[Math] Proof that $\text{det}(AB) = \text{det}(A)\text{det}(B)$ without explicit expression for $\text{det}$

determinanteigenvalues-eigenvectorslinear algebramatricesmatrix decomposition

Overview

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.

Best Answer

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 $$ \det(AB)=\widetilde{\det}(AB)=\widetilde{\det}(A)\widetilde{\det}(B)=\det(A)\det(B).$$

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