7. As shown in Figure 1, given that the side length of square $A B C D$ is $3, E$ is the midpoint of $D C$, and $A E$ intersects $B D$ at point $F$. Then $\overrightarrow{F D} \cdot \overrightarrow{D E}=$ $\qquad$
$=7 .-\frac{3}{2}$.
From the given, we know $\frac{D F}{F B}=\frac{D E}{A B}=\frac{1}{2}$.
Therefore, $\overrightarrow{F D} \cdot \overrightarrow{D E}=-D F \cdot D E \cos \angle F D E$
$=-\frac{1}{3} \times 3 \sqrt{2} \times \frac{3}{2} \times \frac{\sqrt{2}}{2}=-\frac{3}{2}$.