^2\otimes \mathbb{F}^2$, composing this with $p$, and then taking the trace.  Considering that $p=0$, we have
$$\operatorname{tr}(p\circ(A\otimes B\otimes C))=0.$$
Perhaps you can convince yourself that, using the expansion of $p$ into six permutations, that this is
$$
\operatorname{tr}A\operatorname{tr}B\operatorname{tr}C
- \operatorname{tr}A\operatorname{tr}(BC)
+ \operatorname{tr}(ABC)
- \operatorname{tr}C\operatorname{tr}(AB)
+ \operatorname{tr}(CBA)
- \operatorname{tr}B\operatorname{tr}(CA)
= 0
$$
which after moving some terms to the other side is exactly the equality we wanted to establish.  Sorry for the lack of details in this last step -- if anyone wants a Penrose diagram proof I can add it.
