In the few linear algebra texts I have read, the determinant is introduced in the following manner;

"Here is a formula for what we call $detA$. Here are some other formulas. And finally, here are some nice properties of the determinant."

For example, in very elementary textbooks it is introduced by giving the co-factor expansion formula. In Axler's "Linear Algebra Done Right" it is defined, for $T\in L(V)$ to be $(-1)^{dimV}$ times the constant term in the characteristic polynomial of $T$.

However I find this somewhat unsatisfactory. Its like the real definition of the determinant is hidden. Ideally, wouldn't the determinant be defined in the following manner:

"Given a matrix $A$, let $detA$ be an element of $\mathbb{F}$ such that x, y and z."

Then one would proceed to prove that this element is unique, and derive the familiar formulae.

So my question is: Does a definition of the latter type exist, is there some minimal set of properties sufficient to define what a determinant is? If not, can you explain why?

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    You might find this thread helpful. http://math.stackexchange.com/questions/668/whats-an-intuitive-way-to-think-about-the-determinant –  Feb 11 '11 at 23:57
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    See also this MO answer/rant http://mathoverflow.net/questions/7584/what-are-the-most-misleading-alternate-definitions-in-taught-mathematics/7952#7952 – JDH Feb 12 '11 at 00:22
  • @JDH an, I had found the property that detA is the scaling factor of area when transforming the unit square, and I liked the geometric intuition, but he says that this is the "right" definition. Would everyone agree with him on that? – AnonymousCoward Feb 12 '11 at 00:29
  • It depends on what you want to focus. If you want to *understand* the concept, this is definitely the right definition. But I'm not sure you can say about a definition it is the "right" one. – Sam Feb 12 '11 at 00:37
  • Gottfried, all the other formulaic definitions of the determinant follow easily from the volume definition. For example, that definition immediately implies the right transforms of the determinant under row operations, from which the familiar definitions can be derived. (See the comments to the anser on MO.) – JDH Feb 12 '11 at 00:42
  • @sivaram Thanks, that was very nice and elementary. – AnonymousCoward Feb 12 '11 at 00:43
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    @Gottfried: the volume definition is absolutely the "right" definition for a first course on linear algebra: it is intuitive and powerful. But the determinant is more general than this: it makes sense in situations where volumes don't (e.g. over a finite field, or something even weirder) and requires one to make fewer choices than the volume definition would imply, and in mathematics one needs to apply the determinant to these general situations where there are no volumes and it would be a bad idea to make choices. – Qiaochu Yuan Feb 12 '11 at 01:17
  • Qiaochu, I agree with you, but still the determinant is rarely defined by the volume definition in introductory linear algebra texts, and often it is not even mentioned in such texts. I find myself often having to supplement the text with such ideas when teaching linear algebra. – JDH Feb 12 '11 at 01:54
  • @JDH it certainly wasn't brought up in my first linear algebra course. Im glad I asked this question now. – AnonymousCoward Feb 12 '11 at 02:38
  • @Qiaochu: the volume definition of the determinant is, in principle, completely general. This is an application of the principle of permanence of algebraic identities (I think Keith Conrad has some nice notes on this). At one point I even batted around the idea of a presentation of linear algebra taking this as a first principle...and then I got distracted by mathematical real life. But I think "determinant as volume" is a powerful idea that the geometers seem to understand very well and the algebraists tend to undervalue. (I am an algebraist who would in some ways like to be a geometer.) – Pete L. Clark Feb 12 '11 at 04:26
  • Maybe I am too much of an algebrist, but the "definition" using volumes (the quotes indicate my not being sure how exactly we define the volume of a parallelepiped without determinants) does not tell me anything! – Mariano Suárez-Álvarez Feb 12 '11 at 05:34
  • @Mariano: "(the quotes indicate my not being sure how exactly we define the volume of a parallelepiped without determinants)": Are you perhaps being a little disingenuous here? The volume is, for instance, the integral (Riemann or Lebesgue) of the characteristic function of the parallelepiped. I strongly doubt this is news to you... – Pete L. Clark Feb 12 '11 at 08:04
  • (But honestly, I think part of the point of being a geometer -- rather than an algebraist or an analyst -- is that you feel like you have a good intuition for what "volume" means and although you could give a precise definition suitable for any given situation, you would feel strange saying "By the volume of a parallelepiped, I mean..." By the volume, you mean how much space it takes up! Again, my own perspective is more algebraic / formalist than this, but I have observed this type of thinking in others and they seem to do well with it...) – Pete L. Clark Feb 12 '11 at 08:10
  • @Mariano: by Cavalieri's principle. The area of a parallelogram is the sum over the lengths of its constituent lines. The volume of a paralleletope is the sum over the areas of its constituent parallellograms. Etc. Are you really saying you think the definition of, say, differential forms is completely formal and you have no intuition for it at all?! @Pete: the volume definition still requires one to fix an isomorphism from the top exterior power to the base field, at least the way JDH phrases it. – Qiaochu Yuan Feb 12 '11 at 09:10
  • My opinion is that talking about the determinant in an introductory linear algebra course without using the volume of the image as a principal or defining idea is like introducing Lebesgue measure without using the idea that you are trying to measure the "size" of a set. – JDH Feb 12 '11 at 12:26
  • @Pete: Surely I know what the volume is. But doesn't having the definition of determinants depend on something as complicated as Riemann or Lebesgue integration strike anyone as weird? Even on purely aesthetical grounds! @Qiaochu, I am more familiar in C's principle as a way to *compare* volumes, and then you are right that it can be used to compute the area of a parallelepided by comparing it to a straight one (otherwise of doing the computation are ugly or depend on changing variables!) One can also cut it in pieces and reassemble to obtain a straight one---the actual proof is probably fun. – Mariano Suárez-Álvarez Feb 12 '11 at 16:38
  • @Qiaochu, What I am saying is that the connection with volumes, while there, is not significant in in they that I think about determinants. It does not help me much with Vandermonde's determinant or the identity of Jacobi-Trudi. The idea that Lebesgue measure measures size is, on the other hand, inseparable from pretty much any appearance of Lebesgue measure I'm aware of! (I would love to hear of a geometric interpretation of the determinant in, say, Jacobi-Trudi!) – Mariano Suárez-Álvarez Feb 12 '11 at 16:46
  • @Mariano: yes, I precisely mean comparison to a unit parallelotope. You don't need to know anything about integration to do this. – Qiaochu Yuan Feb 12 '11 at 16:53
  • @Mariano: yes, of course, I'm not advocating that one delay linear algebra until after learning about Riemann or Lebesgue integration. Your previous comment, taken literally, seemed to express that you thought that it wasn't possible to define volumes without determinants, so I was indicating both that it was and that you already knew this. I think Joel is saying what I am trying to say better than I am: giving clean, formal definitions often has the effect of pushing away important geometric concepts like "volume", which are more sophisticated to define analytically/algebraically. – Pete L. Clark Feb 12 '11 at 17:56
  • @Qiaochu: I'm not sure if I understand all the nuances of what you're saying, but -- absolutely volume is a basis-dependent quantity: that's the whole point of the determinant. The determinant measures the *change* in volume of the linear transformation $T$: pick any parallelepiped $P$ in $V$: then, up to sign, $\operatorname{det}(V) = \operatorname{vol}(T(P))/\operatorname{vol}(P)$. This is basis-invariant, right? – Pete L. Clark Feb 12 '11 at 18:02
  • @Pete: yes, but to even write down this fraction requires a definition of Vol(P) as a number (so that one may take the quotient). My point is that this is useful, but unnecessary. (If you can't tell, I am not a fan of making choices and then proving that a construction does not depend on those choices.) – Qiaochu Yuan Feb 12 '11 at 18:11
  • @Qiaochu: okay, it seems we understand each other perfectly. – Pete L. Clark Feb 12 '11 at 18:39

5 Answers5


Let $V$ be a vector space of dimension $n$. For any $p$, the construction of the exterior power $\Lambda^p(V)$ is functorial in $V$: it is the universal object for alternating multilinear functions out of $V^p$, that is, functions

$$\phi : V^p \to W$$

where $W$ is any other vector space satisfying $\phi(v_1, ... v_i + v, ... v_p) = \phi(v_1, ... v_i, ... v_p) + \phi(v_1, ... v_{i-1}, v, v_{i+1}, ... v_p)$ and $\phi(v_1, ... v_i, ... v_j, ... v_p) = - \phi(v_1, ... v_j, ... v_i, ... v_p)$. What this means is that there is a map $\psi : V^p \to \Lambda^p(V)$ (the exterior product) which is alternating and multilinear which is universal with respect to this property; that is, given any other map $\phi$ as above with the same properties, $\phi$ factors uniquely as $\phi = f \circ \psi$ where $f : \Lambda^p(V) \to W$ is linear.

Intuitively, the universal map $\psi : V^p \to \Lambda^p(V)$ is the universal way to measure the oriented $p$-dimensional volumes of paralleletopes defined by $p$-tuples of vectors in $V$, the point being that for geometric reasons oriented $p$-dimensional volume is alternating and multilinear. (It is instructive to work out how this works when $n = 2, 3$ by explicitly drawing some diagrams.)

Functoriality means the following: if $T : V \to W$ is any map between two vector spaces, then there is a natural map $\Lambda^p T : \Lambda^p V \to \Lambda^p W$ between their $p^{th}$ exterior powers satisfying certain natural conditions. This natural map comes in turn from the natural action $T(v_1, ... v_p) = (Tv_1, ... Tv_p)$ defining a map $T : V^p \to W^p$ which is compatible with the passing to the exterior powers.

The top exterior power $\Lambda^n(V)$ turns out to be one-dimensional. We then define the determinant $T : V \to V$ to be the scalar $\Lambda^n T : \Lambda^n(V) \to \Lambda^n(V)$ by which $T$ acts on the top exterior power. This is equivalent to the intuitive definition that $\det T$ is the constant by which $T$ multiplies oriented $n$-dimensional volumes. But it requires no arbitrary choices, and the standard properties of the determinant (for example that it is multiplicative, that it is equal to the product of the eigenvalues) are extremely easy to verify.

In this definition of the determinant, all the work that would normally go into showing that the determinant is the unique function with such-and-such properties goes into showing that $\Lambda^n(V)$ is one-dimensional. If $e_1, ... e_n$ is a basis, then $\Lambda^n(V)$ is in fact spanned by $e_1 \wedge e_2 \wedge ... \wedge e_n$. This is not so hard to prove; it is essentially an exercise in row reduction.

Note that this definition does not even require a definition of oriented $n$-dimensional volume as a number. Abstractly such a notion of volume is given by a choice of isomorphism $\Lambda^n(V) \to k$ where $k$ is the underlying field, but since $\Lambda^n(V)$ is one-dimensional its space of endomorphisms is already canonically isomorphic to $k$.

Note also that just as the determinant describes the action of $T$ on the top exterior power $\Lambda^n(V)$, the $p \times p$ minors of $T$ describe the action of $T$ on the $p^{th}$ exterior power $\Lambda^p(V)$. In particular, the $(n-1) \times (n-1)$ minors (which form the matrix of cofactors) describe the action of $T$ on the second-to-top exterior power $\Lambda^{n-1}(V)$. This exterior power has the same dimension as $V$, and with the right extra data can be identified with $V$, and this leads to a quick and natural proof of the explicit formula for the inverse of a matrix.

As an advance warning, the determinant is sometimes defined as an alternating multilinear function on $n$-tuples of vectors $v_1, ... v_n$ satisfying certain properties; this properly defines a linear transformation $\Lambda^n(V) \to k$, not a determinant of a linear transformation $T : V \to V$. If we fix a basis $e_1, ... e_n$, then this function can be thought of as the determinant of the linear transformation sending $e_i$ to $v_i$, but this definition is basis-dependent.

Qiaochu Yuan
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    Nice! (...15 char minimum) – George Lowther Feb 12 '11 at 00:23
  • This is my favorite definition. – Cheerful Parsnip Feb 12 '11 at 00:25
  • Oh yeah ! I like to compare this definition to the "Down with determinants" one xD – Sam Feb 12 '11 at 00:29
  • Thanks for the exposition, I really know nothing about exterior powers and functorials so the explanation is much appreciated. – AnonymousCoward Feb 12 '11 at 00:51
  • It get's even better with *a quick and natural proof of the explicit formula for the inverse*. Something I was about to ask, but see its already been added. – George Lowther Feb 12 '11 at 00:56
  • By the way, does anyone know a textbook on linear algebra that actually uses this approach? I have never seen this material written up anywhere at that level and have only picked all of it up from the aether, so to speak. – Qiaochu Yuan Feb 12 '11 at 00:58
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    There is a natural bilinear map $V\times\Lambda^{n-1}(V)\to\Lambda^n(V)$. So, the action $T$ induces on $\Lambda^{n-1}(V)$ leads to an adjoint action on $V$, which I assume is the "matrix of cofactors" (or adjugate), and showing that multiplying this by $T$ gives the determinant is a simple algebraic step. Neat! – George Lowther Feb 12 '11 at 01:05
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    @Qiaochu there is the book "Linear Algebra via Exterior Products" by Sergei Winitzki which is written somewhat like this (there doesn't seem to be any category theory), although I have only kind of started to read it. – Harry Stern Feb 12 '11 at 01:16
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    @Harry: ah, I'd forgotten about Winitzki's book. I haven't taken a close look at it either but he defines the exterior product as a subspace of the tensor product instead of a quotient, and this is a slightly different definition. For developing the basic theory it's largely equivalent but there are some subtle differences. – Qiaochu Yuan Feb 12 '11 at 01:23
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    @QiaochuYuan: for posterity's sake, you asked several years ago for a book that uses this approach. Birkhoff and Mac Lane do it in Algebra (even the 1st edition). In fact, they do it *both* the ways you described it (first as a function on tuples, then as the top-degree multiplier). – symplectomorphic Jan 09 '16 at 08:50
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    I feel it should be mentioned that the real hard work in this approach is verifying that $e_1 \wedge e_2 \wedge ... \wedge e_n$ is _nonzero_. In particular, this is considerably harder than showing that $e_1 \wedge e_2 \wedge ... \wedge e_n$ spans $\Lambda^n(V)$. – Eric Wofsey Nov 09 '18 at 21:02

Let $B$ a basis of a vector space $E$ of dimension $n$ over $\Bbbk$. Then $det_B$ is the only $n$-alternating multilinear form with $det_B(B) = 1$.

A $n$-multilinear form is a map of $E^n$ in $\Bbbk$ which is linear for each variable.

A $n$- alternated multilinear form is a multilinear form which verify for all $i,j$ $$ f(x_1,x_2,\dots,x_i,\dots, x_j, \dots, x_n) = -f(x_1,x_2,\dots,x_j,\dots, x_i, \dots, x_n) $$ In plain english, the sign of the application change when you switch two argument. You understand why you use the big sum over permutations to define the determinant with a closed formula.

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  • Could you expand on what you mean by $n$-alternating multilinear form here? – AnonymousCoward Feb 12 '11 at 00:19
  • A multilinear form is a map $E^n$ in $\Bbbk$ (for instance $\mathbb{R}$ or $\mathbb{C}$) which is linear for each variable. Think for instance of bilinear application (related to product scalar for example). Then, an alternate one is one where $f(x_1,x_2,\dots,x_i,\dots, x_j, \dots, x_n) = -f(x_1,x_2,\dots,x_j,\dots, x_i, \dots, x_n)$ for any $i,j$. – Sam Feb 12 '11 at 00:22
  • I think maybe I will need to read more into this to really understand it. I will ask my linear algebra professor sometime. – AnonymousCoward Feb 12 '11 at 00:44
  • As an advice, try to make relations between this definition and the geometric one for dim 2 and 3. – Sam Feb 12 '11 at 00:48

Determinants are best understood in the context of exterior algebra, which goes back to the work of Hermann Grassmann. Here a down-to-earth description of the intuition behind it.

Consider an $n$-dimensional vector space $V = K^n$ where $K$ is some field. Assume that the dimension $n$ is as large as we need it for our examples. Let $e_1,\dots,e_n$ denote a basis of the space $V$.

As you know, any (say) three linearly independent vectors $a,b,c$ span a 3-dimensional subspace of $V$. Wouldn't it be nice if we could somehow calculate with subspaces? I.e. if we had some notion of algebraic product that is geometrically meaningful? Like, if the product of two 3-dimensional subspaces is zero, then something geometric has happened. It turns out that — with some limitations — there is something like that, namely the exterior algebra or Grassmann algebra.

The key point that the exterior product of two "subspaces" corresponds to their geometric span. The drawback of the exterior product is that it's definition is entirely formal, but it's usefulness shall serve as motivation. Namely, the exterior product $\wedge$ of three vectors $a,b,c,$ is defined as the formal expression

$$ a \wedge b \wedge c$$

which is subject to the following three rules

  1. The product is linear in each factor, $(\alpha x + y) \wedge z = \alpha(x \wedge z) + (y \wedge z)$.
  2. The two factors are the same, then the product vanishes, $x \wedge x = 0$.
  3. The product is antisymmetric, $x \wedge y = - y \wedge x$. (This actually follows from 1 and 2)

For instance, we have

$$ (a+b) \wedge b \wedge d = (a \wedge b \wedge d) + (b \wedge b \wedge d) = a \wedge b \wedge d + 0 = a \wedge b \wedge d$$

As you can see, we also consider sums of these products, though they don't necessarily carry geometric meaning.

The beauty of these rules is that they allow us to represent the subspace spanned by the three vectors $a,b,c$ by their exterior product $a \wedge b \wedge c$. In particular, it is an easy exercise to show that the rules imply that

  1. We have $a \wedge b \wedge c = \lambda (a' \wedge b' \wedge c')$ for some scalar multiple $\lambda$ if and only if the three vectors $a,b,c$ span the same subspace as the three vectors $a',b',c'$.
  2. We have $a \wedge b \wedge c = 0$ if and only if the three vectors are linearly dependent.

Isn't that awesome? The question of whether vectors are linearly dependent has been transformed into a question about an algebraic product being zero, something that is easy to calculate with!

In particular, if you have a matrix $A \in K^{n\times n}$, the exterior product of the column vectors

$$ Ae_1 \wedge Ae_2 \wedge \dots \wedge Ae_n $$

is zero if and only if the matrix is singular. It is not difficult to show that in an $n$-dimensional vectors space, the only non-zero products of $n$ factors must be of the form $\lambda(e_1\wedge e_2 \wedge \dots \wedge e_n)$. The scalar multiple $\lambda$ can be interpreted as the volume of the parallelepiped. The determinant of a matrix is defined to be this factor:

$$ (\det A)(e_1\wedge e_2 \wedge \dots \wedge e_n) := Ae_1 \wedge Ae_2 \wedge \dots \wedge Ae_n .$$

It is an instructive exercise to show that this coincides with the standard formulas for the determinant.

For more about using the exterior product to do calculations with subspaces, consider a book by John Browne, though I think it's a bit long-winded.

These techniques feature prominently in differential geometry as differential forms. The cup-product in cohomology is another instance of these ideas. Schubert calculus is a natural extension of these ideas to problems from algebraic geometry.

Greg Graviton
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This paper: http://www.axler.net/DwD.html "Down with determinants" by Sheldon Axler, is probably what you are looking for. He proves many of the familiar properties of finite-dimensional vector spaces without the use of determinants - and in the end he gives an easy definition of the determinant as the product of the eigenvalues of the matrix, and finally proves the usual defining formula for the determinant.

Fredrik Meyer
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In Artin's book, Algebra, the determinant is defined as the only function $\det: \mathbb{R}^{n \times n} \rightarrow \mathbb{R}$ such that

1 - $\det(I) = 1$
2 - $\det$ is an $n$-multilinear function
3 - if two rows (columns) of $A$ are equal, then $\det(A) = 0$

Which, I think, can be proven using Qiaochu's answer (but I'm not sure).

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    As a function of the rows (columns) of $A$, that it equivalent to saying thatit is an alternating multilinear function taking $(e_1,e_2,\ldots,e_n)$ to one. By the universal property of $\Lambda^n(V)$, this is equivalent to a map $\Lambda^n(V)\to\mathbb{R}$ taking $e_1\wedge e_2\wedge\cdots\wedge e_n$ to 1. Uniqueness of this is equivalent to $\Lambda^n(V)$ being one-dimensional, so it is related to Qiaochu's answer. – George Lowther Feb 19 '11 at 00:22
  • And, as Qiaochu mentions at the end of his answer, this definition is basis dependent. – George Lowther Feb 19 '11 at 00:26