# Let $A,B \in \mathcal{B}(\mathcal{H})$ with $A$ self-adjoint and $B$ positive. Prove that if $BAB + A = 0$, then $A = 0$.

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?

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 $$\left=\left=\|Ax\|^{2}\geq 0$$, so $$A^{2}$$ is positive, so we must have $$A^{2}=0$$, this means that $$\|Ax\|^{2}=0$$ and we can conclude that $$A=0$$.