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Suppose there are three linearly independent vectors in $\mathbb R^4$. The 3d space they span will also contain a 3d parallelepiped formed by these three vectors. I am interested in finding the volume of this parallelepiped.

My attempt for this was based on imagining this in terms one less dimension. When there are two linearly independent vectors in $\mathbb R^3$, the area of the parallelogram will be the cross product which I can compute for three dimensional vectors. But I could not extend this to 4 dimensional vectors as I don't know how do get cross product now.

Based on some search and a hunch I stumbled on this answer which seems to suggest that the square-root of the $|X'X|$, where X is $4 \times 3$ matrix with columns as my three linearly independent vectors, will get me the volume I need. Is this correct (the linked answer does not give a proof of this)?

If so, can I naturally extend this to higher dimensions, i.e., volume of k-dimensional parallelepiped in $\mathbb R^n$ is given by the above formula?

Dayne
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  • Cross product only works in $\mathbb{R}^3$ and $\mathbb{R}^7$. You might be interested in the wedge product. – Enrico M. May 05 '23 at 18:01
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    yes. If you made a 4 by 4 matrix with the last column unit length and perpendicular to your three, the Gram matrix would give the square of the 4-volume of the thing. But the thing has volume 1 times your 3d volume – Will Jagy May 05 '23 at 18:06
  • @WillJagy: thanks! That sounds perfectly logical. I did not know about gram matrix so thanks a lot. Please feel free to post your comment as answer and I will accept it. – Dayne May 05 '23 at 18:12
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    I recommend a little model; take your 4 by 3 matrix $X$ with a bunch of numbers, maybe 1 through 12. Find the squared volume your way. Then, find a vector perpendicular to all three vectors; the final step is to rotate that fourth vector to the $e_4$ direction. The original three vectors are now in the $e_1, e_2, e_3$ subspace, just like $x,y,z$ coordinates. And now you can find the volume again, but in three dimensions. – Will Jagy May 05 '23 at 18:24
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    Oh, there is always a product of $n-1$ vectors in $R^n$ Take your 4 by 3 matrix, make the final column the symbols $a,b,c,d$ This determinant is an expression, linear in $a,b,c,d$ and the coefficients of these, in order, give the desired orthogonal vector. – Will Jagy May 05 '23 at 18:28
  • @WillJagy: thanks again. For your second last comment, I was thinking on similar lines but didn't know how to rotate in 4d. In any case, I get the idea and just to confirm: after doing what you propose, the 4th coordinate for all the three original vectors will become zero? – Dayne May 06 '23 at 02:32
  • Also, is your comment about how to find the 4th vector orthogonal to original three? – Dayne May 06 '23 at 02:33
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    Relevant https://math.stackexchange.com/questions/981238/why-is-the-volume-of-a-parallelepiped-equal-to-the-square-root-of-sqrtdetaa/982191#982191 – rych May 06 '23 at 10:10
  • @rych: thanks for the reference. – Dayne May 07 '23 at 06:43

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