[example: Commutators As The First Source]
Let $A$ be an associative algebra over $F$, and define
\begin{align*}
[x,y]=xy-yx
\end{align*}
for $x,y\in A$. We show that this bracket is bilinear, skew-symmetric, and satisfies the Jacobi identity.
For $a,b\in F$ and $x_1,x_2,y\in A$, distributivity and compatibility of scalar multiplication with multiplication in the associative algebra give
\begin{align*}
[ax_1+bx_2,y]
&=(ax_1+bx_2)y-y(ax_1+bx_2)\\
&=a x_1y+b x_2y-a yx_1-b yx_2\\
&=a(x_1y-yx_1)+b(x_2y-yx_2)\\
&=a[x_1,y]+b[x_2,y].
\end{align*}
The same calculation in the second variable gives
\begin{align*}
[x,ay_1+by_2]
&=x(ay_1+by_2)-(ay_1+by_2)x\\
&=a xy_1+b xy_2-a y_1x-b y_2x\\
&=a[x,y_1]+b[x,y_2],
\end{align*}
so the bracket is bilinear. Skew-symmetry follows from
\begin{align*}
[y,x]=yx-xy=-(xy-yx)=-[x,y].
\end{align*}
It remains to verify the Jacobi identity. Using associativity to remove parentheses in products of three elements, we compute
\begin{align*}
[x,[y,z]]
&=x(yz-zy)-(yz-zy)x\\
&=xyz-xzy-yzx+zyx,\\
[y,[z,x]]
&=y(zx-xz)-(zx-xz)y\\
&=yzx-yxz-zxy+xzy,\\
[z,[x,y]]
&=z(xy-yx)-(xy-yx)z\\
&=zxy-zyx-xyz+yxz.
\end{align*}
Adding these three displayed identities term by term gives
\begin{align*}
[x,[y,z]]+[y,[z,x]]+[z,[x,y]]
&=(xyz-xzy-yzx+zyx)\\
&\quad +(yzx-yxz-zxy+xzy)\\
&\quad +(zxy-zyx-xyz+yxz)\\
&=0,
\end{align*}
because $xyz$ cancels with $-xyz$, $xzy$ with $-xzy$, $yzx$ with $-yzx$, $zyx$ with $-zyx$, $yxz$ with $-yxz$, and $zxy$ with $-zxy$. Thus the commutator bracket makes every associative algebra into a Lie algebra.
[/example]