Prove, that the vector Space Hat (n; F) with the multipliсation A⋅B=AB−BA is a F-algebra...

To prove that the vrctor space Hat(n,f) := {set of ${\displaystyle n\times n}$ matrices over F} is a F-algebra
Note that Hat(n, F) is said to be F-algebra if for any elements x, y, z ${\displaystyle \in }$ Hat(n, F) and all elements a, b ${\displaystyle \in }$ F.
Right distribulity, left distribulity and compatibility with scalars followed.
Note that
1) ${\displaystyle (x+y)\cdot z=(x+y)z-z(x+y)}$
${\displaystyle =xz+yz-zx-zy}$
${\displaystyle =(xz-zx)+(yz-zy)}$
${\displaystyle =x\cdot z+y\cdot z}$
(Right distribulity)
2) ${\displaystyle z\cdot (x+y)}$
${\displaystyle =z(x+y)-(x+y)z}$
${\displaystyle =(zx-xz)+(zy-yz)}$
${\displaystyle =z\cdot x+z\cdot y}$
(Left distribulity)
3) ${\displaystyle \left(ax\right)\cdot \left(by\right)=axby-byax}$
${\displaystyle =ab(xy-yx)}$
${\displaystyle =ab(x\cdot y)}$
(Compatibility with scalars) So, Hat(n,f) is an F-algebra for x, y, z ${\displaystyle \in }$ Hat(n, F) and a, b ${\displaystyle \in }$ F

That is not full answer, here is full:
Note that,
${\displaystyle (x\cdot y)\cdot z=(xy-yx)\cdot z}$
${\displaystyle =(xy-yz)z-z(xy-yx)}$
${\displaystyle =xyz-yxz-zxy+zyx}$
and
${\displaystyle x\cdot (y\cdot z)=x\cdot (yz-zy)}$
${\displaystyle =x(yz-zy)-(yz-zy)x}$
${\displaystyle =xyz-xzy-yzx+zyx}$
We can check ${\displaystyle x\cdot y\cdot x\ne (x\cdot y)\cdot z}$
So (Not associative)
Comm? Note that
${\displaystyle x\cdot y=xy-yz}$
${\displaystyle y\cdot x=yx-xy}$
${\displaystyle \therefore x\cdot y\ne y\cdot x}$
${\displaystyle \therefore }$ Not Commutative
Unitary? Mean it sholud have identity element operation *
Let T be identity element
Then ${\displaystyle x\cdot T=x\mathrm{\forall }x\in }$ H(n, F)
${\displaystyle \Rightarrow xT-Tx=x}$ ---- (1)
and ${\displaystyle T\cdot x=x\Rightarrow Tx-xT=x}$ ---- (2)
${\displaystyle \therefore }$ from (1) + (2)
${\displaystyle \to 2x=0\Rightarrow x=0}$
So there are no identity elements for all x ${\displaystyle \in }$ H(n, f)
Not Unitary