TITLE: Why is the volume of a parallelepiped equal to the square root of $\sqrt{det(AA^T)}$
QUESTION [5 upvotes]: Why is the $\sqrt{det(AA^T)}$ equal to the volume of a parallelepiped?
Is is somehow related to the fact that $det(A) = det(A^T)$?
EDIT: To clarify, the parallelepiped is spanned by the columns of A.

REPLY [4 votes]: Let us define the area of a parallelogram as $\lVert a_1\rVert\lVert a_2\rVert\sin\theta$ and see that it equals $\lVert a_1\rVert\lVert a_2\rVert\sqrt{1-\cos^2\theta}=\sqrt{\lVert a_1\rVert^2 \lVert a_2\rVert^2-\langle a_1, a_2\rangle^2}$. The expression under the square root is the determinant of the Gramian matrix $A^TA$, where $A=\big(a_1 \enspace  a_2\big)$.
The same expression $\sqrt{\det(A^TA)}$ with $A=\big( a_1\; a_2\; ...\;a_k\big)$ can be used to find/define the volume of parallelepiped and k-parallelotope in higher n-dimensional spaces. Note that while $A$ may not be a square matrix, $A^TA$ always is.