There seems to be some sort of misunderstanding here. Making a measurement of observable $A$ of a system in the state $|\psi\rangle$ does *not* mean we need to get a number from the calculation $A|\psi\rangle$. The issue here is that $A|\psi\rangle$ is still a vector. If you are expecting the measurement to give a value of $a$ for $A|\psi\rangle=a|\psi\rangle$ then this is still incorrect, as in general $|\psi\rangle$ will not be an eigenvector of $A$.

So how does the operator $A$ relate to the measurement of the observable associated with $A$? Well, all you have to do is express $|\psi\rangle$ in the eigenbasis of $A$
$$|\psi\rangle=\sum_nc_n|a_n\rangle$$
Quantum theory tells us that if we were to measure $A$ of our system that all can cen determine is the *probability* of measuring some value $a_n$. This probability is equal to $|c_n|^2=|\langle a_n|\psi\rangle|^2$.

So, from the operator we can determine two things:

 1. Its eigenvalues (possible measurement outcomes)
 2. Its eigenvectors (what we can use as basis vectors).

And from these two things we can then determine the probability of our system to have a value of $a_n$ when we measure $A$.