What I need to prove: Let $A,B \in \mathcal{B}(\mathcal{H})$ with $A$ self-adjoint and $B$ positive. Prove that if $BAB + A = 0$, then $A = 0$.
Since the definition of a positive operator relies on inner products, I've been trying to use the fact that $\langle (BAB + A)x, (BAB + A)x \rangle = 0$ to show that $\langle Ax, Ax \rangle = 0$. I started by using the fact that the inner product is linear to split it into four inner products.
I know that $\langle BABx, BABx \rangle$ and $\langle Ax, Ax \rangle$ are nonnegative, but I don't know how to show that $\langle ABx,BAx \rangle$ and $\langle BAx,ABx \rangle$ are nonnegative, which would force all four inner products to be zero. Could I have a hint?
Best Answer
Use the fact that if $B$ is positive, then $ABA$ is positive for self-adjoint $A$.
Then $(ABA)B=-A^{2}=(BAB)A=B(ABA)$, so $ABA$ and $B$ commute, so both $BABA$ and $ABAB$ are positive, as a consequence, $-A^{2}$ is positive.
But then $\left<A^{2}x,x\right>=\left<Ax,Ax\right>=\|Ax\|^{2}\geq 0$, so $A^{2}$ is positive, so we must have $A^{2}=0$, this means that $\|Ax\|^{2}=0$ and we can conclude that $A=0$.