# Is the exponential speedup and output $\langle x|M|x\rangle$ in contradiction in HHL algorithm?

Isn't the exponential speedup and the output $$\langle x|M|x\rangle$$ in contradiction in HHL algorithm? How can we print the solution vector $$|x\rangle$$ without losing the exponential speedup?

• The beginning of page 2 of the paper provides the answer. "Clearly, to read out all the components of $\vec x$ would require one to perform the procedure at least $N$ times. However, often one is interested not in $\vec x$ itself, but in some expectation calue $\vec x^T M\vec x$, where $M$ is some linear operator (our procedure also accommodates nonlinear operators as described below)." Nov 19, 2021 at 20:31