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Let $T$ be a linear operator on a finite-dimensional vector space $V$ over the field $K$, with $\dim V=n$. Is there a definition of the determinant of $T$ that (1) does not make reference to a particular basis of $V$, and (2) does not require $K$ to be a particular field?


As motivation, if $K=\mathbb C$, I know of three ways to define $\det(T)$, two of which refer to a choice of basis, and the other of which relies on $\mathbb C$ being algebraically closed:

  • Choose an ordered basis $B$ of $T$, and let $\mathcal M(T)$ denote the matrix of $T$ with respect to this basis. Then apply any of the formulas/algorithms for calculating a determinant to $\mathcal M(T).$ [This works for any field, but requires choosing a basis to express $\mathcal M(T)$.]
  • Choose an ordered basis $B$ of $T$, and let $\det_n$ be an alternating multilinear map from $V^n\to K$. Then the determinant of $T$ can be defined as $\det_n(TB)/\det_n(B)$. [This works for any field, but requires choosing a basis to extract "column vectors of the matrix of $T$."]
  • Define $\det(T)$ as the product of eigenvalues of $T$, repeated according to their algebraic multiplicity. [This makes no reference to a basis, but only works because $\mathbb C$ is algebraically closed.]
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    $\begingroup$ The wikipedia page mentions it, under "exterior algebra": en.wikipedia.org/wiki/Determinant#Abstract_algebraic_aspects $\endgroup$ Jun 19 at 17:43
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    $\begingroup$ @Vercassivelaunos, Thanks for that reference. It seems like the basis-independent version of the second definition I gave above. $\endgroup$
    – WillG
    Jun 19 at 18:14
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    $\begingroup$ Another idea I was considering: What if we take the algebraic closure of $K$ and apply the third definition? Does this work, and always result in an element of $K$ (rather than $\overline{K})$? I might make a new post about this. $\endgroup$
    – WillG
    Jun 19 at 18:15
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    $\begingroup$ @mr_e_man $\dim\ker[(T-\lambda I)^{\dim V}]$. See e.g. Axler, Linear Algebra Done Right. $\endgroup$
    – WillG
    Jun 20 at 18:57
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    $\begingroup$ One thing that almost works is: the abelianization of $GL_n(K)$ is the determinant map $GL_n(K) \to K^*$, except in one exceptional case $GL_2(\mathbb{F}_2)$. That determines the determinant up to an automorphism of $K^*$. $\endgroup$ Jun 20 at 19:37

5 Answers 5

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This answer was edited quite a few times after receiving valuable input from several users. While its present form reflects quite faithfully the process that led to it, the patchy nature of the text is perhaps not especially pleasing to read. I thus decided to add a (hopefully) last edit down at the bottom, with a substantially streamlined and complete proof which could only be written in hindsight. The reader might therefore prefer to jump straight into it. You will find it under "EDIT 4".


The determinant is the unique multiplicative map $$ \varphi : \text{End}(V)\to K $$ such that

  1. $\varphi (I+T)=1$, when $T^2=0$,

  2. $\varphi (\alpha P+I-P) = \alpha $, when $P$ is idempotent and has rank one.

(See EDIT 2, below, for a proof of the fact that condition (1), above, is superfluous).


EDIT: Now that I have some free time, let me give some justification for my perhaps a bit too blunt (sorry!) answer above.

It is not so hard to see that the determinant satisfies the above properties, so I will only prove uniqueness.

The whole point of the question is to get rid of coordinates but I believe it doesn't hurt if the proof is based on coordinates. In other words, let us speak of $n\times n$ matrices.

So we suppose that $$ \varphi : M_n(K)\to K $$ is a multiplicative map satisfying the above conditions and let us prove that $\varphi $ coincides with the determinant.

For every $i$ and $j$, consider the $n\times n$ matrix $E_{i,j}$ whose entries are all zero except for the $(i,j)$ entry, which is equal to 1.

We then observe that if $A$ is any $n\times n$ matrix, $\lambda $ is any scalar, and $i\neq j$, then $$ (I+\lambda E_{i,j})A $$ is the matrix one gets by applying to $A$, the elementary operation of adding $\lambda $ times the $j^{th}$ row of $A$ to the $i^{th}$ row.

Since $i\neq j$, one has that $(\lambda E_{i,j})^2=0$, so the hypothesis gives $$ \varphi \big ((I+\lambda E_{i,j})A\big ) = $$$$=\varphi (I+\lambda E_{i,j} ) \varphi (A) = \varphi (A) . $$

This implies that the value of $\varphi (A)$ remains unchanged no matter how many elementary row operations we apply to $A$.

As observed in "EDIT 3" below, we are also able to swap any two rows of $A$ by means of a sequence of elementary operations, as long as we change the sign of one of the rows.

We are therefore able to bring $A$ to it's reduced row echelon form, keeping the value of $\varphi (A)$ unchanged, except that we cannot make the leading entries of each row equal to 1, since this requires multiplying a row by a scalar, an operation under which $\varphi$ is certainly not invariant.

Letting $A'$ be this quasi reduced row echelon form of $A$, (with leading entties not necessarily equal to 1), we consequently have that $\varphi (A)= \varphi (A')$.

Case 1: $A$ is invertible and hence $A'$ is diagonal.

Letting $a_i$ denote the $i^{th}$ diagonal entry of $A'$, we then have that $$ A'=\prod_{i=1}^n(a_iE_{i,i}+I-E_{i,i}), $$ whence $$ \varphi (A)= \varphi (A')=\prod_{i=1}^n\varphi (a_iE_{i,i}+I-E_{i,i}) = $$$$ = \prod_{i=1}^na_i=\text{det}(A')=\text{det}(A). $$

Case 2: $A$ is not invertible and hence the last row of $A'$ is identically equal to zero.

In this case $E_{n,n}A' =0$, so $$ A' = (I-E_{n,n})A' = (0E_{n,n}+I-E_{n,n})A', $$ whence $$ \varphi (A)=\varphi (A') = \varphi \big ((0E_{n,n}+I-E_{n,n})A'\big )= $$$$= \varphi (0E_{n,n}+I-E_{n,n})\varphi (A') = 0\varphi (A') = 0 = \text{det}(A). $$


EDIT 2: Notice that the hypothesis "$\varphi (I+T)=1$, when $T^2=0$" in the above proof was used exclusively to argue that $\varphi (I+\lambda E_{i,j}) = 1$. Here we will prove that this hypothesis is superfluous.

I thank user @math54321 for a comment which led to the proof of this result without any special hypothesis on $K$.

Theorem. The determinant is the unique multiplicative map $\varphi :M_n(K)\to K$ such that

  • $\varphi (\alpha P+I-P) = \alpha $, when $P$ is idempotent and has rank one.

Proof. Given any such $\varphi $, and in view of the discussion above, it is enough to show that $\varphi (I+\lambda E_{i,j}) = 1$, whenever $i\neq j$.

Given $a\in K$, nonzero, a simple computation shows that $$ (1+aE_{i,j})\Big (aE_{i,i}+1-E_{i,i}\Big ) = \Big (aE_{i,i}+1-E_{i,i}\Big )(1+E_{i,j}), $$ and since $$ \varphi \Big (aE_{i,i}+1-E_{i,i}\Big ) = a \neq 0, $$ we get $$ \varphi (1+aE_{i,j})=\varphi (1+E_{i,j}). \qquad (*) $$

The proof will then be concluded once we prove that $\varphi (1+E_{i,j})=1$, which we do by considering two cases:

Case 1) The characteristic of $K$ is 2.

In this case notice that $$ (1+E_{i,j})^2 = 1+2E_{i,j}=1, $$ so $\varphi (1+E_{i,j})^2 = 1$, and we see that $\varphi (1+E_{i,j})$ is the unique solution of the polynomial equation $x^2=1$, namely 1 (recall that $1=-1$ here).

Case 2) The characteristic of $K$ is not 2.

In this case we have that $$ (1+E_{i,j})^2 = 1+2E_{i,j}, $$ and since $2\neq 0$, we have $$ \varphi (1+E_{i,j})^2 = \varphi (1+2E_{i,j}) \mathrel{\buildrel (*)\over =} \varphi (1+E_{i,j}). $$ Since $1+E_{i,j}$ is invertible (with inverse $1-E_{i,j}$), and hence $\varphi (1+E_{i,j})\neq 0$, we deduce that $\varphi (1+E_{i,j})$ is the unique nonzero solution of the polynomial equation $x^2=x$, namely 1. $\qquad$ QED


EDIT 3: As pointed out by user @math54321, in order to bring a matrix to its reduced row echelon form one also needs to be able to swap rows. However, since swapping rows causes a change of sign in the determinant, it is not reasonable to expect $\varphi (A)$ to be invariant under such an elementary operation. Instead, we will show invariance of $\varphi (A)$ under a row swap, followed by a change of sign of one of the rows involved. Clearly this is equally effective in the task of bringing a matrix to its reduced row echelon form.

We will soon see that the key computation to support this claim is that, defining $$ \Sigma _{i, j} := (1+E_{i,j})(1-E_{j,i})(1+E_{i,j}), $$ one has $$ \Sigma _{i, j} = 1 - E_{j,j} - E_{i,i} + E_{i,j} - E_{j,i}. \qquad (**) $$ For example, in case $n=3$, $i=2$, and $j=1$, this becomes $$ \Sigma _{2, 1} = \pmatrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 } \pmatrix{ 1 & -1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 } \pmatrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 } = \pmatrix{ 0 & -1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 }. $$ The computation in $(**)$ amounts to saying that $\Sigma _{i,j}$ is the $n\times n$ matrix that coincides with the identity matrix everywhere outside the $2\times 2$ submatrix formed by the rows and columns with indices $i$ and $j$, where it instead looks like $ \pmatrix{ 0 & -1\cr 1 & 0 }. $

Moreover, given any matrix $A$, the matrix $\Sigma _{i,j}A$ is easily seen to be matrix obtained from $A$ by swapping the $i^{th}$ and $j^{th}$ rows, followed by a change of sign of the $i^{th}$ row (which was formerly known as the $j^{th}$ row).

A glance into the definition of $\Sigma _{i,j}$ is then enough to convice the reader that $\varphi (\Sigma _{i,j})=1$, and hence that $\varphi (\Sigma _{i,j}A)=\varphi (A)$. This shows that $\varphi $ is invariant under our row swapping/sign changing operation.

We thank user @math54321 for pointing out the need to verify this extra point.


EDIT 4. A streamlined proof.

Theorem. The determinant is the unique multiplicative map $\varphi :M_n(K)\to K$ such that $$ \varphi (a P+I-P) = a , $$ for every $a$ in $K$, and every idempotent matrix $P$ with rank one.

Proof. It is clear that the determinant satisfies the above property, so we move on to the proof of uniqueness. Thus, supposing that $$ \varphi : M_n(K)\to K $$ is a multiplicative map satisfying the above condition, we must prove that $\varphi $ coincides with the determinant.

For every $i$ and $j$, consider the $n\times n$ matrix $E_{i,j}$ whose entries are all zero except for the $(i,j)$ entry, which is equal to 1. We then claim that $$ \varphi (I+a E_{i,j}) = 1, \tag{1} $$ for every $a\in K$, and every $i\neq j$. To prove this, and supposing first that $a$ is nonzero, a simple computation shows that $$ (1+aE_{i,j})\Big (aE_{i,i}+1-E_{i,i}\Big ) = \Big (aE_{i,i}+1-E_{i,i}\Big )(1+E_{i,j}), $$ and since $$ \varphi \Big (aE_{i,i}+1-E_{i,i}\Big ) = a \neq 0, $$ we get $$ \varphi (1+aE_{i,j})=\varphi (1+E_{i,j}). \tag{2 } $$

The proof will then be concluded once we prove that $\varphi (1+E_{i,j})=1$, which we do by considering two cases:

Case 1) The characteristic of $K$ is 2.

In this case notice that $$ (1+E_{i,j})^2 = 1+2E_{i,j}=1, $$ so $\varphi (1+E_{i,j})^2 = 1$, and we see that $\varphi (1+E_{i,j})$ is the unique solution of the polynomial equation $x^2=1$, namely 1 (recall that $1=-1$ here).

Case 2) The characteristic of $K$ is not 2.

In this case we have that $$ (1+E_{i,j})^2 = 1+2E_{i,j}, $$ and since $2\neq 0$, we have by $(2) $ that $$ \varphi (1+E_{i,j})^2 = \varphi (1+2E_{i,j}) = \varphi (1+E_{i,j}). $$ Noticing that $1+E_{i,j}$ is invertible (with inverse $1-E_{i,j}$), and hence that $\varphi (1+E_{i,j})\neq 0$, we deduce that $\varphi (1+E_{i,j})$ is the unique nonzero solution of the polynomial equation $x^2=x$, namely 1.

This takes care of claim $(1)$ for any nonzero $a$, but if $a=0$, the claim simply states that $\varphi (1)=1$, which follows immediately from the hypothesis (choosing $P$ to be any rank one projection and $a=1$).

Next consider the subgroup $H\subseteq GL_n(K)$ generated by the union of the following two sets $$ \big \{a E_{i, i}+I-E_{i, i}: a\in K^\times , \ 1\leq i\leq n\big \}, $$ and $$ \big \{ 1+aE_{i,j}: a\in K, \ 1\leq i,j\leq n,\ i\neq j\big \}. $$

Observe that the hypothesis together with $(1)$ imply that $\varphi $ coincides with the determinant on the generators of $H$, and hence $$ \varphi (U)=\text{det}(U), \quad\forall \, U\in H.\tag{3} $$

We then claim that if $A$ is any $n\times n$ matrix, and $A'$ is the matrix obtained from $A$ by any one of the following so called elementary row operations, then there exists some $U\in H$ such that $UA=A'$.

The operations are:

a) Replacing the $i^{th}$ row of $A$ with itself plus $\lambda $ times the $j^{th}$ row, where $i\neq j$, and $\lambda \in K$.

b) Multiplying the $i^{th}$ row of $A$ by a nonzero $\lambda \in K$.

c) Swapping the $i^{th}$ row of $A$ with the $j^{th}$ row.

In order to verify the claim under (a), it is enough to take $U=I+\lambda E_{i,j}$. Under (b) one takes the diagonal matrix $U=\lambda E_{i,i}+I-E_{i,i}$, so it remains to check the claim under (c).

Defining $$ \Sigma _{i, j} = (1+E_{i,j})(1-E_{j,i})(1+E_{i,j}), $$ a simple computation gives $$ \Sigma _{i, j} = 1 - E_{i,i} - E_{j,j} + E_{i,j} - E_{j,i}. \tag{4} $$

For example, in the case of $3\times 3$ matrices, if $i=2$, and $j=1$, this becomes $$ \Sigma _{2, 1} = $$$$ = \pmatrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 } \pmatrix{ 1 & -1 & 0 \cr 0 & 1 & 0 \cr 0 & 0 & 1 } \pmatrix{ 1 & 0 & 0 \cr 1 & 1 & 0 \cr 0 & 0 & 1 } = $$$$ = \pmatrix{ 0 & -1 & 0 \cr 1 & 0 & 0 \cr 0 & 0 & 1 }. $$ The computation in $(4)$ amounts to saying that $\Sigma _{i,j}$ is the $n\times n$ matrix that coincides with the identity matrix everywhere outside the $2\times 2$ sub-matrix formed by the rows and columns with indices $i$ and $j$, where it instead looks like $ \pmatrix{ 0 & -1\cr 1 & 0 }. $

Moreover, given any matrix $A$, the matrix $\Sigma _{i,j}A$ is easily seen to be matrix obtained from $A$ by swapping the $i^{th}$ and $j^{th}$ rows, followed by a change of sign of the $i^{th}$ row (which was formerly known as the $j^{th}$ row). This unwanted change of sign can clearly be undone by further multiplying $A$ on the left by $$ -E_{i,i}+I-E_{i,i}, $$ so the claim is proved.

We then see that all of the steps needed to bring $A$ to its reduced row echelon form can be performed by multiplying $A$ on the left by some member of the subgroup $H$. This implies that, if $A'$ is now the reduced row echelon form of $A$, then there exists some $U$ in $H$ such that $UA=A'$.

This allows us to conclude the proof that $\varphi $ coincides with the determinant, as follows:

Case 1) $A$ is invertible and hence $A'$ is the identity.

As seen above, there is some $U$ in $H$ such that $UA=I$, whence $A=U^{-1}\in H$, so the conclusion follows from $(3)$.

Incidentally, it is interesting to observe that we have just shown that $H=GL_n(K)$!

Case 2) $A$ is not invertible and hence the last row of $A'$ is identically equal to zero.

In this case $E_{n,n}A' =0$, so $$ A' = (I-E_{n,n})A' = (0E_{n,n}+I-E_{n,n})A', $$ whence $$ \varphi (U)\varphi (A)=\varphi (UA)=\varphi (A') = \varphi \big ((0E_{n,n}+I-E_{n,n})A'\big )= $$$$= \varphi (0E_{n,n}+I-E_{n,n})\varphi (A') =$$$$= 0\varphi (A') = 0, $$ so $$ \varphi (A)=0 = \text{det}(A). $$ QED.

I'd like to thank all users who gave important feedback to earlier versions of this result, including, but not limited to, @math54321 and @Aaratrick.

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Pedro
    Jun 24 at 2:52
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An endomorphism $T$ of a vector space $E$ yields endomorphisms on various vector spaces deduced from $E$. For instance on the dual space $E'$, the space of linear forms on $E$, the endomorphism $T$ induces the transposition $\hbox{}^tT$ of $T$ defined by $(\hbox{}^tTf)(x) = f(T(x))$. Similar constructs are possible on cartesian powers of $E$, tensor products, symmetric and alternate powers of $E$ and there are a great many functorial constructions besides the few ones I quoted. Look up Schur's polynomial and plethysm in a representation theory book to find more of them.

When $E$ has finite dimension $n$, the $n$-th external power $\Lambda^n E$ of $E$ is a line and the induced endomorphism $\Lambda^n T$ on $\Lambda^n E$ is therefore a dilatation. The determinant of $T$ is (definition) the coefficient of this dilatation.

With a little bit of care, this definition works well when $E$ is a module (when coefficients are in a ring rather than in a field). This is pretty much standard algebra, it's the way determinants are introduced in Algèbre générale by Denis Allouch and Bernard Charles and I am confident there are plenty of good references in other languages.

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    $\begingroup$ Yes, this is the most explanatory description of determinant, in my opinion. Of course, this viewpoint requires proving that the $n$th exterior of an $n$-dimensional vector space is not $\{0\}$. :) $\endgroup$ Jun 20 at 21:43
  • $\begingroup$ @paulgarrett IIRC Algèbre générale exhibits a basis for $\Lambda^kE$ associated to a basis of $E$ (I mean, the basis we are used to :-) ) and deduce its dimension. This requires a bit of preliminary work on the symmetric group and tensor products. $\endgroup$ Jun 23 at 21:40
  • $\begingroup$ Yes, and, again, the spanning is easy, but the linear independence is not completely trivial. Far easier than the analogous Poincare-Birkhoff-Witt linear independence, yes, but, still, not trivial. :) $\endgroup$ Jun 23 at 21:47
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This makes no reference to a basis, but only works because C is algebraically closed.

Not really. Every field has an extension that's algebraically closed, so you can do the math in the extension, and the result is guaranteed to be in the original field (this can be proven either by noting that the result is the same as the basis-dependent result, or by noting that the result must be unchanged by all of the members of the Galois group of the extension).

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    $\begingroup$ Can you elaborate a little on your last point? In particular, why must the result be unchanged by members of the Galois group? $\endgroup$
    – WillG
    Jun 21 at 3:40
  • $\begingroup$ If you can describe something using only terms available in field K, then the Galois group of the closure acts on the set of these “something”. If furthermore this something is unique, then it is stabilised by the Galois group and in favourable cases, its coefficients are in K. $\endgroup$ Jun 21 at 21:04
  • $\begingroup$ @MichaëlLeBarbier Let $\overline{K}$ be the algebraic closure of $K$, and define $\det{T}$ to be the product of the eigenvalues of $T$, viewed as a linear map on the vector space over $\overline{K}$ instead of $K$. Now how do we show $\det{T}\in K$? One route is to prove that this definition is equivalent to any of the others, for which automatically $\det{T}\in K$. But I don't understand this other route. How do we know the product of these eigenvalues is stabilized by the Galois group? $\endgroup$
    – WillG
    Jun 22 at 5:16
  • $\begingroup$ My knowledge of Galois theory is pretty sketchy, by the way, so maybe this is just something I can't see until learning more of it. I guess the argument is that (for some reason) the Galois group only permutes the eigenvalues of $T$, hence their product is unaffected by the Galois group, hence (for some reason) their product lies in $K$? $\endgroup$
    – WillG
    Jun 22 at 5:39
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    $\begingroup$ @WillG Yes. “For some reason #1” is a generic argument when working with group actions: if I can construct Y speaking only about X and a group operates on my situation, the the stabiliser of X operates on the set of possible Y's (they are stabilsied too). “For some reason #2” is the Galois correspondance, which has a few gotchas in finite characteristic. Stewart's book on Galois theory is a very good and accessible reference for this topic! $\endgroup$ Jun 22 at 6:33
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  • Giving an endomorphism $T:V\to V$ of a dimensional vector space over a field $k$ is the same as making $V$ a module over the polynomial ring $k[t]$. (Edit/Clarification: A polynomial $p(t)$ acts on $V$ by $p(T):V\to V$ under this identification.)

  • This is a finitely generated torsion module if (and only if) $V$ is finite dimensional.

  • By the structure theorem of finitely generated modules over $k[t]$ we see that $V$ is uniquely expressible as direct sum $\oplus_i k[t]/(p_i(t))$ where $p_i(t)$ divides $p_{i+1}(t)$ and all are monic polynomials.

  • The determinant of $T$ is defined as $(-1)^n\prod_i p_i(0)$ where $n=\dim(V)$.

The proof that this is the "usual" determinant can be given by applying @Ruy's argument to the matrix $t\cdot 1-A$ and using the elementary operations (over $k[t]$) to reduce this to the Smith Normal Form.

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  • $\begingroup$ Can you elaborate on the first point? I understand the objects you reference, but I don't understand how endomorphisms on $V$ make $V$ into a module over $k[t]$. $\endgroup$
    – WillG
    Jun 23 at 20:12
  • $\begingroup$ Thanks for the edit. I haven't studied module theory, but now I feel motivated to learn the structure theorem you mentioned. Two questions: (1) Is $\prod_ip_i(t)$ the characteristic polynomial? (2) Does this definition break if $k$ is only a commutative ring? (I think yes, because it the structure theorem requires $k[t]$ to be a principle ideal domain, and it is claimed here that this is true iff $k$ is a field.) $\endgroup$
    – WillG
    Jun 24 at 22:44
  • $\begingroup$ @WillG Indeed $\prod_i p_i(t)$ is the characteristic polynomial. The structure theorem does not work for $k$ a commutative ring. So this method does not work to define the determinant of an endomorphism of a free/projective module over a commutative ring. $\endgroup$
    – Kapil
    Jun 25 at 1:38
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See my question here: Does any normalized function $D$ other than determinant of matrix satisfy $D(AB) = D(A)D(B)$?

I think it might be the case that the determinant is the unique function satisfying the following. Suppose $V$ is an $n$-dimensional vector field over $\mathbb{K}$ and $M_{n\times n}(\mathbb{K}$) is the set of $n\times n$ matrices over $\mathbb{K}$. Let $D:M_{n\times n}(\mathbb{K}) \to \mathbb{K}$. Let $k\in \mathbb{K}$.

(1) $D(I)=1$

(2) $D(AB) = D(A)D(B)$

(3) $D(kA) = k^n D$

(4) $D$ is continuous

Unfortunately, similar to the answer by Ruy, I think a basis-dependent version of the determinant is required to prove existence of such a function.

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    $\begingroup$ What do you mean by "continuous", when $\mathbb K$ is an arbitrary field? $\endgroup$
    – mr_e_man
    Jun 21 at 16:23
  • $\begingroup$ Did you notice that if you restrict your application $D$ to GLn(K) you get a character of the group? The character theory of GLn is quite well-known nowadays. :-) (Beware of the Frobenius morphism on finite fields!) $\endgroup$ Jun 21 at 17:45
  • $\begingroup$ @mr_e_man I don’t know. I don’t know if that constraint is required even for fields where it makes sense. Thanks for pointing that out. $\endgroup$
    – Jagerber48
    Jun 21 at 18:08
  • $\begingroup$ @MichaëlLeBarbier thanks for pointing that out. I’ll research more about that. Is The determinant a unique extension of the character to all matrices or something? $\endgroup$
    – Jagerber48
    Jun 21 at 18:10
  • $\begingroup$ @Jagerber48 In the most favorable cases the group of characters of GLn is isomorphic to Z and has therefore two generators. One can be extended to the matrix algebra and this is the determinant. Another way to look at this is to study the operation of GLn on n-homogeneous forms on M(nxn). All of these ideas are revolving around the one I exposed in my answer, only I removed all the representation theory context to describe it with elementary algebra. $\endgroup$ Jun 21 at 19:34

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