Why is this determinant positive?

Question

I have seen that the $k$-dimensional volume of an parallelepiped in $\mathbb{R}^n$, i.e., $$P(v_1, \ldots, v_k) = \{t_1v_1 + \dotsb + t_kv_k : 0 \le t_i \le 1 \}$$ is $\sqrt{\det(T^{\top}T)}$, where $T$ is the $n\times k$ matrix with columns $v_1, \ldots, v_k$.

How do we know that $\det(T^{\top}T)$ is non-negative?

Answer 1

$T^{\top}T$ is positive semidefinite, so all the eigenvalues are non-negative.

The determinant of $T^{\top}T$ is the product of the eigenvalues; hence, it is non-negative.