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There is one very interesting thread at mathoverflow about "What are the most misleading alternate definitions in taught mathematics?" This highest voted answer presents the case of determinant, there are so many ways to define it. It is proposed to use "that det(A) is the volume of the image of the unit cube after applying the transformation determined by A." From this alone, everything supposedly follows. I would be interested in some nice book where this program is implemented.

I like the Axler book Linear Algebra Done Right, but they adopt the definition of det as a product of eigenvalues. This is close, but lacks some geometric interpretation.

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yarchik
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    I think it's much harder than it sounds. Could you use that definition to prove that $\det(AB) = \det(A)\det(B)$ for $2\times2$ matrices? Now what about $3\times 3$? – Matthew Leingang Dec 14 '22 at 16:21
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    I am glad that no book adopts this volume-based definition, as it only works over $\mathbb R$. It is plain nonsense over a finite field, not to mention a commutative ring. – user1551 Dec 14 '22 at 16:28
  • Check out my video series presenting this approach to the determinant and the references listed at the end of the videos. – blargoner Dec 14 '22 at 16:40
  • @blargoner I would be nice if you can post it as an answer adding a small summary of advantages/disadvantages of this approach and maybe a comment on how easy it is, for instance, to prove the product formula (see comment above). – yarchik Dec 14 '22 at 16:44
  • @yarchik See this post – Ben Grossmann Dec 14 '22 at 16:49
  • @user1551 To be fair, even an interpretation that "only" works over $\Bbb R$ (and over $\Bbb C$ with a slight modification) can be very useful. It is similarly true that any result regarding real symmetric, Hermitian, or normal matrices "only" applies over $\Bbb R$ and $\Bbb C$. – Ben Grossmann Dec 14 '22 at 16:54
  • @blargoner I took a peek at your video series and those slides and diagrams are gorgeous. I would love it if you could tell me how you made them – Ben Grossmann Dec 14 '22 at 16:58
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    @BenGrossmann I don't think the circumstances are similar. There isn't any notion of Hermitian matrix over a finite field to begin with, but we may speak of determinant. Anyway, “volume” is an interpretation that is good to know. I've never opposed to teaching that to students. But it shouldn't be taken as a definition. I am not sure, however, whether “volume” should be used as a motivation for the concept of determinant, because the volume approach often seems rather hand-wavy to me. – user1551 Dec 14 '22 at 17:06
  • @MatthewLeingang Are you talking about the volume definition? To me, the multiplicativity of the determinant makes the most sense from the volume definition. If $AB$ is a composition of linear transformations, then $B$ transforms the unit cube into a parallelepiped with volume $\det(B)$, and $A$ transforms this parallelepiped into another one with its volume scaled by $\det(A)$. So $\det(AB) = \det(A)\det(B)$. – Frank Dec 14 '22 at 17:43
  • @BenGrossmann Thanks, I usually just hear "why don't you animate your videos like 3b1b ?" :) Feel free to gmail me and I can share deets. – blargoner Dec 14 '22 at 18:07
  • @blargoner Sure, that would be great! I'm not sure what your email address is though... – Ben Grossmann Dec 14 '22 at 20:25
  • @blargoner I should have guessed... thanks for spelling it out – Ben Grossmann Dec 14 '22 at 23:28
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    I've just learnt that the late geometer Daniel Pedoe had authored a (now out-of-print) book entitled “Geometric Introduction to Linear Algebra”. I don't know if this book adopts a volume-based approach to determinant or not, but it may worth a look. – user1551 Dec 17 '22 at 13:57

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First of all, the "best" definition of a determinant depends on what you are doing. If you are trying to prove the change of variables formula for Riemann integration, then the volume definition is precisely what you need. If you are doing algebraic geometry, it is useful to know that the determinant is a polynomial in the matrix entries. If you are trying to compute a determinant, it might be useful to consider Laplace expansion.

I don't know of any linear algebra books that define the determinant as a volume, and I'm not sure there are many out there. If the book is focused on the computational side of linear algebra, they may motivate the determinant this way, but the actual definition will likely be something that is easy to compute directly. If the book deals with vector spaces (as in Axler's case), it isn't even possible to define the determinant as a volume because there isn't really a notion of "volume" in a vector space. (Though you may be able to define a notion of volume by using the determinant.)

That doesn't mean interpreting the determinant as volume isn't useful. Actually, I think it can be extremely useful for developing intuition. To demonstrate, here are some properties you can quickly "prove" using the volume definition:

  • $\det(AB) = \det(A)\det(B)$. Think of $AB$ as the composition of two linear transformations. If we apply $AB$ to the unit cube, $B$ transforms the cube into a parallelepiped with its volume scaled by $\det(B)$, and $A$ transforms this parallelepiped into another parallelepiped with its volume scaled by $\det(A)$. So $AB$ transforms the unit cube into a parallelepiped with volume $\det(A)\det(B)$.
  • $\det(A) = 0$ iff $A$ is not invertible. Think of $A$ as the linear transformation that sends the unit basis of $\mathbb{R}^n$ to the columns of $A$. The determinant is just the volume of the parallelepiped formed by the columns of $A$. It is intuitively clear that $\det(A) = 0$ if and only if the columns of $A$ are linearly dependent.
  • The determinant is multilinear and alternating. If you scale an individual column vector of $A$, then the volume of the corresponding parallelepiped scales by the same factor. If you switch two column vectors, then the "sign" of the determinant switches.
  • If a matrix is diagonalizable, then the determinant is the product of its eigenvalues. Think of how the matrix transforms the parallelepiped formed by its eigenbasis.

Though these arguments are not rigorous, there is a way to make this idea of "volume" rigorous. We define the exterior power $\bigwedge^k V$ as the vector space such that any multilinear, alternating map $V^k \to W$ factors through $\bigwedge^k V$. The elements of $\bigwedge^k V$ are written as products $v_1 \wedge \dots \wedge v_k$ of vectors $v_1, \dots, v_k$. The idea is that the element $v_1 \wedge \dots \wedge v_k$ somehow represents the "volume" of the parallelepiped formed by the vectors $v_1, \dots, v_k$, taken after projecting onto some $k$-dimensional subspace of $V$. The upshot: if we take $n = \dim V$, then $\bigwedge^n V$ is the vector space consisting of products $v_1 \wedge \dots \wedge v_n$, where this product represents the volume of the associated parallelepiped. We can then define the determinant of a linear transformation $T$ as the unique scalar such that $$ Tv_1 \wedge \dots \wedge Tv_n = (\det T)(v_1 \wedge \dots \wedge v_n) $$ for all vectors $v_i$. See this answer for more details.

The above discussion is purely algebraic, and the only idea we have really used is that the determinant is the unique multilinear and alternating map on matrices (up to scaling). But the exterior algebra is somehow the "nicest" context in which to think about these things. For example, the multiplicativity of the determinant is pretty much immediate: $$ ABv_1 \wedge \dots \wedge ABv_n = (\det A)(Bv_1 \wedge \dots \wedge Bv_n) = (\det A)(\det B)(v_1 \wedge \dots \wedge v_n). $$ All the other properties above are just as easy to prove, which isn't surprising because the elements of $\bigwedge^n V$ can be thought of as volumes. In fact, the volume operation is the unique alternating and multilinear map on matrices (up to scaling), which is why there is a purely algebraic way of encoding it.

If you haven't seen exterior powers before, the previous two paragraphs may not have made much sense. I haven't read it, but I believe a good book that takes this approach is Linear Algebra via Exterior Products by Sergei Winitzki.

Frank
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