I am seeking an approach to linear algebra along the lines of Down with the determinant! by Sheldon Axler. I am following his textbook Linear Algebra Done Right. In these references the author takes an approach to linear algebra that avoids using the explicit expression for determinants as much as possible. The motivation for this is that the the explicit expressions for the determinant, while useful for calculating things, are hard to glean intuition from. This question is not about debating the merits of this approach however!

I use Axler's terminology which is a little different from what I am used to. What he calls an isometry I would call unitary and what he calls positive I would call positive semi-definite.

Further Motivation

I would like to provide further information for why I am seeking this proof. I know that in highly general settings (differential geometry for example) it is possible to perform integrals on manifolds using differential $N$-forms from $\Lambda^N$. The essential feature of these objects is that they are alternating and multilinear. I want to understand why there is a deep relationship between alternating and multilinear forms. The suggestion is that this relationship occurs because volumes can be characterized, fundamnetally, as something that transforms in an alternating and multilinear way.

Unfortunately I don't find this to be a compelling statement about volumes. See my searching here. Help proving that signed volume of n-parallelepiped is multilinear. Rather, the fact that volumes transform in a multilinear and alternating way is actually hard to prove directly and seems to be a little bit particular to the properties of parallelepipeds specifically. And then of course we can tile $\mathbb{R}^N$ using parallelepipeds so the general reesult follows.

What seems more intuitive to me is to "fundamentally" characterize the signed volume as something scales linearly with coordinate axis scalings (multiplication by a diagonal matrix) and either stays the same or flips sign under a proper or improper rotation (multiplication by a unitary matrix with positive or negative product of eigenvalues).

These two properties SEEM like they can proven by only looking at the eigenvalues of transformation matrices but a proof does not seem evident without relying in some way on multilinear/alternating functions. This would imply that the multilinear/alternating characterization of volumes is somehow more fundamental than the one I am proposing.

This is just very surprising to me, especially given how tricky it is to prove directly for volumes... Hence my search for a proof about the product of eigenvalues of a product of matrices.

Problem Setup

My question:

Every complex $N\times N$ matrix $A$ has $N$ eigenvalues $\lambda_1, \ldots, \lambda_N$. Some of these eigenvalues may be repeated and some of them may be equal to zero. Define the determinant of $A$ as

$$ \text{det}(A) = \prod_{i=1}^N\lambda_i $$

It is well known that for two $N\times N$ matrices $A$ and $B$ that

$$ \text{det}(AB) = \text{det}(A)\text{det}(B) $$

However, this proof typically (see below) relies on knowing an explicit expression for $\text{det}$ in terms of the matrix elements of $A$ and $B$ such as

$$ \text{det}(A) = \sum_{\sigma \in S_N}\prod_{i=1}^N A_{i, \sigma(i)} = \sum_{i_1, \ldots, i_N=1}^N \epsilon_{i_1\ldots i_N} \prod_{j=1}^N A_{j, i_j}. $$

I am curious if there is a proof of $\text{det}(AB) = \text{det}(A)\text{det}(B)$ using the definition I give above and which does not rely on these explicit alternating summations, e.g. in the vein of the Axler approach.

One Attempt at a Solution

In his textbook, Axler proves a few key theorems, all without using the alternating expressions above. These include facts about eigenvalues of matrices and various spectral and singular value decomposition theoerems. I think polar decomposition may be useful here. Something like:

\begin{align} A =& S_A P_A\\ B =& P_B S_B\\ AB =& S_A P_A P_B S_B \end{align}

With $S_{A,B}$ isometries (unitary) and $P_A$ and $P_B$ are positive (positive semi-definite) matrices.

With two facts the proof would be complete: If we could prove $\text{det}(PS) = \text{det}(PS) = \text{det}(S)\text{det}(P)$ for isometry $S$ and positive $P$ (or arbitrary $P$) then we would have

$$ \text{det}(AB) = \text{det}(S_A)\text{det}(S_B)\text{det}(P_A P_B) $$

If we could then also show $\text{det}(P_A P_B) = \text{det}(P_A)\text{det}(P_B)$ then the proof would be complete.

Other Attempts at a Solution

Some other useful things I've learned: It is clear that multiplying a matrix by a unitary matrix does not change the singular values. Also, if I could prove that the product of eigenvalues or singular values of $C = \Sigma_1 U \Sigma_2$ with $U$ unitary and $\Sigma_{1, 2}$ diagonal and positive, is equal to the product of the product of eigenvalues/singular values of $\Sigma_1$ and $\Sigma_2$ I would be able to show that multiplying $A$ and $B$ arbitrary results in multiplying the product of their singular values. The $\text{det(A)}$ could then alternately be defined as something like $A = U \Sigma V^*$ and $\text{det}(A)$ is equal to the product of diagonal elements on $\Sigma$ times the product of eigenvalues of $U$ and $V$.

Finally another approach. All of my approaches so far have basically relied on properties of matrices and matrix manipulations. It might be that this apporach is misguided. Eigenvalues are related to the characteristic polynomial of a matrix (which Axler defines without needing the determinant) and the fundamental theorem of algebra. Perhaps one needs to return to this algebraic domain to get the answer to what I am looking for. I'm pretty inexperienced in that domain, so any ideas would be greatly helpful.

Explicit Questions

My questions are:

  • Can someone provide a proof for $\text{det}(PS)=\text{det}(SP) = \text{det}(S)\text{det}(P)$ for isometry $S$ and positive (or arbitary) $P$?
  • Can someone provide a proof for $\text{det}(P_A P_B) = \text{det}(P_A)\text{det}(P_B)$ for positive $P_A$ and $P_B$?
  • Let $C = \Sigma_1 U \Sigma_2$ with $\Sigma_{1, 2}$ diagonal (with positive non-zero entries) and $U$ unitary. Let $\Pi(A)$ equal the product of the singular values of $A$. Knowing that $\Pi(UA)=\Pi(A)$ for unitary $U$, Can someone provide a proof that $\Pi(C) = \Pi(\Sigma_1)\Pi(\Sigma_2)$?
  • Alternatively, maybe there's an entirely different approach to proving this without the calculational alternating formula?
  • If someone has reason to believe what I am asking for is impossible for some reason I would appreciate comments or answers explaining why it might be impossible.

Proofs involving explicit alternating expressions for $\text{det}$

  • This website builds up the fact that $\text{det}(A)\text{det}(B)$ by expanding $A$ and $B$ into elementary matrices. It then proves that multiplying a matrix by an elementary matrix multiplies the determinant by a known value. The result follows. The issue is that the effect of multiplying by elementary matrix on the determinant is proven using an expansion by minors definition or property of the determinant. https://sharmaeklavya2.github.io/theoremdep/nodes/linear-algebra/matrices/determinants/elementary-rowop.html

  • Pretty much all the proofs on this MSE use some alternating formula or another How to show that $\det(AB) =\det(A) \det(B)$?

  • Wikipedia and Hubbard and Hubbard give a nice proof which (1) uses knowledge that the determinant is the unique alternating multilinear normalized $N$-form on the columns of a matrix shows that the function $D(A) = \text{det}(AB)$ is alternating and multilinear so that $D(A)$ must be a multiple of $\text{det}(A)$. We also have $D(I) = \text{det}(B)$ so $D(A) =\text{det}(AB) = \text{det}(A)\text{det}(B)$. This proof is nice in that it doesn't rely on the explicit alternating expression for $\text{det}$, but it has the problem that I don't know how to prove the product of eigenvalues satisfies the alternating multilinear property without providing the explicit alternating expression for $\text{det}$. Other example of this sort of proof can be found on MSE such as Determinant of matrix product

  • Here is a proof that uses an identity on the levi-civita symbol: proving the determinant of a product of matrices is the product of their determinants in suffix / index notation

Recent Insights

After thinking more about this problem I think I am becoming convinced that what I am asking for is not possible, but I'm realizing a more convincing way to think about thing. There are three related concepts. (1) How a linear transformation scales volumes, $\text{svol}$ and $\text{vol}$, (2) The product of eigenvalues or singular values of a linear transformation, $\Pi_e$ and $\Pi_s$, and (3) the alternating multilinear form, $\text{det}$ and $|\text{det}|$.

Above, I have been hoping to give a treatment in which $\Pi_e$ is somehow the most fundamental property and to derive all properties about $\text{svol}$ and $\text{det}$ from the properties of eigenvalues. As I said above, it's starting to look like this may not be possible.

Instead I think a more satisfactory approach may be the following. Take $\text{vol}$ or $\text{svol}$ to be fundamental and work from there. The interesting insight I've had is that there are two ways to break down the properties of $\text{vol}$ or under linear transformations.

In the first approach, we can see that (a1) when scaling by a diagonal matrix $\text{vol}$ scales by the absolute value of the product of the entries and $\text{svol}$ scales by the product, (b1) when multiplying by a reflection on a single axis that $\text{vol}$ is unchanged while $\text{svol}$ flips sign, and (c1) both $\text{vol}$ and $\text{svol}$ are unchanged by proper rotations.

In the second approach we have that (a2) when scaling by a diagonal matrix $\text{vol}$ scales by the absolute value of the product of the entries and $\text{svol}$ scales by the product, (b2) when two axes are swapped $\text{vol}$ is unchanged while $\text{svol}$ flips sign and (c2) $\text{vol}$ and $\text{svol}$ are both unchanged by shear operations.

The first approach lends itself to a polar decomposition of the form

$$ A = RUP $$

Where $R$ is either the identity or a reflection across a single axis, $U$ is a special orthogonal matrix, and $P$ is positive-semi-definite. This decomposition lends itself to show the relationship between the volume scaling properties of a transformation and the product of singular values or eigenvalues.

The second approach lends itself do a decomposition of the form

$$ A = E_1\ldots E_k $$

Where $E_i$ are elementary matrices. This decomposition lends itself to show the relationship between the volume transforming properties of a transformation and the alternating multilinear form.

One can see that properties (a1), (a2) are the same, (b1) and (b2) are essentially the same, and that (c1) and (c2) are similar and equivalent when considered in light of the other properties. So the first approach focuses on rotations while the second focuses on shears.

Considering positive definite matrices for the moment, I think one way to think about the relationship between the product of eigenvalues, $\Pi_e$ and the alternating multilinear form $\text{det}$ is that they are equivalent because (1) $\Pi_e$ is related to rotations and axis stretching of volumes and (2) $\text{det}$ is related to shears and axis stretching of volumes.

It is still a little bit unsatisfying that we would need to make this relationship with volumes and shears to explain why $\Pi_e(AB) = \Pi_e(A) \Pi_e(B)$, but one reason I could see for it is that $\Pi_e$ is a very complicated object algebraically. The existence of eigenvalues is only guaranteed by the very non-constructive fundamental theorem of algebra and triangularizability of complex matrices also feels a bit abstract. We know how to work with a transformation and a single eigenvalue or eigenvector, but there is not much machinery to work with ALL the eigenvalues at once, i.e. $\Pi_e$. Rather, we can use the fact that the product of eigenvalues stretch volumes, and the fact that signed volume are alternating and multilinear, to get at an explicit expression for the product of eigenvalues that allows us to derive properties we wouldn't be able to otherwise, such as $\Pi_e(AB) = \Pi_e(A)\Pi_e(B)$.

Curious for others' thoughts on this. I may be asking (and answering perhaps) a new question related to this last point. I have some partial proofs and will hopefully have a full, hopefully intuitively satisfying, explanation a bit later but not sure.

  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Xander Henderson
    Dec 8, 2021 at 19:28
  • $\begingroup$ The main problem with defining det as the product of eigenvalues is perhaps that the product of two nilpotent endomorphisms need not be nilpotent. $\endgroup$ Dec 9, 2021 at 7:09
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    $\begingroup$ @SayanDutta I want a proof where we don't even touch the formula for determinants. If possible, I'd like a proof that the product of eigenvalues of $AB$ is equal to the product of eigenvalues of $A$ times the product of eigenvalues of $B$ which does not rely on the fact that product of eigenvalues of $A$ is multilinear and alternating. $\endgroup$
    – Jagerber48
    Dec 9, 2021 at 14:43
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    $\begingroup$ @Desura the algebraic multiplicities are the dimensions of the generalized eigenspaces. $\endgroup$ Dec 11, 2021 at 15:49
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    $\begingroup$ @Jagerber48 RE your update: This isn't really an issue since we can just pass to the algebraic closure, but you might want to contemplate that your proposed definition is at least a bit weird because eigenvalues of matrices over $\mathbb{k}$ need not lie in $\mathbb{k}$ itself. $\endgroup$ Dec 13, 2021 at 19:20

6 Answers 6


Disclaimer: I am cheating a little bit here because

  1. In some sense, I just proved that two different definitions of the determinant are equal (namely the one that the OP gave and the one using multilinear forms).

  2. One would still have to prove that the determinant as defined by the OP is continuous (which must be true). Perhaps one can use that the characteristic polynomial varies continuously as you move continuously through $\text{End}(V)$.

Let $\Lambda^n(V)$ denote the space of alternating $n$-multilinear transformations. For each $T\in\text{End}(V)$ there is a pull-back map $T^*:\Lambda^n(V)\rightarrow\Lambda^n(V)$ defined by $$ T^*(\omega)(v_1,\ldots,v_n)=\omega(T(v_1),\ldots,T(v_n)) ,$$ for each $\omega\in\Lambda^n(V)$ and $v_1,\ldots,v_n\in V$. Since $\Lambda^n(V)$ is $1$-dimensional, one can define $$\widetilde{\det}:\text{End}(V)\rightarrow \mathbb{C}$$ by setting $$ T^*(\omega)=\widetilde{\det}(T)\omega,$$ for each $\omega\in\Lambda^n(V)$. Then, by @hm2020's answer, it is clear that $\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)$ (it follows from the definition at once). Now, for each diagonalizable $A\in\text{End}(V)$, take a basis of eigenvectors $\{e_i\}_{i=1}^n$ with eigenvalues $\{\lambda_i\}_{i=1}^n$ and $n$ vectors $v_k=\displaystyle\sum_{i=1}^n v^i_ke_i\in V$, $k\in\{1,\ldots,n\}$. It follows that \begin{equation*}\begin{split}T^*(\omega)(v_1,\ldots,v_n)&=\omega(T(v_1),\ldots,T(v_n))\\&=\omega(\sum v^i_1\lambda_ie_i,\ldots,\sum v^i_n\lambda_ie_i)\\ &=\left(\displaystyle\prod_{i=1}^n\lambda_i\right)(\omega(v_1,\ldots,v_n))\\ &=\det(T)\omega(v_1,\ldots,v_n), \end{split}\end{equation*} where we used that $\omega$ is alternating. It follows that $\det=\widetilde{\det}$ on the set of diagonalizable matrices, which is dense in $\text{End}(V)$. Since both of them agree on a dense set and are continuous (and the codomain is Hausdorff), we have $\det(T)=\widetilde{\det}(T)$ for every $T\in\text{End}(V)$. In particular, for every $A,B\in\text{End}(V)$, there holds $$ \det(AB)=\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)=\det(A)\det(B).$$

  • 3
    $\begingroup$ @Jagerber48 I understand your point. However, I would definitely say that the concept of signed volume is deeply connected to the idea of alternating $n$-multilinear functions. In some sense, the properties in the definition of such an element are exactly the ones (and perhaps are the CRUCIAL ones) you would expect from a function assigning signed volumes. I mean, what do you integrate on manifolds to find volume? Differential forms! (which are just sections of alternating $n$-multilinear maps). $\endgroup$
    – amnesiac
    Dec 9, 2021 at 14:58
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    $\begingroup$ It EXACTLY doesn't satisfy me to say volumes are related to differential $n$-forms because those are what we integrate to find volumes. Rather, it feels to me (or I am trying to answer) that the REASON we integrate $n$-forms to find volume is because there is some, to be understood, relationship between alternating multilinear maps and volumes. It can be proven that a normalized alternating multilinear map is proportional to the product of eigenvalues of a map. $\endgroup$
    – Jagerber48
    Dec 9, 2021 at 15:07
  • 1
    $\begingroup$ Your doubt is noted. Indeed it has hard to impossible to find a proof without relying on alternating multilinear maps. But you say "This is exactly what you would want from a function $\text{svol}(v_1, \ldots, v_n)$..." Why is this obvious? In my answers I take a messy page just calculating the unsigned volume of a parallelepiped, I then need to inductively define the $n$-volume from the $(n-1)$-volume to bootstrap Cavalieri's principle and then somehow (I don't know exactly how yet) I need to figure out how to define the sign of the volume and finally prove multilinearity. $\endgroup$
    – Jagerber48
    Dec 9, 2021 at 15:36
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    $\begingroup$ I can see that volumes transform in a multilinear way, but it is not obvious to me, from first principles, that we should have always expected them to be. $\endgroup$
    – Jagerber48
    Dec 9, 2021 at 15:37
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    $\begingroup$ @Jagerber48 For what it's worth, defining determinants as the product of eigenvalues doesn't really fit my intuition of volume either, because you would then have to make sense of what "stretching by $i$" means in the case of, say, $\begin{pmatrix}0 & 1 \\ -1 & 0\end{pmatrix}$ and I think an answer to that, while entirely possible, could feel just as unsatisfying as explaining it through the multilinearity (additivity, really) of parallelepiped volumes. $\endgroup$
    – Elliot Yu
    Dec 9, 2021 at 16:32

Question: Proof that $det(AB)=det(A)det(B)$ without explicit expression for $det$.

Answer: A "Bourbaki-proof": You may use the exterior product to prove the formula for any two matrices $A,B$ (of the same rank) over any commutative unital ring:

Example: Let $k$ be any commutative unital ring. If $\phi$ is the matrix

\begin{align*} \phi= \begin{pmatrix} a & b \\ c & d \end{pmatrix} \end{align*}

and you view $\phi:V:=k\{e_1,e_2\} \rightarrow k\{e_1,e_2\}$ as a $k$-linear endomorphism, it follows by the definition of the exterior product that you get an endomorphism

$$\wedge^2(\phi): \wedge^2 V \rightarrow \wedge^2 V$$

defined by

$$\wedge^2 \phi(e_1\wedge e_2)=\phi(e_1)\wedge \phi(e_2)= (ae_1+ce_2)\wedge (be_1+de_2)=(ad-bd)e_1 \wedge e_2:= $$

$$det(\phi)e_1 \wedge e_2.$$

Here we have used the property that $e_2 \wedge e_1=-e_1 \wedge e_2$ in the exterior product $\wedge^2 V$. More generally if $\phi$ is multiplication with $A:=(a_{ij})$ - an $n\times n$-matrix with coefficients in $k$, and $V:=k\{e_1,..,e_n\}$ with $\phi: V \rightarrow V$ it follows we get the formula

$$\wedge^n \phi(e_1\wedge \cdots \wedge e_n):=\phi(e_1)\wedge \cdots \wedge \phi(e_n)=det(A)e_1\wedge \cdots \wedge e_n.$$

Here we define $det(A)$ as in your thread.

Principle: Taking the $n$'th exterior power of an $n \times n$-matrix $A$ is an "abstract" construction of the determinant $det(A)$.

Example: Let $E:=k\{e_1,..,e_n\}$ be a free $k$-module of rank $n$ and let $\def\End{\operatorname{End}} A,B\in \End_R(E)$ be $n \times n$-matrices with coefficients in $k$ ($k$ any commutative unital ring). It follows $$\wedge^n A, \wedge^n B \in \End_R(\wedge^n E)$$ are $k$-linear endomorphisms of the rank one free $k$-module $\wedge^n E \cong k e_1\wedge \cdots \wedge e_n:=k e$ with $e:=e_1\wedge \cdots \wedge e_n$. It has the property that

$$\wedge^n A(ue) = \det(A) ue$$

is "multiplication with the element $\det(A)\in k$".

From this it follows since $\wedge^n$ is a functor (see the below link) that

$$ \det(AB) := \wedge^n (A \circ B) = \wedge^n A \circ \wedge^n B := \det(A) \det(B). $$

With the formulation using the exterior product, it follows $\wedge^n(\phi)(u)=det(A)u$ is a "theorem". Here we define (as in your thread)

$$D1.\text{ }det(A) = \sum_{\sigma \in S_N}\prod_{i=1}^N a_{i, \sigma(i)}.$$

Hence if we define the determinant using the exterior product, it follows formula D1 must be proved. Conversely if we define the determinant using D1, it follows the multiplicativity of the determinant must be proved.

Prove that $\det(T) = \det(T_{w_{1}})\det(T_{w_{1}})\cdots \det(T_{w_{k}})$ where $W_{i}$ are $T$-invariant subspaces of $V$ - Trouble at last step.

  • 2
    $\begingroup$ This proof uses a little higher math than I'm readily familiar with. How are you defining $\text{det}(A)$? Is it connected with the product of eigenvalues of a matrix? I am essentially seeking a proof that the product of eigenvalues of $AB$ is the product of the product of eigenvalues of $A$ and the product of eigenvalues of $B$. I can't tell if your proof shows that. $\endgroup$
    – Jagerber48
    Dec 8, 2021 at 19:31
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    $\begingroup$ Is this any more general than the previous answers that seem to be almost duplicates? The condition that $A$ and $B$ can be simultaneously diagonalized is pretty limiting. $\endgroup$
    – robjohn
    Dec 8, 2021 at 20:43
  • 1
    $\begingroup$ It's not true at all that this proof requires the matrix to be diagonalized (simultaneously or not). This is the conceptually correct general proof. Now it is true that it uses exterior powers, which can be a little high-level for beginners (but they are not that scary either). $\endgroup$ Dec 9, 2021 at 11:11
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    $\begingroup$ @CaptainLama If you define the determinant as the OP did (product of eigenvalues), then the determinant in this answer could have been called another thing (say, $\widetilde{\det}$). Then, this answer proves that $\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)$. Now it remains to establish $\det=\widetilde{\det}$. $\endgroup$
    – amnesiac
    Dec 9, 2021 at 11:46
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    $\begingroup$ @hm2020 I seek a proof that the product of the eigenvalues of $AB$ is the product of the product of eigenvalues of $A$ and the product of eigenvalues of $B$ which does not rely on the fact that the determinant (or product of eigenvalues) is an alternating and multilinear function. Your approach relies on the determinant being a member of the exterior algebra which means it implicitly uses the fact that it is alternating and multilinear. $\endgroup$
    – Jagerber48
    Dec 13, 2021 at 17:35

The gist of it

Suppose $A,\,B$ act on an $n$-dimensional vector space. Let $X_A$ denote the region enclosed by a hypercuboid spanned by eigenvectors of $A$. Its hypervolume is $\int_{X_A}\mathrm d^nx>0$. Since $\det A$ is the product of $A$'s eigenvalues, $A$ transforms $X_A$ to a region whose oriented hypervolume, now possibly negative, can be expressed in two ways:$$\det A\int_{X_A}\mathrm d^nx=\int_{X_A}\mathrm d^nAx.$$Since $x\mapsto Ax$ is a linear coordinate transformation, its Jacobian is a constant; of course, that constant must be $\det A$ for the above equation to work. We now know $\mathrm d^nAx=\det A\mathrm d^nx$ works in general on the space, not just in the above problem. Indeed, the eigenvectors of $A$ are irrelevant. So now consider what happens if you linearly transform twice, once with $A$, once with $B$:$$\det(AB)\mathrm d^nx=\mathrm d^nABx=\det A\mathrm d^nBx=\det A\det B\mathrm d^nx\implies\det(AB)=\det A\det B.$$In particular, this calculation doesn't require $A,\,B$ to be simultaneously diagonalizable, because it is simply a comparison of $d^nx$ coefficients in an arbitrary small region of the vector space.

The footnote

The reader may as an exercise respond to their own objections of "doesn't this require the eigenvalues to be a complete basis, the matrix to be invertible or whatever?" with a fleshed-out version of "we've at least proven $\det(AB)=\det A\det B$ for enough of a special case that, since we know now determinants are Jacobians and hence degree-$n$ polynomials in $n^2$ matrix entries, the known solution set is too dense in the general case for the latter to not also hold".

  • $\begingroup$ How is $d^n Ax$ defined? Perhaps relatedly how is $d^n x$ defined? I think this answer is pretty similar to the one I gave that address singular values rather than eigenvalues. Basically: the volume of a single transformation is equal to product of eigenvalues of that transformation, and volumes are multiplicative over transformations so products of eigenvalues must be multiplicative over transformations. $\endgroup$
    – Jagerber48
    Dec 13, 2021 at 19:58
  • $\begingroup$ And your footnote says: I showed $\text{det}(AB) = \text{det}(A)\text{det}(B)$ for $n\times n$ matrices which have $n$ eigenvectors (basically non-defective matrices) and non-defective matrices are dense in all matrices, so by continuity of $\text{det}$ we have that $\text{det}(AB) = \text{det}(A)\text{det}(B)$ holds for all matrices? Is this correct? $\endgroup$
    – Jagerber48
    Dec 13, 2021 at 20:02
  • $\begingroup$ @Jagerber48 It's not even about using continuity; it's more coefficient matching in the polynomials. To your other question, $\mathrm d^nx:=\prod_{i=1}^n\mathrm dx_i$. $\endgroup$
    – J.G.
    Dec 13, 2021 at 21:07

We can start by defining the determinant as the only n-linear alternating map such that $\det(I) = \det(\vec e_1, \cdots, \vec e_n) = 1$.

Under this scenario, you already pointed out that it is easy to prove: $\det(AB) = \det(A) \det(B)$. Thus, I'll assume this is known, instead of also proving this as well.

If I understand correctly what you are asking, if we manage to prove that $\displaystyle\det(A) = \prod_{i=1}^n \lambda_i$, the product of the eigenvalues, then we are done.

Proof: The determinant of the inverse, is the inverse of the determinant. $$ AA^{-1} = I\quad\implies\quad \det(A)\det(A^{-1}) = \det(I) = 1 $$

Proof: The determinant of a diagonal matrix is the product of the diagonal elements. $$ \det(diag(a_1, a_2,\cdots a_n)) = \det(a_1\vec e_1,\cdots,a_n \vec e_n) = a_1a_2\cdots a_n\det(\vec e_1,\cdots,\vec e_n) = a_1a_2\cdots a_n $$

Proof: The determinant of a triangular matrix, is the product of the diagonal elements.

$$\det(a_{11}\vec e_1, \vec v_2, \vec v_3,\cdots\vec v_n) = a_{11}\det(\vec e_1, \vec v_2, \vec v_3,\cdots\vec v_n) = a_{11}\det(\vec v_2', \vec v_3',\cdots\vec v_n')$$

$$ \left|\begin{matrix} a_{11} & a_{12} & \cdots & a_{1n} \\ 0 & a_{22} & \cdots & a_{2n} \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & a_{nn} \\ \end{matrix}\right| = a_{11} \left|\begin{matrix} a_{22} & \cdots & a_{2n} \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a_{nn} \\ \end{matrix}\right| = a_{11} a_{22} \left|\begin{matrix} a_{33} & \cdots & a_{3n} \\ \vdots & \ddots & \vdots \\ 0 & \cdots & a_{nn} \\ \end{matrix}\right| = a_{11}a_{22}\cdots a_{nn} $$

Proof: The determinant of a matrix $M$, is the product of its eigenvalues. $$M = JDJ^{-1}\implies\det(M) = \det(J)\det(D)\det(J^{-1}) = \det(D)$$

If $M$ is diagonalizable, then $D = diag(\lambda_1,\cdots\lambda_n)$, and we are done.

If $M$ is not diagonalizale, then, $D$ is in Jordan-Canonical form, an upper triangular matrix where the diagonal elements are the eigenvalues of $M$.

Thus, for both cases: $\det M = \det D = \lambda_1\lambda_2\cdots\lambda_n$ the product of its eigenvalues.


  • $\begingroup$ Doesn't the upper-triangular case fully subsume the diagonal case, and hence all references to the diagonal case can just be deleted? $\endgroup$ Dec 12, 2021 at 12:55
  • $\begingroup$ Yes. Though I still like the diagonal case because it is very easy to see. =]. $\endgroup$ Dec 12, 2021 at 13:11
  • $\begingroup$ Thanks for this answer, but I'm hoping for an answer that doesn't rely on the connection between the product of eigenvalues and an alternating multilinear map. $\endgroup$
    – Jagerber48
    Dec 13, 2021 at 14:46

Quick sketch of an answer:

We define the volume of a set $S\subset \mathbb{R}^n$ by

$$ \text{vol}(S) = \int_{\mathbb{R}^n} 1_S $$

Where $1_S$ is the indicator function on $S$. $1_S(\boldsymbol{x}) =1$ for $\boldsymbol{x}\in S$ and 0 otherwise. We can take this to be the Riemann integral.

If $T$ is a linear transformation we define

$$ TS = \{T\boldsymbol{x}:\boldsymbol{x} \in S\} $$

It can be proven, non-trivially, that

$$ \text{vol}(TS) = \text{vol}(TQ)\text{vol}(S) $$

Where $Q$ is the unit cube. Define $\text{vol}(T) = \text{vol}(TQ)$. The basic idea is that $\int 1_S$ is approximated by covering $S$ by cubes, If we act $T$ on $S$ then each of the cubes volumes gets scaled by $\text{vol}(T)$ so the overall volume of $S$ does as well.

Consider set $S$ and linear transformation $A$ and $B$. $$ \text{vol}(ABS) = \text{vol}(AB)\text{vol}(S) = \text{vol}(A)\text{vol}(BS) =\text{vol}(A)\text{vol}(B)\text{vol}(S) $$ So that $$ \text{vol}(AB) = \text{vol}(A)\text{vol}(B) $$ The proof essentially relies on the fact that the transformation $B$ scales the volume of the cubelets that cover $S$ by $\text{vol}(B)$, and then the volume of those transformed cubelets can be covered by more cubelets which are transformed by $A$ by $\text{vol}(A)$. So essentially we rely on the possibility to cover volumes with cubes to prove $\text{vol}(AB) = \text{vol}(A)\text{vol}(B)$.

We can then proceed a couple of ways to make connection with the products of eigenvalues. I'll use the singular value decomposition to make connection with singular values. First unitary matrices.

A unitary matrix $U$ is any matrix that preserves the norm of vectors:

$$ ||U\boldsymbol{x}|| = ||\boldsymbol{x}|| $$

Unitary matrices are invertible and satisfy $U^{-1} = U^{\dagger}$. Define the unit ball in $\mathbb{R}^n$ as $B_n = \{\boldsymbol{x}\in\mathbb{R}^n:\boldsymbol{x}\le 1\}$. It can be proven that $UB_n = B_n$. That is unitary matrices map the unit ball to itself. We then have

$$ \text{vol}(UB_n) = \text{vol}(U)\text{vol}(B_n) = \text{vol}(B_n) $$

So $\text{vol}(U) = 1$. Unitary transformations don't change volumes. (this is to be expected since they don't change norms of vectors).

We then have that any matrix can be decomposed using a singular value decomposition:

$$ A = W\Sigma V^{\dagger} $$

With $W$ and $V$ unitary and $\Sigma$ diagonal with non-negative numbers on the diagonal. We then have

$$ \text{vol}(A) = \text{vol}(W) \text{vol}(\Sigma) \text{vol}(V^{\dagger}) = \text{vol}(\Sigma) $$

Since $W$ and $V^{\dagger}$ are unitary. It can be shown that

$$ \text{vol}(\Sigma) = \prod_{i=1}^n \sigma_i $$

Where $\sigma_i$ are the singular values of $A$.

Then, let $C = AB$. We then have

$$ \text{vol}(C) = \prod_{i=1}^n \sigma_{C,i} = \text{vol}(A)\text{vol}(B) = \prod_{i=1}^n \sigma_{A,i} \prod_{i=1}^n \sigma_{B, i} $$

So I have proven that the product of the singular values of $AB$ is the product of the product of singular values of $A$ and the product of singular values of $B$.

I have not yet shown the desired relationship for eigenvalues. To do so would require a good definition of the signed volume of a set. For this we need to specify both a set as well as a basis. The set determines $\text{vol}(S)$ and the basis determines the sign of the signed volume. I believe the definition can involve whether the unitary matrices are special or not.

I think this answer is a good step in the right direction. I will note that I have not been able to prove $\prod_{i=1}^n \sigma_{AB,i} = \prod_{i=1}^n \sigma_{A,i} \prod_{i=1}^n \sigma_{B, i}$ directly using matrix mechanics alone. Rather, I had to foray into analysis where I could cover volumes with nested sets of cubes. It may be the case that this property is just too abstract to prove using matrix/transformation algebra alone.

I think for eigenvalues the proof will involve the facts that (1) any complex matrix can be triangularized, (2) that the eigenvalues of the matrix are the diagonal elements of the similar upper triangular matrix and (3) the upper triangular matrix transforms a cube into a parallelepiped.

  • $\begingroup$ This approach only works for real matrices. $T$ must be a real matrix if $T\boldsymbol{x}$ is to be in $\mathbb{R}^n$. $\endgroup$
    – Jagerber48
    Dec 14, 2021 at 8:12

This is in response to: "A proof that $\det AB=\det A\cdot\det B$ that does not involve any explicit formulae." I admit that no geometric eigenvalue intuition is appealed to, but to me it is already geometrically intuitive that $\det A\cdot\det B=\det AB$, and as far as a formal proof goes this is the simplest way to go about it of which I am aware. The meat of the proof is hidden in the powerful definition I use; seeing that this definition is indeed the determinant requires more work, some of which is shown here.

To motivate this a bit more: As far as I know, the historical motivation for the determinant was not in geometry, but in determining the solvability of linear systems, i.e. their linear independence. From this, it is clear that the alternating property is essential. The multilinear rule is to extend alternation to cover cases where a third equation is some linear combination of two others; geometrically I suppose you could say we want the determinant to be zero when one of the sides of the parallelotope falls on the others (its dimension is "squished"). The beautiful thing is that alternation and multilinearity are sufficient to completely characterise the determinant as done below, if one also demands that the determinant is non-trivial, i.e. the identity matrix has non-zero determinant, which is geometrically and logically and obvious thing to demand.

I will simply use this characterisation of the determinant:

Let $\Bbb K$ be the base field, and $V$ an $n$-dimensional vector space over $\Bbb K$. Let also $\{e_1,e_2,\cdots,e_n\}$ be a basis for $V$. Let $T$ be a linear map $T:V\to V$, and let $\omega:V^n\to\Bbb K$ be any alternating multilinear form such that $\omega(e_1,e_2,\cdots,e_n)\neq0$. Then: $$\det T\overset{\triangle}=\frac{\omega(Te_1,Te_2,\cdots,Te_n)}{\omega(e_1,e_2,\cdots,e_n)}$$

That this is well defined, independent of the form $\omega$ follows from the uniqueness (up to a constant) property of such forms. That this is "actually" the determinant follows from the explicit Leibniz formula for the determinant, however the proof never uses an explicit formula.

Now: Let the $A$ represent such a linear map, and let $B$ represent another such linear map. If $B$ is singular, so too is $AB$ and trivially $\det AB=0=\det A\cdot\det B$. Then assume now that $B$ is non-singular, so that we may divide by $\omega(Be_1,\cdots,Be_n)$ without trouble. Important is the observation that if $B$ is non-singular, and $\{e_i\}$ is a basis, then $B\left(\{e_i\}\right)$ is also a basis for $V$.

Using the above definition:

$$\begin{align}\det(A\circ B)&=\frac{\omega((A\circ B)e_1,\cdots,(A\circ B)e_n)}{\omega(e_1,\cdots,e_n)}\\&=\frac{\omega((A\circ B)e_1,\cdots,(A\circ B)e_n)}{\omega(Be_1,\cdots,Be_n)}\cdot\frac{\omega(Be_1,\cdots,Be_n)}{\omega(e_1,\cdots,e_n)}\\&=\frac{\omega(A\color{red}{(Be_1)},\cdots,A\color{red}{(Be_n)}}{\omega(\color{red}{Be_1},\cdots,\color{red}{Be_n})}\cdot\det B\\&=\det A\cdot\det B\end{align}$$

The red colouring is to emphasise that $\{Be_i\}$ forms a basis, and in the definition of determinant used above the basis used may be arbitary. Geometrically, this is intuitive since the volume of a transformation should not change even if the coordinate system changes.


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