Does any normalized function $D$ other than determinant of matrix satisfy $D(AB) = D(A)D(B)$?

Let $$M(n, \mathbb{C})$$ be the set of $$n\times n$$ matrices over $$\mathbb{C}$$. Let

$$D:M(n, \mathbb{C}) \rightarrow \mathbb{C}$$

Suppose $$D$$ satisfies

• $$D(I) = 1$$
• $$D(AB) = D(A)D(B)$$

Is it possible to prove that $$D = \text{det}$$ where $$\text{det}$$ is the usual determinant?

If not then

(1) can I see a counter-example and

(2) can we add additional, coordinate-free, conditions to $$D$$ to make it so?

edit: comments quickly pointed out some counter-examples including $$D(A) = \text{det}(A^k)$$ for positive integer $$k$$.

What if we add the condition

(3) $$D(cA) = c^nD(A)$$ for $$c \in \mathbb{C}$$

• No. Consider e.g. $D(A):=\det A^k$ for any positive integer $k$. Commented Jan 5, 2022 at 16:29
• Another example: $D(A) = 1$. (actually the same as the example above for $k=0$)
– Snaw
Commented Jan 5, 2022 at 16:31
• More examples: $D(A)=\overline{\det A}$, or $D(A)=0$ or $1$ according to whether $A$ is singular or not. Commented Jan 5, 2022 at 16:32
• Whew, thanks for the quick counter examples. It looks like even adding the condition that $D(A) = 0$ iff $A$ is not invertible still wouldn't be enough. Commented Jan 5, 2022 at 16:39
• @aschepler your definition is not a counter-example. Let $v=e_1$ so that $D(A) = A_{11}^n$ per your definition of $D$ then let $A = \begin{pmatrix}1 & 1\\1 & 1\end{pmatrix}$ and $B = \begin{pmatrix}1 & 2 \\ 3 & 4\end{pmatrix}$. $D(A)=D(B)=1$ but $D(AB) = 16$. Commented Jan 5, 2022 at 20:04

The comments contain a few counterexamples, including $$A\mapsto \mathrm{det}(A)^k$$ for $$k\ge 0$$. If you assume that the map $$\varphi:M_n(\mathbb{C})\to \mathbb{C}$$ is continuous in the entries of the matrix, every possibility will be of the form $$f\circ \mathrm{det}$$, where $$f$$ is a group homomorphism $$\mathbb{C}^* \to \mathbb{C}^*$$, extended continuously to $$\mathbb{C}$$.
To see this, observe that $$\mathrm{GL}_n(\mathbb{C})$$ is dense in $$M_n(\mathbb{C})$$, so determining the map $$\varphi$$ amounts to determining it on $$\mathrm{GL}_n(\mathbb{C})$$. Now $$\psi = \varphi_{|\mathrm{GL}_n(\mathbb{C})}$$ is a group homomorphism to $$\mathbb{C}^*$$. Since the latter is abelian and the derived subgroup of $$\mathrm{GL}_n(\mathbb{C})$$ is $$\mathrm{SL}_n(\mathbb{C})$$, the map $$\psi$$ factors through $$\mathrm{det}:\mathrm{GL}_n(\mathbb{C})\to\mathrm{GL}_n(\mathbb{C})/\mathrm{SL}_n(\mathbb{C})=\mathbb{C}^*$$.
Note that there are many group automorphisms $$f:\mathbb{C}^*\to\mathbb{C}^*$$, including $$x\mapsto x^k$$ and $$x\mapsto \overline{x}$$, among many others (in a sense, most cannot be written down).
• This seems like a nice deep answer, I'm continuing to parse it. I've updated my question to add the constraint that $D(cA) = c^n D(A)$ for $c \in \mathbb{C}$. How does this update affect your answer? Commented Jan 5, 2022 at 17:23
• Well now $f(c^n)=f(\mathrm{det}(cI))=D(cI)=c^n$ for all $c\in \mathbb{C}$, so $f$ is the identity and $D$ is indeed the determinant. Commented Jan 5, 2022 at 17:38
• It seems like uniqueness of $\varphi$ could be proven even without knowing $\text{det}$ exists? Commented Jan 5, 2022 at 21:04