# Exercise 2.25

Suppose $A$ is an arbitrary operator and $|v\rangle$ is arbitrary vector, we can compute the inner product of $|v\rangle$ and $A^{\dagger}A|v\rangle$ as

$$
\left(|v\rangle, A^{\dagger}A|v\rangle\right) = \langle v|A^{\dagger}A|v\rangle 
$$(eqn:2.25.1)

Suppose $|u\rangle = A|v\rangle$ then eq. {eq}`eqn:2.25.1` becomes $\langle v|A^{\dagger}A|v\rangle = \langle u|u\rangle$. According to the properties of inner product, we have $\langle u|u\rangle \geq 0$. That is, for arbitrary operator $A$ and arbitrary vector $|u\rangle$, we always have

$$
\langle v|A^{\dagger}A|v\rangle = \langle u|u\rangle \geq 0
$$(eqn:2.25.2)

Thus, we could conclude that $A^{\dagger}A$ is positive.