I will give a semi mathematical point of view, and a completely intuitive point of view.

Mathematical(ish) POV

We have $$\langle \alpha |A|\alpha\rangle = 0 \ \ \forall |\alpha\rangle$$

Switching to the matrix view of operators we can say

$$A = \sum \limits_n a_n |a_n\rangle\langle a_n|$$

Thus $$|a_n|^2 ||\langle \alpha|a_n\rangle||^2 = 0 $$

Besides the trivial case the inner product will not be 0, but we also have the condition from the axioms of QM that $|a_n|^2 \geq 0 $. Thus the only way for this to be true for an arbitrary $|\alpha\rangle$ is for the coefficients to be $0$ which is equivalent to saying $A = 0$.

(Another way of seeing this is since $|\alpha \rangle$ is arbitrary, choose it to be one of the eigenvectors, $|a_n\rangle$. Each time you do this the only solution is that the coefficient is 0. Repeat $\forall n$ and you'll see that $A$ must be the $0$ operator.

I will give a semi mathematical point of view, and a completely intuitive point of view.

Mathematical(ish) POV

We have $$\langle \alpha |A|\alpha\rangle = 0 \ \ \forall |\alpha\rangle$$

Switching to the matrix view of operators we can say

$$A = \sum \limits_{i,j} a_{ij} |a_i\rangle\langle a_j|$$

Thus $$|a_n|^2 ||\langle \alpha|a_n\rangle||^2 = 0 $$

Besides the trivial case the inner product will not be 0, but we also have the condition from the axioms of QM that $|a_n|^2 \geq 0 $. Thus the only way for this to be true for an arbitrary $|\alpha\rangle$ is for the coefficients to be $0$ which is equivalent to saying $A = 0$.

(Another way of seeing this is since $|\alpha \rangle$ is arbitrary, choose it to be one of the eigenvectors, $|a_n\rangle$. Each time you do this the only solution is that the coefficient is 0. Repeat $\forall n$ and you'll see that $A$ must be the $0$ operator.

I will give a semi mathematical point of view, and a completely intuitive point of view.

Mathematical(ish) POV

We have $$\langle \alpha |A|\alpha\rangle = 0 \ \ \forall |\alpha\rangle$$

This is the magnitude of the vector $A|\alpha \rangle$ within our Hilbert space.

i.e. $||A|\alpha \rangle ||^2 = 0$ regardless of the vector it is acting on.

Switching to the matrix view of operators we can say

$$A = \sum \limits_n a_n |a_n\rangle\langle a_n|$$

Thus $$|a_n|^2 ||\langle \alpha|a_n\rangle||^2 = 0 $$

Besides the trivial case the inner product will not be 0, but we also have the condition from the axioms of QM that $|a_n|^2 \geq 0 $. Thus the only way for this to be true for an arbitrary $|\alpha\rangle$ is for the coefficients to be $0$ which is equivalent to saying $A = 0$.

(Another way of seeing this is since $|\alpha \rangle$ is arbitrary, choose it to be one of the eigenvectors, $|a_n\rangle$. Each time you do this the only solution is that the coefficient is 0. Repeat $\forall n$ and you'll see that $A$ must be the $0$ operator.

Intuitive POV

Limit yourself to 3D for a moment. Imagine a vector floating about in space. Take some other random vector and take it's inner product. It equals 0! Well, that's okay. They're just perpendicular. Do this again, in fact be very methodical and do it with EVERY vector in $\mathbb{R}^3$. You just can't get it to not equal 0. What this means is that this vector that you're dotting everything with has to be of 0 magnitude, otherwise there would be a whole set of vectors that wouldn't give a 0. In fact, this is precisely the definition of the $0$ operator.

I will give a semi mathematical point of view, and a completely intuitive point of view.

Mathematical(ish) POV

We have $$\langle \alpha |A|\alpha\rangle = 0 \ \ \forall |\alpha\rangle$$

Switching to the matrix view of operators we can say

$$A = \sum \limits_n a_n |a_n\rangle\langle a_n|$$

Thus $$|a_n|^2 ||\langle \alpha|a_n\rangle||^2 = 0 $$

Besides the trivial case the inner product will not be 0, but we also have the condition from the axioms of QM that $|a_n|^2 \geq 0 $. Thus the only way for this to be true for an arbitrary $|\alpha\rangle$ is for the coefficients to be $0$ which is equivalent to saying $A = 0$.

(Another way of seeing this is since $|\alpha \rangle$ is arbitrary, choose it to be one of the eigenvectors, $|a_n\rangle$. Each time you do this the only solution is that the coefficient is 0. Repeat $\forall n$ and you'll see that $A$ must be the $0$ operator.