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A well-known definition of the determinant is:

The determinant is the only function of a vector space of dimension $n$ to its underlying field which is multilinear, alternating and normalized.
See e.g. here for a nice outline.

Another one, which seems to be less known, is:

The determinant is the unique multiplicative map $\varphi :M_n(K)\to K$ such that $ \varphi (\alpha P+I-P) = \alpha $ for every $\alpha\in K$ and every idempotent matrix $P$ with rank one.
See here, scroll down to "Edit 4".

Are there other definitions by a unique collection of features?

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    $\begingroup$ Any generating set for $\operatorname{GL}_n$ will give a uniqueness-type statement for the determinant. $\endgroup$
    – LSpice
    Jun 28 at 10:08

2 Answers 2

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Let $M_n$ be the "affine monoid scheme" of $n \times n$ matrices under multiplication (like an affine group scheme but no inverses).

Claim: Every polynomial monoid homomorphism $M_n \to M_1$ is a non-negative integer power of the determinant; in particular the determinant is the unique generator of the monoid of such homomorphisms.

The same is true for the more familiar affine group scheme $GL_n$ (with "integer power") but then we need to distinguish between the determinant and its inverse, which we can do using the fact that the determinant extends to $M_n$.

Proof. Let $\varphi : M_n \to M_1$ be such a homomorphism. Passing to any algebraically closed field of characteristic zero, the diagonalizable matrices are Zariski dense and $\varphi$ must be invariant under conjugation, so $\varphi$ is determined by its restriction to diagonal matrices and therefore must be a polynomial in the eigenvalues. Since diagonal matrices are conjugate to all their diagonal permutations, $\varphi$ must be a symmetric polynomial in the eigenvalues. By restricting to each diagonal entry separately $\varphi$ restricts to a polynomial homomorphism $M_1 \to M_1$ which must be the same for each diagonal entry (because each of the corresponding copies of $M_1$ is conjugate in $M_n$), and it's not hard to show with a direct computation that such a homomorphism (concretely, a polynomial $f(x) \in \mathbb{Z}[x]$ satisfying $f(xy) = f(x) f(y)$ and $f(1) = 1$) must be of the form $f(x) = x^k$ for some non-negative integer $k$. By writing a diagonal matrix $\text{diag}(\lambda_1, \dots \lambda_n)$ as a product of diagonal matrices of the form $\text{diag}(1, \dots \lambda_i, \dots 1)$ we conclude that $\varphi = (\lambda_1 \dots \lambda_n)^k$ for some non-negative integer $k$ as desired. (Unfortunately this argument doesn't construct the determinant; we need to show that it exists separately.) $\Box$

An alternative way to state the conclusion is that the bialgebra $\mathcal{O}_{M_n}$ of functions on $M_n$, which concretely is $\mathbb{Z}[x_{ij}], 1 \le i, j \le n$ equipped with the comultiplication $\Delta x_{ij} = \sum_k x_{ik} \otimes x_{kj}$ encoding matrix multiplication, has the property that its monoid of grouplike elements (elements satisfying $\Delta x = x \otimes x$) is generated by the determinant.

This is indirect, plus it doesn't immediately tell you that the determinant is invariant under change of basis (although that's easy to prove once you know it's a homomorphism), but it has the pleasant property of specifying a single universal polynomial over $\mathbb{Z}$ which specializes to the determinant over every field (even every commutative ring), without needing to work over any particular field.

Instead of talking about generators we can also single out the determinant using the fact that, among all the powers $\det^k$, the determinant itself is the only one which is multilinear.

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In what follows, I will work over $\mathbb{C}$.

Let $d_1,\ldots,d_n$ be positive integers. Consider generic homogeneous polynomials $F_1(x_1,\ldots,x_n),\ldots,F_n(x_1,\ldots,x_n)$ of degrees $d_1,\ldots,d_n$ respectively. The resultant of format $(d_1,\ldots,d_n)$ is the unique polynomial in variables given by the coefficents of the polynomials $F_1,\ldots,F_n$ which satisfies:

  1. It vanishes iff $\exists x\in\mathbb{C}^n\backslash\{0\}$, $\forall i, 1\le i\le n, F_i(x)=0$.

  2. It takes the value 1 when $\forall i, F_i(x)=x_{i}^{d_i}$.

  3. It is irreducible.

The determinant is just the resultant of format $(1,1,\ldots,1)$.

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