What does "volume" mean in "signed volume"?

Question

I've read that the determinant of a matrix gives the signed volume of the parallelotope spanned by the matrix column vectors. I can see how that works for Cartesian coordinates using standard unit basis vectors, but how does it work if, for example, the basis vectors aren't unit and are skewed, as in this example:

The determinant of the two vectors $\mathbf{w}=1e_{1}+3e_{2}$ and $\mathbf{v}=3e_{1}+2e_{2}$ gives the area of the big parallelogram as $\pm7$ of the little parallelograms spanned by $e_{1}$ and $e_{2}$. But that isn't the true area (in units of square something or other) of the parallelogram. So does the "volume" in "signed volume" only refer to Cartesian coordinates and standard unit basis vectors?

Answer 1

I know that it is a very popular way to introduce determinants, and it definitely has its merits for forging basic intuition, but I always deplore how much emphasis is put in basic courses on the "volume aspect" of the determinant. The determinant is much more fundamental than a notion of volume. It makes perfect sense for real vector spaces with no euclidean structure, and most of all makes perfect sense over arbitrary fields (even commutative rings), which makes the volume analogy shaky at best.

I have two ways to answer your question:

• First, do not interpret the determinant of a matrix by the volume of the parallelotope spanned by its column vectors. Because a matrix is, at heart, the representation of a linear transformation, not of a family of vectors. Interpret the determinant as the "expansion factor" of the matrix, in the sense that for any (say) parallelotope $X$ of volume $V$, applying the matrix $A$ to $X$ gives a parallelotope $AX$ of volume $|\det(A)|V$. This is something that is unit-independent: for any volume unit you want to use, the "expansion factor" of the matrix is the same.

• If you really want to use the volume of the parallelotope interpretation, then it all comes down to a choice of unit. Because then you are really using the definition of the determinant of a family of vectors (here the column vectors). But this is not well-defined in itself. It is only defined with respect to a fixed basis, which basically amounts to saying that you can't give a volume without first fixing a unit. In your example, if the volume unit you fix is this small parallelogram spanned by $e_1$ and $e_2$, then indeed the volume you are looking for is $7$. The confusion comes from the fact that since you are working in $\mathbb{R}^n$ and not an abstract vector space, you have for free a canonical euclidean structure, and thus a notion of "actual" volume (for instance given by Lebesgue measure). This amounts to saying that in a euclidean space, there are some bases that are "better" than others: the orthonormal ones. So if you want to relate the determinant to the "actual" volume in euclidean space, it's simple: make the correct choice of units, and pick an orthonormal basis instead of an arbitrary basis. But it's the same result. Just in different units.