Determinants (version 6)

For now only for real regular (invertible) matrices.

Throughout, $R$ denotes the real numbers and $R^* = R \setminus \{0\}$, regarded as a multiplicative group.

1. Homomorphisms $R^* \rightarrow R^*$¶

Definition. $R^* = R \setminus \{0\}$, regarded as a multiplicative group.

A homomorphism $R^* \rightarrow R^*$ is any real function $g$ such that

Task. We want to determine the form of all homomorphisms $R^* \rightarrow R^*$.

Theorem 1. Let $g$ be a homomorphism $R^* \rightarrow R^*$ that is regular (continuous, or merely measurable / monotone on $R^*$). Then either

where $\alpha \in R$.

Proof. See the separate text, chapters 2 and 3. (This should eventually be merged into a single proof for $R^*$ as a multiplicative group, without reference to the additive notation of chapter 2.)

Remark on regularity. The conclusion of Theorem 1 requires a regularity hypothesis. Without it, a Hamel-basis construction produces pathological multiplicative homomorphisms of $R^*$ that are not of the stated form. We assume regularity throughout. $\qquad\Box$

2. Restriction of homomorphisms $GL(n) \rightarrow R^*$ to $GD(n)$¶

Definitions.

• $GL(n)$ is the group of all real regular (invertible) matrices.

• $GD(n)$ is the group of all real regular diagonal matrices; it is a subgroup of $GL(n)$.

A homomorphism $GL(n) \rightarrow R^*$ is any function $f$ such that

Proposition. Let $g$ be a homomorphism $GL(n) \rightarrow R^*$. Then $g(I) = 1$ and $g(A^{-1}) = g(A)^{-1}$.

Proof.

Remark (conjugation invariance). For any homomorphism,

where we used that the codomain $R^*$ is commutative and that $f(I)=1$ from the previous Proposition.

Theorem 2. Let $f : GD(n) \rightarrow R^*$ be a homomorphism. Then $f$ can be expressed through a homomorphism $g : R^* \rightarrow R^*$ via

Proof. Use the permutation matrices $P_i$ that swap rows/columns 1 and $i$. $P_i$ arises from $I$ by swapping rows 1 and $i$ (equivalently columns 1 and $i$ — the result is the same). We have $P_i^{-1} = P_i$; left multiplication by $P_i$ swaps rows, right multiplication swaps columns, so for a diagonal matrix $D$ the operation $P_i\,D\,P_i^{-1}$ swaps the 1st and $i$th diagonal entries.

where the third equality uses the conjugation invariance (2), and the last equality uses the homomorphism property together with

The map $g$ is a homomorphism: it is defined by $g(x) = f\big(\operatorname{diag}(x, 1, \dots, 1)\big)$, so

Corollary. If $f$ is a homomorphism $GD(n) \rightarrow R^*$, then by Theorem 2 it can be written as $f(D) \equiv g(\prod_i d_i)$ with $g$ a homomorphism $R^* \rightarrow R^*$. Hence, by Theorem 1, either

or

where $\alpha \in R$.

3. Homomorphisms $GL(n) \rightarrow R^*$¶

Task. Find all homomorphisms $GL(n) \rightarrow R^*$.

Theorem 3. Every homomorphism on $GL(n)$ is uniquely determined by its restriction to the subgroup $GD(n)$.

Proof.

(a) If $C$ is diagonalizable over $R$, then $C = P\,D\,P^{-1}$ with $D$ diagonal and $P$ invertible, and by (2)

The diagonal matrix $D$ is determined up to the order of its diagonal entries (which are the eigenvalues of $C$), so the number $\prod_i d_i$ is uniquely determined. Thus $f(C)$ depends only on $f\restriction_{GD(n)}$.

(b) For a general $A \in GL(n)$ use the polar decomposition $A = H\,U$, where $H = (A A^{\mathsf T})^{1/2}$ is symmetric positive definite and $U = H^{-1}A$ is orthogonal. Then

• $H$ is symmetric, hence orthogonally diagonalizable over $R$: $H = Q\,D_H\,Q^{\mathsf T}$ with $Q$ orthogonal and $D_H \in GD(n)$ (positive diagonal entries). By part (a), $f(H) = f(D_H)$, which is determined by $f\restriction_{GD(n)}$.

• $U$ is orthogonal. In general an orthogonal matrix is not diagonalizable over $R$ (e.g. a rotation has complex eigenvalues), so we do not diagonalize $U$ directly. Instead, by the Cartan–Dieudonné theorem, $U$ is a product of at most $n$ reflections in hyperplanes,

  $$U = S_1\,S_2 \cdots S_k.$$

  (15)

  Each reflection $S_j$ is symmetric and orthogonal with eigenvalues $(1,\dots,1,-1)$, hence orthogonally diagonalizable over $R$:

  $$S_j = Q_j\,\operatorname{diag}(1,\dots,1,-1)\,Q_j^{-1}.$$

  (16)

  By part (a), $f(S_j) = f\big(\operatorname{diag}(1,\dots,1,-1)\big) =: r$, a value fixed by $f\restriction_{GD(n)}$. Since $\operatorname{diag}(1,\dots,1,-1)^2 = I$ we have $r^2 = f(I) = 1$, so $r = \pm 1$. Therefore

  $$f(U) = r^{\,k},$$

  (17)

  which depends only on the parity of $k$, i.e. only on $\det U = (-1)^k = \pm 1$, and hence is determined by $f\restriction_{GD(n)}$.

Combining the two factors,

which is completely determined by the restriction $f\restriction_{GD(n)}$ (through the two values $f(D_H)$ and $r = f(\operatorname{diag}(1,\dots,1,-1))$). This argument stays entirely within the real numbers; it does not require complex diagonalization.

Hence the extension of a homomorphism from $GD(n)$ to $GL(n)$ is unique. $\qquad\Box$

Conclusion (using Theorem 1). Write $\prod_i d_i := \det A$ (the common value of the diagonal product in any diagonal form obtained above; note $\det H = |\det A|$ and $\det U = (-1)^k$). All homomorphisms $GL(n) \rightarrow R^*$ then have the form: either

or

where $\alpha \in R$. Indeed, with $f(D_H) = g(|\det A|)$ and $r = g(-1)$:

• case $g(x) = |x|^{\alpha}$ gives $r = 1$ and $f(A) = |\det A|^{\alpha}$, i.e. (h11);

• case $g(x) = \operatorname{sign}(x)|x|^{\alpha}$ gives $r = -1$ and $f(A) = |\det A|^{\alpha}(-1)^k = \operatorname{sign}(\det A)\,|\det A|^{\alpha}$, i.e. (h22),

using $\operatorname{sign}(\det A) = \operatorname{sign}(\det U) = (-1)^k$.

4. Homogeneous homomorphisms $GL(n) \rightarrow R^*$¶

Definition. A homomorphism $f : GL(n) \rightarrow R^*$ is called homogeneous (of degree $n$) if for all $\lambda \in R^*$

Remark. From this definition, $f(\lambda A) = f((\lambda I) A) = f(\lambda I)\,f(A) = \lambda^{n} f(A)$.

Definition. A homomorphism $f : GL(n) \rightarrow R^*$ is called positively homogeneous if equality (homo) holds for all $\lambda > 0$.

Remark.

• For even $n$ the two definitions coincide.

• For odd $n$ positive homogeneity is only a necessary condition for homogeneity.

Theorem 4.1. Both homomorphisms (h11) and (h22) are positively homogeneous on $GL(n)$ if and only if $\alpha = 1$.

Proof (verifying (homo) for $\lambda > 0$, where $\prod_i d_i = \lambda^n$):

respectively

For all $\lambda > 0$ we have $(\lambda^{n})^{\alpha} = \lambda^{n\alpha}$, which equals $\lambda^{n}$ for all $\lambda$ iff $\alpha = 1$. $\qquad\Box$

Theorem 4.2. Every homogeneous homomorphism $f : GL(n) \rightarrow R^*$ has, for even $n$, either the form

or

and for odd $n$ only the form (h222).

Proof.

(a) Without Theorem 1. For any $\lambda$, Theorem 2 gives

At the same time, by homogeneity $f(\lambda D) = \lambda^{n} f(D)$. Since $f(D) \neq 0$, we get

• For odd $n$, the map $x \mapsto x^n$ is injective on $R$, so $g(\lambda) = \lambda$.

• For even $n$, this gives $g(\lambda) = \pm\lambda$ pointwise. Writing $g(\lambda) = \lambda\,s(\lambda)$ with $s(\lambda) = \pm 1$, the homomorphism property forces $s(\lambda\mu) = s(\lambda)s(\mu)$, i.e. $s$ is a homomorphism $R^* \to \{\pm 1\}$. There are exactly two of these — $s \equiv 1$ and $s = \operatorname{sign}$ — and no regularity hypothesis is needed to see this: $R_{>0}$ is divisible (every positive real is a square), so any homomorphism $R_{>0} \to \{\pm 1\}$ is trivial; hence $s$ is determined by $s(-1) \in \{\pm 1\}$. This gives $g(\lambda) = \lambda$ or $g(\lambda) = |\lambda|$.

Hence $f(D) = \prod_i d_i$ or $f(D) = |\prod_i d_i|$.

(b) Using Theorem 1. For even $n$ the claim follows from Theorem 4.1 together with the fact that for even $n$ positive homogeneity implies homogeneity. For odd $n$ the case $\lambda^{n} < 0$ can occur (when $\lambda < 0$), which excludes variant (h111) (its values are always non-negative, whereas $f(\lambda I) = \lambda^n$ must be negative); homogeneity again forces $\alpha = 1$ via Theorem 4.1, leaving only (h222). $\qquad\Box$

5. Definition of the determinant on $GL(n)$¶

On $GD(n)$ we define the determinant as a homogeneous homomorphism $GD(n) \rightarrow R^*$.

By Theorem 4.2:

• For odd $n$ the definition is unique, and

  $$\det(D) = \textstyle\prod_i d_i. \tag{detD}$$

  (28)

  This agrees with the standard definition of the determinant.

• For even $n$ the homogeneous homomorphism is not unique: Theorem 4.2 leaves the two options (h111) and (h222). To fix the standard determinant we add one further condition.

Additional condition (reflections, fixing the even-$n$ case). We require that a pure reflection in a hyperplane has determinant -1. A reflection in a hyperplane is a diagonal matrix with all entries equal to 1 except a single -1, i.e. $\operatorname{diag}(1,\dots,1,-1)$ (any hyperplane reflection is orthogonally conjugate to this one, so by the conjugation invariance (2) the value does not depend on the chosen hyperplane). We impose

This selects the correct branch:

• option (h111) gives $\big|\prod_i d_i\big| = |-1| = +1$;

• option (h222) gives $\prod_i d_i = -1$.

So the condition picks out (h222) $= \prod_i d_i$, the standard determinant, and makes the definition on $GD(n)$ unique for even $n$ as well.

Odd vs. even $n$. For odd $n$ this condition is automatically satisfied: it follows from homogeneity (not merely from being a homomorphism). Indeed, a bare homomorphism may give +1 on a reflection even for odd $n$ — for instance $f = |\det|^{\alpha}$, or the trivial $f \equiv 1$ — but Theorem 4.2 shows that the unique homogeneous homomorphism for odd $n$ is $\prod_i d_i$, which already gives -1. For even $n$ both options (h111) and (h222) are homogeneous, so the reflection condition is a genuine extra assumption needed to single out the determinant.

On $GL(n)$ we define the determinant as the extension from the subgroup $GD(n)$, which by Theorem 3 is unique and given by

where $D$ is any diagonal form obtained in (AD).

Theorem 5. The above definition of the determinant on $GL(n)$ agrees with the standard definition.

Proof. With the reflection condition fixing the even-$n$ branch, the standard determinant and our determinant agree on $GD(n)$ (both equal $\prod_i d_i$ there). Both the standard determinant and our determinant are homomorphisms $GL(n) \rightarrow R^*$ extending this common restriction from $GD(n)$ to $GL(n)$. By Theorem 3, the extension of a homomorphism from $GD(n)$ to $GL(n)$ is unique. Hence the two definitions must coincide. $\qquad\Box$

6. Passing from $R^*$ to $C^*$¶

We now redo the construction with complex matrices $GL(n,C)$ and codomain $C^*$. The pleasant surprise is that the theory becomes simpler: the even/odd dichotomy and the reflection condition of Section 5 both disappear, and the determinant becomes the unique homogeneous homomorphism for every $n$.

Why the even/odd split disappears¶

It is tempting to attribute this to “complex continuity being a stronger assumption”, but that is not the reason. The regularity hypothesis (continuity / measurability) plays exactly the same role over $C$ as it does over $R$ in Theorem 1: it only serves to exclude the pathological Hamel-basis homomorphisms. What really changes is the algebra and topology of the group $K^*$ itself ($K = R$ or $C$).

Recall where the ambiguity came from (Theorem 4.2): homogeneity forces, for the underlying homomorphism $g : K^* \to K^*$,

If $g_1, g_2$ are two solutions, their ratio $\chi = g_1/g_2$ satisfies $\chi(\lambda)^n = 1$, i.e. $\chi$ is a homomorphism into the group $\mu_n(K)$ of $n$-th roots of unity in $K$. Hence the set of homogeneous-compatible $g$ is a torsor under

and the determinant is unique iff this group is trivial. Two equivalent ways to see what it is:

• Algebraically: $g(\lambda)^n = \lambda^n$ only constrains $g$ on the subgroup $(K^*)^n$ of $n$-th powers, and leaves it free on $K^*/(K^*)^n$.

  • $K = R$: $(R^*)^n = R^*$ for odd $n$ (odd powers are surjective), but $(R^*)^n = R_{>0}$ for even $n$ — a proper subgroup. So $R^*/(R^*)^n$ is trivial for odd $n$ and $\cong \{\pm 1\}$ for even $n$. The leftover freedom is precisely the sign character, i.e. the choice $g(\lambda)=\lambda$ vs. $g(\lambda)=|\lambda|$, equivalently $\det$ vs. $|\det|$.

  • $K = C$: every nonzero complex number has an $n$-th root, so $(C^*)^n = C^*$ and $C^*/(C^*)^n$ is trivial for every $n$. No freedom remains.

• Topologically: $\operatorname{Hom}_{\mathrm{cont}}(K^*, \mu_n)$ consists of continuous maps from $K^*$ into a discrete finite group $\mu_n$, which must be constant on connected components. So it is nontrivial only when $K^*$ is disconnected.

  • $R^*$ has two connected components (±), and the nontrivial character is exactly $\operatorname{sign}$ — present only when $\mu_n \ni -1$, i.e. for even $n$.

  • $C^*$ is connected, so the only such character is trivial, for all $n$.

In one sentence: the even/odd ambiguity is the sign character of the disconnected group $R^*$; over the connected group $C^*$ there is no sign to choose, so it vanishes. This is a statement about -1 being a square in $C$ but not in $R$, not about the strength of the continuity assumption.

The construction over $C$¶

(i) Diagonalization now works directly. Over $C$, the polar decomposition $A = H\,U$ has $H = (A A^{*})^{1/2}$ Hermitian positive definite and $U = H^{-1}A$ unitary. Both are normal, hence unitarily diagonalizable over $C$:

So the diagonalization proof of Theorem 3 goes through verbatim — the very step that failed over $R$ (a real orthogonal matrix need not be $R$-diagonalizable) is fine over $C$. This is exactly what was meant by “checking that $R^*$ can be replaced by $C^*$”; it is an independent alternative to the Cartan–Dieudonné fix used in Section 3.

(ii) Classification of homomorphisms $C^* \rightarrow C^*$ (analog of Theorem 1). Writing $z = |z|\,e^{i\theta}$, the continuous homomorphisms are exactly

If one additionally requires $g$ to be holomorphic, then $c = k$ and $g(z) = z^{k}$, $k \in \mathbb{Z}$.

(iii) Conceptual shortcut. For any infinite field $K$ one has $[GL(n,K), GL(n,K)] = SL(n,K)$, so the determinant induces an isomorphism $GL(n,K)^{\mathrm{ab}} \cong K^*$. Consequently every homomorphism $f : GL(n,K) \rightarrow K^*$ factors uniquely as $f = g \circ \det$ with $g : K^* \rightarrow K^*$. The elementary “restrict to $GD(n)$, classify, extend” program of Sections 2–3 simply rebuilds this factorization with the regularity needed to write down $g$ explicitly.

Theorem 6. Over $C$ the determinant is the unique homogeneous homomorphism $GL(n,C) \rightarrow C^*$, for every $n$.

Proof. By (iii) (or, elementarily, by (i) together with Theorem 2), $f = g \circ \det$ for some homomorphism $g : C^* \rightarrow C^*$. Homogeneity gives $g(\lambda^n) = g(\det(\lambda I)) = f(\lambda I) = \lambda^n$ for all $\lambda \in C^*$. Since $\lambda \mapsto \lambda^n$ is surjective onto $C^*$, every $w \in C^*$ is of the form $\lambda^n$, so $g(w) = w$ for all $w$, i.e. $g = \mathrm{id}$ and $f = \det$. No reflection normalization is needed.

Note this argument is purely algebraic: it uses neither the classification (ii) nor any regularity (continuity / measurability) hypothesis on $g$. The same is in fact true over $R$ — the homogeneous characterization never needs regularity, because homogeneity already pins $g$ to the identity on the $n$-th powers $(K^*)^n$: for any $w$ admitting an $n$-th root $\lambda = w^{1/n}$ one has $g(w) = g(\lambda^n) = \lambda^n = w$, with no appeal to regularity. (Regularity enters only in Theorem 1, which classifies the non-homogeneous homomorphisms, where $g$ on $R_{>0}$ would otherwise range over a wild Hamel-basis family.)

The genuine $C$-versus-$R$ difference is therefore not regularity but whether $\lambda \mapsto \lambda^n$ is surjective, i.e. whether $K^*/(K^*)^n$ is trivial. Over $C$ every $w$ — including $-1 = i^2$ — has an $n$-th root, so $g(w) = w$ is forced everywhere and the determinant is unique even among abstract homomorphisms. Over $R$ with even $n$, the value -1 has no real $n$-th root, so $g(-1) = \pm 1$ escapes the homogeneity constraint; that single escaped sign is exactly the $\det$-versus-$|\det|$ ambiguity, removed by the reflection condition of Section 5. $\qquad\Box$

Example ($n = 2$). Take a homogeneous homomorphism $f$ of degree 2, so $f(\lambda I) = \lambda^2$, and write $f = g \circ \det$.

• Positive values are forced (over $R$ and $C$ alike). For $w > 0$, choosing $\lambda = \sqrt{w}$ gives $g(w) = g(\lambda^2) = \lambda^2 = w$. So $g$ is the identity on $R_{>0}$ no matter what.

• Over $R$ the value at -1 is free. No real $\lambda$ satisfies $\lambda^2 = -1$, so homogeneity says nothing about $g(-1)$; the homomorphism law only gives $g(-1)^2 = g(1) = 1$, hence $g(-1) = \pm 1$. Both choices are genuine homogeneous homomorphisms:

  • $g(-1) = +1 \Rightarrow g = |\cdot| \Rightarrow f = |\det|$, and indeed $f(\lambda I) = |\lambda^2| = \lambda^2$; $\checkmark$

  • $g(-1) = -1 \Rightarrow g = \mathrm{id} \Rightarrow f = \det$, and $f(\lambda I) = \lambda^2$. $\checkmark$

  A reflection (e.g. $\operatorname{diag}(1,-1)$) has $\det = -1$ but $|\det| = +1$, so the reflection condition $f = -1$ picks out $f = \det$.

• Over $C$ the value at -1 is forced. Now $\lambda = i$ satisfies $\lambda^2 = -1$, so

  $$g(-1) = g(i^2) = i^2 = -1$$

  (35)

  is no longer a free choice — the existence of $\sqrt{-1}$ collapses the two real options into one. Hence $f = \det$ is the unique degree-2 homogeneous homomorphism over $C$, with no extra condition.

Remark (the role of the homomorphism property — why $C$ has fewer solutions, not more). A common confusion: if one merely solves the pointwise equation $g(\lambda)^n = \lambda^n$ for an arbitrary function $g$, then writing $g(\lambda) = \lambda\,\omega(\lambda)$ one needs $\omega(\lambda)^n = 1$, i.e. $\omega(\lambda)$ is some $n$-th root of unity — and over $C$ one may pick any of the $n$ roots independently at each point $\lambda$, giving an enormous (mostly discontinuous) solution set. From this angle $C$ seems to have more solutions than $R$ (where only $\pm1$ are available, or just +1 for odd $n$).

The resolution is that $g$ is not an arbitrary function: it is a group homomorphism, $g(\lambda\mu) = g(\lambda)g(\mu)$. Feeding $g(\lambda) = \lambda\,\omega(\lambda)$ into this law gives

So $\omega$ is itself a homomorphism $K^* \to \mu_n$ — the per-point root-of-unity choices are forced to vary multiplicatively, not freely. The number of homogeneous homomorphisms is therefore exactly $\#\operatorname{Hom}(K^*, \mu_n)$, and this is where the divisibility of $K^*$ takes over:

Lemma. If $G$ is $n$-divisible (every element is an $n$-th power) and $A$ has exponent dividing $n$, then the only homomorphism $G \to A$ is trivial. Proof: any $g = h^n$, so $\varphi(g) = \varphi(h)^n = 1$. $\;\Box$

• $C^*$ is $n$-divisible (every nonzero complex number has an $n$-th root), so $\operatorname{Hom}(C^*, \mu_n) = \{1\}$: the only $\omega$ is trivial, $g = \mathrm{id}$, unique.

• $R^*$ with even $n$ is not 2-divisible (negatives are not squares), so $\operatorname{Hom}(R^*, \mu_n) = \operatorname{Hom}(R^*, \{\pm1\}) = \{\text{trivial}, \operatorname{sign}\}$: two solutions, with $\omega = \operatorname{sign}$ giving $g(\lambda) = |\lambda|$ (the $|\det|$ branch).

• $R^*$ with odd $n$ is $n$-divisible, so again only the trivial $\omega$, $g = \mathrm{id}$.

So the apparent abundance of complex solutions is an artifact of dropping multiplicativity. With the homomorphism law reinstated, the $n$-divisibility of $C^*$ rigidifies $\omega$ to the constant 1, and $C$ yields exactly one solution — fewer than $R$, not more.

7. Extension to singular matrices¶

The obstruction. The homomorphism framework cannot, by itself, reach singular matrices: the codomain is the group $R^*$ (or $C^*$), and a group homomorphism never takes the value $0 \notin R^*$. To include singular matrices we must change the setting — either algebraically (enlarge to a monoid) or topologically (use density). Both routes agree, and both single out the genuine polynomial determinant.

(A) Algebraic route: the multiplicative monoid $M_n$¶

Drop invertibility and ask for $f(AB) = f(A)\,f(B)$ for all $A, B \in M_n$ (the full multiplicative monoid of $n \times n$ matrices). The standard determinant satisfies this, since $\det(AB) = \det(A)\det(B)$ holds for all square matrices.

Theorem 7. Let $f : M_n \rightarrow R$ be multiplicative on all of $M_n$, restricting on $GL(n)$ to a homomorphism that is non-constant on the diagonal matrices (e.g. any homogeneous determinant). Then $f$ vanishes on every singular matrix, and this extension is unique.

Proof. Any matrix of rank $r$ has the normal form $A = P\,E_r\,Q$ with $P, Q \in GL(n)$ and $E_r = \operatorname{diag}(1, \dots, 1, 0, \dots, 0)$ ($r$ ones), so $f(A) = f(P)\,f(E_r)\,f(Q)$; thus $f$ on singular matrices is determined by the idempotents $E_r$, $r < n$. Each $E_r$ is idempotent, $E_r^2 = E_r$, so $f(E_r)^2 = f(E_r)$ and $f(E_r) \in \{0, 1\}$. Finally, from $E_{n-1}\,\operatorname{diag}(1, \dots, 1, t) = E_{n-1}$ we get

As $f$ is non-constant on these diagonals (e.g. $f(\operatorname{diag}(1,\dots,1,t)) = t \neq 1$ for the determinant), this forces $f(E_{n-1}) = 0$, and then $f(E_r) = f(E_r)\,f(E_{n-1}) = 0$ for every $r < n$. Hence $f \equiv 0$ on all singular matrices. $\qquad\Box$

So for the determinant the value 0 on singular matrices is forced — no continuity argument is even required.

(B) Topological route: density of $GL(n)$¶

The set of singular matrices is the zero set of the polynomial $\det$, a closed subset of measure zero; for any $A$, the matrix $A + \varepsilon I$ is invertible for all but finitely many $\varepsilon$. Hence $GL(n)$ is dense in $M_n$. The genuine determinant is a polynomial, therefore the unique continuous extension of $\det\restriction_{GL(n)}$ to $M_n$, and it equals 0 on singular matrices. This is the “continuity argument”, and it agrees with the algebraic one.

Remarks¶

• Only the genuine determinant extends polynomially. Among the homomorphisms $|\det A|^{\alpha}$ ($\alpha \ne 1$) some extend continuously to 0 (those with $\alpha > 0$) and some blow up ($\alpha < 0$); the trivial $g \equiv 1$ even admits two monoid extensions (the constant 1, or the indicator of $GL(n)$). Only $\alpha = 1$, $k = 1$ — the genuine, sign-preserving determinant — extends as a polynomial. This singles it out canonically.

• Cleanest picture (combining Sections 6 and 7). Over $C$, the determinant is the unique homogeneous multiplicative map $M_n(C) \rightarrow (C, \cdot)$ that vanishes on singular matrices, for all $n$ — no even/odd split, no reflection normalization, no separate continuity hypothesis needed.