Why is the volume of a parallelepiped equal to the square root of $\sqrt{det(AA^T)}$ 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.
 A: 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.
A: Denote the vectors corresponding to the parallelepiped edges by $\vec a,\vec b,\vec c$. The volume is equal to the absolute value of the mixed product of these three vectors
$$\operatorname{Vol}(\mathcal P)=\left|\,\vec a\cdot\left(\vec b\times\vec c\right)\right|.\tag{1}$$
On the other hand, the mixed product can be computed in cartesian coordinate system:
since 
$$\vec b\times \vec c= \left(b_yc_z-b_z c_y\right)\vec e_x+\left(b_zc_x-b_x c_z\right)\vec e_y+\left(b_xc_y-b_y c_x\right)\vec e_z,$$ 
we have
\begin{align}\vec a\cdot\left(\vec b\times\vec c\right)&=a_x\left(b_yc_z-b_z c_y\right)+a_y\left(b_zc_x-b_x c_z\right)+a_z\left(b_xc_y-b_y c_x\right)=\\&=
\operatorname{det}\left(\begin{array}{ccc}
a_x & b_x & c_x \\ a_y & b_y & c_y \\ a_z & b_z & c_z\end{array}\right)=\operatorname{det} A.
\end{align}
A: Write $A= QR$ with $Q$ orthogonal and $R$ upper triangular, so that
$$
vol(A) = vol(R)
$$
Then, using the usual formula for the volume of the triangle, the volume of $R$ is the (absolute value of the) product of its diagonal values. Then:
$$
vol(R) = \prod_{i=1}^d |R_{ii}| = |\det R|
$$
Hence
$$
vol(A) = vol(R) = |\det R| = |\det A| = \sqrt{\det AA^T}
$$
A: This exercise is not interesting except if $A$ is a $n\times m$ matrix with $n<m$. Then $\det(AA^T)$ is the sum - over every choice of $n$ distinct columns $C_{i_j}$ of $A$ - of the squares of the volumes of the parallelepipeds spanned by the $C_{i_1},\cdots,C_{i_n}$.
