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As described here, the volume of the n-dim. parallepiped with $v_i$ in $\mathbb R^n$ as common edges from the originb is the abs. value of the determinant of the linear transformation taking the standard basis to the vectors $v_i$.

Now according to Keith Conrad here, the volume of this parallepiped is $\sqrt{|det(v_i.v_j|}$, where the dot indicates vector dot product.

How to show that both descriptions agree?

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    $\begingroup$ $\det(A^TA)=(\det A)^2$ $\endgroup$ – user8268 Dec 6 '14 at 11:21
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One-liner in the comments shows the equivalence for a maximal parallelepiped, $\sqrt{\det(A^TA)}=\vert\det A\vert$. The expression $\sqrt{\det(A^TA)}$ is more general. 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 and k-parallelepiped (parallelotope) in higher n-dimensional spaces. Note that while $A$ may not be a square matrix, $A^TA$ always is.

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