Determinant: Alternative Definitions

Question

Reference

Problem

Given a vector space $V$.

Consider an endomorphism $T:V\to V$.

The rank of an endomorphism: $$\mathrm{rank}T:=\dim\left(\mathrm{im}T\right)$$ The determinant of an endomorphism: $$\det T:=\text{???}$$

What would be a nice definition not relying on representations by matrices?

(I assume basic knowledge of Differential Geometry and Functional Analysis.)

If you're comfortable with group theory and the classical groups, I believe that the multiplicative group of nonzero scalars is (isomorphic to) the abelianization of $GL(V)$, the group of invertible endomorphisms. The map from $GL(V)$ to its abelianization is canonical, and the isomorphism with the nonzero scalars is the determinant. (there's probably a natural way to single out the determinant specifically rather than any of the other possible isomorphisms...) — , Dec 19 '14 at 04:21
@Freeze_S: Please consider using more informative titles. In particular, for this question I performed an edit which changed the title to What is a nice definition of the determinant not relying on representations by matrices? I'm not sure why you decided to rollback to the original title. The importance of a descriptive title has been discussed on meta many times, for example here. — Michael Albanese, Dec 19 '14 at 14:36
@MichaelAlbanese: I hope this is a fair compromise. (I try to keep my questions organized, so I prefer titles in bullet point form.) — C-star-W-star, Dec 19 '14 at 14:42
@Freeze_S: I don't think that the current title is much better than the original title. Note that the title is not just for you but for anyone who uses this site. You prefer titles in this form, but such titles can makes the task of finding relevant questions difficult. This can lead to duplicates for example. — Michael Albanese, Dec 19 '14 at 15:01
@Freeze_S: Also, I'm not sure why you think writing your titles this way keeps them more organised. You can access a list of your questions via your profile page, and if necessary, you can search this list, see here for example. If you would like to discuss this more, I suggest opening a question on meta asking if your titling procedure is acceptable. — Michael Albanese, Dec 19 '14 at 15:01
@MichaelAlbanese: Mmh, and is it sufficient using keywords? (I really don't like sentences because this way it looks more clean.) — C-star-W-star, Dec 19 '14 at 15:07
@Freeze_S: I don't think so, but like I said, any further discussion about this should occur on meta. This isn't really the place to discuss it. — Michael Albanese, Dec 19 '14 at 15:10
@MichaelAlbanese: Ok so I delete my comments regarding titles? — C-star-W-star, Dec 19 '14 at 15:14
@Hurkyl: If you want to you can of course put your comment as answer. ;) — C-star-W-star, Dec 19 '14 at 15:23
@Orient: That seems not really abstract though little axiomatic. I thought you meant something like a functor from bla to bla ;) — C-star-W-star, Dec 19 '14 at 20:20

Michael Albanese · Answer 1 · 2014-12-19T04:15:36.857

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Let $V$ be an $n$-dimensional vector space over the field $\mathbb{F}$.

Given a linear map $T : V \to V$, there is an induced linear map $\bigwedge^nT : \bigwedge^n V \to \bigwedge^n V$ given by $\left(\bigwedge^nT\right)(v_1\wedge\dots\wedge v_n) = (Tv_1)\wedge\dots\wedge(Tv_n)$. As $\bigwedge^nV$ is one-dimensional, $\bigwedge^nT = k\operatorname{id}_{\bigwedge^nV}$ for some scalar $k \in \mathbb{F}$. This scalar is precisely $\det T$.

Let me summarise some facts about the vector spaces $\bigwedge^pV$ (see the Wikipedia article on exterior algebras for more information).

Given a vector space $V$ of dimension $n$, there is an associated vector space $\bigwedge^pV$ for any $0 \leq p \leq n$ called the $p^{\text{th}}$ exterior power of $V$. The elements of $\bigwedge^pV$ are linear combinations of terms of the form $v_1\wedge\dots\wedge v_p$ where $v_1, \dots, v_p \in V$. The symbol $\wedge$ is called the wedge product, and it satisfies skew-symmetry, i.e. $v_i\wedge v_j = -v_j\wedge v_i$. If $\{v_1, \dots, v_n\}$ is a basis for $V$, then $\{v_{i_1}\wedge\dots\wedge v_{i_p} \mid i_1 < \dots < i_p\}$ is a basis for $\bigwedge^pV$ and therefore the dimension of $\bigwedge^pV$ is ${n \choose p}$.

edited Dec 19 '14 at 04:15

answered Dec 18 '14 at 15:06

Michael Albanese

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You really have to understand the tensor product and multilinear algebra to understand this definition.This is why,if we're determined to give a matrix-free definition,I feel more comfortable defining it as a product of eigenvalues for a linear transformation T. – Mathemagician1234 Dec 19 '14 at 04:15
@Mathemagician1234 I have tried to simplify this idea in my answer. – Pedro Dec 19 '14 at 04:29
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@Mathemagician1234 I think the way the answer is notated and written, that may be the case, but the core idea can be seen geometrically: elements of $\bigwedge^n V$ can be interpreted as oriented subspaces corresponding to $n$-volumes in $V$. The determinant is the single scalar the quantifies how a given $n$-volume is dilated or shrunk under a transformation, and whether that transformation reverses the volume's orientation. This definition of the determinant makes those properties readily apparent. – Muphrid Dec 19 '14 at 04:39
Remark. The definition is due to N. Bourbaki. Also what Muphrid says can be found in that book. – Mister Benjamin Dover Dec 21 '14 at 01:12

score 3 · Answer 2 · answered Dec 18 '14 at 15:54

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The nicest definition I know without getting into the symmetric group and matrix representation is the following:

Let $\;\lambda_1,...,\lambda_r\;$ be all the eigenvalues of $\;T\;$ (probably with repetitions and probably in some extension of the original definition field), then

$$\det T=\prod_{i=1}^r\lambda_i$$

answered Dec 18 '14 at 15:54

Timbuc

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How do you know the correct multiplicities without the characteristic polynomial? You can use Jordan form, but that's just as difficult as determinants. – Ryan Reich Dec 19 '14 at 08:12
@RyanReich I usually know the eigenvalues and their multiplicities (algebraic and geometric) by means of the char. and minimal polynomials. I just stated a rather easy definition without getting into how to obtain the eigenvalues. – Timbuc Dec 19 '14 at 12:37
Well, if that's your attitude you can just declare the determinant to be "known" in the same way as the eigenvalue multiplicities, and not have to define it at all! – Ryan Reich Dec 19 '14 at 22:13
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@RyanReich My attitude towards questions here is usually based on what the OP wants. The OP asked some "definition" not relying on matrix representations. The above is one. How you get the eigenvalues is another matter. The most upvoted answer here so far talks about exterior product and relies heavily on multilinear maps. I sincerely doubt many of the upvoters even understand what this is, but the fact is that he also did not use matrix representation, and neither developed the required theory to understand the answer. Ta-raa...just like my answer, imo. – Timbuc Dec 19 '14 at 22:38

Pedro · Answer 3 · 2014-12-19T04:29:10.113

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This is an attempted simplification of the answer by Michael. The determinant of degree (?) $n$ is the unique multilinear alternating form $f:V^n\to k$ such that $f(\rm id)=1$. Here multilinear alternating means $f$ is linear in each coordinate and if we let $S_n$ act on $V^n$ by permuting the order of the vectors, $f((ij)x)=-f(x)$ for any $x\in V^n$. By $\rm id$ I mean $(e_1\mid e_2\mid \ldots \mid e_n)$. In particular, there exists a unique multilinear alternating form $f:V^n\to k$ such that $f(\rm id)=\alpha$ given $\alpha\in k$, which is $\alpha\det$. One can prove the uniqueness and existence by simple induction, although some cumbersome calculations are in order.

Note that given any endomorphism $T:V\to V$, the mapping $T^*:V^n\to k$ defined by $T^*(v_1,\ldots,v_n)=\det(Tv_1,\ldots,Tv_n)$ is also multilinear alternating. Thus, there is a unique $\alpha$ such that $T^*=\alpha\det$, and we set $\alpha=\det T$.

edited Dec 19 '14 at 04:29

answered Dec 19 '14 at 04:21

Pedro

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I was just about to comment on the link between your answer and mine. As $T^*$ is an alternating form on $V^n$, it gives rise to a well-defined map $\bigwedge^nV \to k$; this is precisely the association I mention in my answer (i.e. rescaling by the determinant of $T$). – Michael Albanese Dec 19 '14 at 04:32
@PedroTamaroff: What is $\mathrm{id}$ in $f(\mathrm{id})=1$ with $f:V^n\to k$? – C-star-W-star Dec 19 '14 at 20:24
@Freeze_S I clarify that in the post. – Pedro Dec 19 '14 at 20:51
This answers uses non-standard and confusing notation and terms: “id” for an (unspecified) basis, and “det” for $n$-form generated by this basis. One may not write about an abstract vector space thinking it’s $\mathbb{F}^n$. – Incnis Mrsi Apr 28 '15 at 03:26

Incnis Mrsi · Answer 4 · 2015-04-28T09:32:40.640

There is no problem to write a definition of the determinant functional τ using induction by n and avoiding any mention of either matrices or n-forms. But it will not be very convenient since proof of uniqueness should rely on (concealed) decomposition of a matrix to upper and lower triangular factors.

This set of 3 axioms is sufficient to define unique τ on finite-dimensional spaces:

(Multiplicativity) $τ(AB) = τ(A)\,τ(B)$ for any endomorphisms A, B of V.

(Normalization) If $\dim V = 1$, then $τ(λI) = λ$.

(Decomposition) Let $U\subset V$ be a non-trivial (proper and ≠ {0}) A-invariant subspace (i.e. $\operatorname{im}(A\vert_U)\subseteq U$); we can then define the quotient space $W = V / U$ and such unique endomorphism $\hat A: W\to W$ that the diagram
V → W
↓A ↓$\hat A$
V → W
commutes. Then $τ(A) = τ_U(A\vert_U)\,τ_W(\hat A)$ must hold, where right-hand side terms refer to τ defined on the subspace and the quotient space.

Of course, one might prove existence of τ for V of any finite dimension then, that is not a very easy task.

Disintegrating By Parts · Answer 5 · 2014-12-19T19:13:17.883

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Let $V$ be a finite-dimensional linear space over a field $\mathbb{F}$. Let $\mathcal{L}(V)$ denote the linear operators from $V$ into itself. Then I would expect that the determinant to be the unique multiplicative function $\tau : \mathcal{L}(V)\rightarrow\mathbb{F}$ for which $\tau(I)=1$. That is, $\tau(AB)=\tau(A)\tau(B)$. Is it?

edited Dec 19 '14 at 19:13

answered Dec 19 '14 at 19:05

Disintegrating By Parts

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A constant 1 is also multiplicative.. – Peter Franek Dec 19 '14 at 19:15
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This is too good to be truth. – hjhjhj57 Dec 19 '14 at 19:16
@PeterFranek : Yes, I realized that afterwards. I wonder what you have to add to get that, though. I think it may be enough to add is that it is $0$ iff $V$ is non-invertible. – Disintegrating By Parts Dec 19 '14 at 19:18
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Maybe $\tau(A^T) = \tau(A)$? – hjhjhj57 Dec 19 '14 at 19:19
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Aperently, it is solved here, although $(\det A)^k$ satisfies your conditions as well.. http://math.stackexchange.com/questions/727050/is-the-determinant-the-only-group-homomorphism-from-mathrmgl-n-mathbb-r – Peter Franek Dec 19 '14 at 19:21
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@T.A.E. So, it seems to me that if you require that $\det \lambda,I=\lambda^n$ where $n$ is the dimension, it should already be unique. Maybe include this together with the link (and some remarks) to your answer, it would be a nice one! – Peter Franek Dec 19 '14 at 19:24
@PeterFranek : Ah! That's an interesting condition to add. – Disintegrating By Parts Dec 19 '14 at 19:29
@hjhjhj57 : Your adjoint condition would be a good one, but I'm not sure about the formulation. – Disintegrating By Parts Dec 19 '14 at 19:31
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@T.A.E.: As noted by hjhjhj one could also establish a sort of involution: $T^=\mathcal{L}(V^)$ Then the determinant should also satisfy: $\mathrm{det}\left(T^*\right)=\overline{\mathrm{det}T}$ Of course this is only reasonable on real or complex vector spaces but it seems a nice property. – C-star-W-star Dec 19 '14 at 19:32
@Freeze_S : The adjoint map defined in terms of linear functionals is defined and unique, but I think that the involution would be coordinate dependent. The $C^{\star}$ algebra involution most definitely is. – Disintegrating By Parts Dec 19 '14 at 19:36
@T.A.E.: Besides I'm wondering wether it is of interest to consider this rather for bounded operators so one could talk about continuity of the determinant. – C-star-W-star Dec 19 '14 at 19:37
@Freeze_S : You can't get a general determinant on infinite dimensional spaces. So boundedness doesn't buy you much. – Disintegrating By Parts Dec 19 '14 at 19:38
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@T.A.E.: Ah yes you mean as like the trace class operators? – C-star-W-star Dec 19 '14 at 19:39
@Freeze_S : Yes. And trace is related to determinant, too. – Disintegrating By Parts Dec 19 '14 at 19:42
@T.A.E.: Can you outline the dependence on coordinates? (Just to make sure that we're talking of the same thing... Algebraic Dual $V^$ Algebraic Adjoint: $(T^x^)(x):=x^(Tx)$) – C-star-W-star Dec 19 '14 at 19:42
@Freeze_S : Linear space adjoint maps functionals to functionals. Hilbert space adjoint depends on the particular choice of inner product, and there are many equivalent such inner products in general. So the involution is not unique. – Disintegrating By Parts Dec 19 '14 at 19:45
@T.A.E.: I meant the algebraic adjoint. But isn't taking the algebraic adjoint canonical so coordinate independent, is it? – C-star-W-star Dec 19 '14 at 20:15
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@T.A.E.: Interesting thing about Hilbert spaces. Just found an old question of mine adressing these issues. There's a really nice answer by Daniel Fischer, awesome guy btw. – C-star-W-star Dec 19 '14 at 20:17
Rubbish (both answer and hjhjhj57-induced comments). • There is a lot of multiplicative maps $\mathbb{F}→\mathbb{F}$; some fields even have non-trivial automorphisms. • C-algebras (and -algebras in general) have nothing to do with determinant. – Incnis Mrsi Apr 28 '15 at 03:37

Determinant: Alternative Definitions

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