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Let $V$ be a vector space of dimension $n$ and $B:V\times V\rightarrow\mathbb{R}$ be an inner product. Let $\sigma_B:V^n\rightarrow\mathbb{R}$ be the map $$ \sigma_B(v_1,\ldots,v_n)=(\det[b_{i,j}])^{\frac12},$$ where $b_{i,j}=B(v_i,v_j)$. Show that $\sigma_B(v_1,\ldots,v_n)$ is the volume of the parallelepiped $P(v_1,\ldots,v_n)$ having $v_1,\ldots,v_n$ as adjacent edges.

I don't really see how to relate the determinant of the matrix composed of the inner products into the volume of the parallelepiped. What would be the way?

JJ Beck
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2 Answers2

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This is essentially equivalent to showing that a linear operator represented by a matrix $A$ takes shapes to the domain to shapes in the range, and multiplies their volume by the determinant of $A$.

This is the way I convinced myself of this many years ago. First, it is obviously true for diagonal matrices. Next, it is also true for shear matrices, which are of the form identity matrix plus a matrix all of whose entries are zero except one entry. Finally, it is obviously true for permutation matrices. Now you merely need to prove that every matrix is a product of these kinds of matrices, and this can be done, for example, by going through the steps of the Gauss-Jordon method for solving systems of linear equations. And of course, you need to convince yourself that if a matrix $A$ multiplies volumes by a factor $a$, and another matrix multiplies volumes by a factor $b$, then $AB$ multiplies volumes by a factor $ab$. And you can see this by approximating any reasonable shape by lots of little parallelepipeds that fill in the shape.

Stephen Montgomery-Smith
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I am thinking aloud here, but maybe this will be helpful:

Gram schmidt on the vectors $ (v_1,\cdots, v_n) $ corresponds to gauss jordan on $ b_{i,j} $. For instance, let $ v_1 \rightarrow v_1 - c v_k $. Then: $$ b_{1,1} \rightarrow b_{1,1} - 2 c b_{1,k} + c b_{k,k} \\ b_{1,j} \rightarrow b_{1,j} - 2 c b_{k,j} \\ b_{j,1} \rightarrow b_{j,1} - 2 c b_{j,k} $$ One can arrive at this by row/column operations: $ b_{1,j}' = b_{1,j} - c b_{k,j} $ followed by $ b_{j,1}'' = b_{j,1}' - c b_{k,j}' $. So by gauss jordan you should be able to reduce to a diagonal matrix, corresponding to skewing your parallelepiped back into a hyper rectangle.

user110781
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