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 $\det A$. 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 ${(-1)}^{\dim V}$ times the constant term in the characteristic polynomial of $T$.

However I find this somewhat unsatisfactory. It’s 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 $\det A$ 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?

  • 3
    $\begingroup$ You might find this thread helpful. math.stackexchange.com/questions/668/… $\endgroup$
    – user17762
    Commented Feb 11, 2011 at 23:57
  • 3
    $\begingroup$ See also this MO answer/rant mathoverflow.net/questions/7584/… $\endgroup$
    – JDH
    Commented Feb 12, 2011 at 0:22
  • $\begingroup$ @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? $\endgroup$ Commented Feb 12, 2011 at 0:29
  • $\begingroup$ 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. $\endgroup$
    – Sam
    Commented Feb 12, 2011 at 0:37
  • 4
    $\begingroup$ @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. $\endgroup$ Commented Feb 12, 2011 at 1:17

5 Answers 5


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.

  • 1
    $\begingroup$ Nice! (...15 char minimum) $\endgroup$ Commented Feb 12, 2011 at 0:23
  • 4
    $\begingroup$ @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. $\endgroup$ Commented Feb 12, 2011 at 1:16
  • 3
    $\begingroup$ @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. $\endgroup$ Commented Feb 12, 2011 at 1:23
  • 2
    $\begingroup$ @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). $\endgroup$ Commented Jan 9, 2016 at 8:50
  • 4
    $\begingroup$ 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 $\Lambda^n(V)$. $\endgroup$ Commented Nov 9, 2018 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.

  • $\begingroup$ Could you expand on what you mean by $n$-alternating multilinear form here? $\endgroup$ Commented Feb 12, 2011 at 0:19
  • $\begingroup$ 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 $i,j$. $\endgroup$
    – Sam
    Commented Feb 12, 2011 at 0:22
  • $\begingroup$ I think maybe I will need to read more into this to really understand it. I will ask my linear algebra professor sometime. $\endgroup$ Commented Feb 12, 2011 at 0:44
  • $\begingroup$ As an advice, try to make relations between this definition and the geometric one for dim 2 and 3. $\endgroup$
    – Sam
    Commented Feb 12, 2011 at 0:48

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.


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.

  • $\begingroup$ The issue is that he does not use the "common" definition of eigenvalue. $\endgroup$
    – Sam
    Commented Feb 12, 2011 at 0:06
  • 2
    $\begingroup$ The thing here is that he is still defining the determinant in terms of a formula. $\endgroup$ Commented Feb 12, 2011 at 0:33

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).

  • 2
    $\begingroup$ 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. $\endgroup$ Commented Feb 19, 2011 at 0:22
  • $\begingroup$ And, as Qiaochu mentions at the end of his answer, this definition is basis dependent. $\endgroup$ Commented Feb 19, 2011 at 0:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .