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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.

Blackeyes
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    what paralleliliped are you talking about? – mookid Oct 19 '14 at 17:44
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    The product $AA^T$ is there to cater for the possibility that the rows of $A$ are "too long". Or IOW that we are talking about the area of a parallelogram in a higher dimensional space, the 3-volume of a parallelopiped in an $n$-dimensional space and so forth. If $A$ is a square matrix, then $|\det A|$ will give the area/volume/whatever just fine. – Jyrki Lahtonen Oct 19 '14 at 17:46
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    http://en.wikipedia.org/wiki/Gramian_matrix#Gram_determinant and the reference at the end in "Journal of Young Investigators" – Will Jagy Oct 19 '14 at 17:47
  • @Blackeyes, as always, suggest you do the 2 by 2 case really, really carefully, using some examples with, say, integer entries for $A,$ and drawing careful pictures and finding the areas of the parallelograms, just by splitting up the picture into a bunch of right triangles I suppose. – Will Jagy Oct 19 '14 at 17:52
  • @Bla, next two vectors in $\mathbb R^3,$ where you can find the area using the cross product http://en.wikipedia.org/wiki/Cross_product#Geometric_meaning – Will Jagy Oct 19 '14 at 17:55
  • @mookid, see the edit. That should clarify what parallelepiped I am talking about. – Blackeyes Oct 19 '14 at 17:58
  • @WillJagy, I understand "The magnitude of the cross product can be interpreted as the positive area of the parallelogram having a and b as sides". I just am failing to connect that to $\sqrt{det(AA^T)}$ – Blackeyes Oct 19 '14 at 18:00
  • @Bla, the important part of this is that you do some (relatively easy) examples. – Will Jagy Oct 19 '14 at 18:02
  • @Blackeyes: here is the proof (for any $d$). It should not be the expected one... – mookid Oct 19 '14 at 18:08
  • I think I got it. $det(AB)=det(A)det(B)$. Thus, $det(AA^T)=det(A)det(A^T)=(det(A))^2$. Since $Vol(P)=|det(A)|$, $Vol(P)=\sqrt{det(AA^T)}$ – Blackeyes Oct 19 '14 at 18:17
  • I think you should edit the title, since right now, it seems you'd need to take the fourth root to get the volume. – Nishant Oct 19 '14 at 18:59

4 Answers4

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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.

rych
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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}

Start wearing purple
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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 $$ vol(A) = vol(R) = |\det R| = |\det A| = \sqrt{\det AA^T} $$

mookid
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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 $C_{i_1},\cdots,C_{i_n}$.