The magnitude of a vector $\mathbf{v}$ is defined as $\|\mathbf{v}\|=\sqrt{v_1^2+v_2^2+\cdots+v_n^2}$,
where $v_1,v_2,\dots,v_n$ are the components of $\mathbf{v}$.
In this case, since the magnitude of $\mathbf{v}$ is 4,
we have $\|\mathbf{v}\|=4$.
The dot product of $\mathbf{v}$ with itself is defined as $\mathbf{v}\cdot\mathbf{v}=v_1^2+v_2^2+\cdots+v_n^2$.
Therefore, the dot product of $\mathbf{v}$ with itself is $\boxed{16}$.The answer is: 16
Below is a math problem. Provide a detailed, step-by-step solution.

### Problem:
Given that the magnitude of vector $\mathbf{v}$ is 4, determine the dot product of $\mathbf{v}$ with itself.

### Answer:
