7. Given that $P$ is a moving point on the ellipse $\frac{x^{2}}{12}+\frac{y^{2}}{4}=1$, and $F_{1} 、 F_{2}$ are the two foci of the ellipse. Then the range of $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}$ is $\qquad$
7. $[-4,4]$.

Let $P\left(x_{0}, y_{0}\right) 、 F_{1}(-c, 0) 、 F_{2}(c, 0)$. Then $\overrightarrow{P F_{1}}=\left(-x_{0}-c,-y_{0}\right), \overrightarrow{P F_{2}}=\left(c-x_{0},-y_{0}\right)$. Therefore, $\overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}}=x_{0}^{2}-c^{2}+y_{0}^{2}=x_{0}^{2}+y_{0}^{2}-c^{2}$.
Notice that $b^{2} \leqslant x_{0}^{2}+y_{0}^{2} \leqslant a^{2}$. Then $b^{2}-c^{2} \leqslant \overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}} \leqslant a^{2}-c^{2}$,
where, $a^{2}=12, b^{2}=4, c^{2}=a^{2}-b^{2}=8$.
Therefore, $-4 \leqslant \overrightarrow{P F_{1}} \cdot \overrightarrow{P F_{2}} \leqslant 4$.