In a negative feedback loop, why does increase in v_out lead to increase in v-?
You might find the following illustration helpful (but might not!).



simulate this circuit &ndash; Schematic created using CircuitLab
Figure 1. Reference circuit for Figure 2.

Figure 2. Picturing the inverting op-amp as a see-saw with it's pivot point being fixed at &quot;virtual ground&quot;.
How it works:

The gain of the see-saw is -2. That means that if we lift the left side by one unit the right side will go down by two units. Note the relative lengths of Ri (input) and Rf (feedback) are in the ration of 1:2.
(a) With *Vin = 0 V the virtual ground and the output are at 0 V.
(b) If we raise the input to 1 V we can see that if the output were not to respond that the inverting input voltage would rise to 2/3 V. This would drive the output towards the negative supply voltage rail.
(c) As the output swings negative balance is restored when Vout = -2 V.

The see-saw analogy works for virtual grounds at non-zero voltages too. e.g. If the non-inverting input is held at 2 V then move the pivot point up to the 2 V line.

Non-inverting case - if you're interested:



simulate this circuit
Figure 3. Non-inverting amplifier with gain of 3.

Figure 4. The non-inverting lever.
How it works:

The left side of Ri is tied to ground.
When Vin is applied to the non-inverting the output must swing high to bring the inverting input to the same level. This will require the output to go to 3 V in the Figure 3 example.
The gain of the non-inverting amplifier is given by \$ 1 + \frac {R_f}{R_i} = 1 + \frac 2 1 = 3 \$ which agrees with our illustration.

All artwork mine.