Lie bracket $[x,x]=0\implies [x,y]=-[y,x]$

Question

Why does $[x,x]=0\implies [x,y]=-[y,x]$ in regard to Lie brackets.

I have tried to play around with bilinearity, but I can't get it to work.

$$[ax+by,z]=a[x,z]+b[y,z],[x,x]=0$$

I have tried subbing in $z=x$ and $z=y$, but I just can't obtain anti-commutativity. Thanks

Answer 1

$$[x+y,x+y]=0=[x,x]+[x,y]+[y,x]+[y,y]=[x,y]+[y,x]\Rightarrow [x,y]=-[y,x]$$

Answer 2

One can easily show that $[x,x]= x \cdot x - x \cdot x = 0$ and \begin{align} [ax + by, cx + dy] &= ac \, [x,x] + ad \, [x,y] + bc \, [y,x] + bd \, [y,y] \\ &= ad \, [x,y] + bc \, [y,x] \end{align} By letting $c = a$, $d = b$ then $0 = [x,y] + [y,x]$. With this then $$[ax + by, cx + dy] = (ad - bc) \, [x,y].$$