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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 $c \in \mathbb{C}$

Jagerber48
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    No. Consider e.g. $D(A):=\det A^k$ for any positive integer $k$. – user1551 Jan 05 '22 at 16:29
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    Another example: $D(A) = 1$. (actually the same as the example above for $k=0$) – Snaw Jan 05 '22 at 16:31
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    More examples: $D(A)=\overline{\det A}$, or $D(A)=0$ or $1$ according to whether $A$ is singular or not. – user1551 Jan 05 '22 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. – Jagerber48 Jan 05 '22 at 16:39
  • I've added an additional constraint $D(cA) = c^nD(A)$ for $c \in \mathbb{C}$ – Jagerber48 Jan 05 '22 at 17:21
  • If $v$ is any vector with $|v|=1$, $D(A) = (v^T A v)^n$ is still a counter-example. The key property which really determines operator $\det$ is its "alternating" nature: swapping any two columns (that is, multiplying by a transformation which swaps two basis vectors and leaves the others the same) negates the result. – aschepler Jan 05 '22 at 17:39
  • @aschepler how do you show $D(AB) = D(A)D(B)$ for your definition? – Jagerber48 Jan 05 '22 at 19:47
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    @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 $D(AB) = 16$. – Jagerber48 Jan 05 '22 at 20:04

1 Answers1

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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).

Alvaro Martinez
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    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? – Jagerber48 Jan 05 '22 at 17:23
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    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. – Alvaro Martinez Jan 05 '22 at 17:38
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    Thanks @AlvaroMartinez. I need to brush up on my group theory to fully understand this answer but it seems to make sense so I'll accept it. Would appreciate if other can debunk if there are any issues! Otherwise thanks for this answer. Conditions (1)-(3) seem like an interesting different characterization of the determinant. – Jagerber48 Jan 05 '22 at 20:26
  • Condition (1)-(3) plus continuous I guess. – Jagerber48 Jan 05 '22 at 20:38
  • It seems like uniqueness of $\varphi$ could be proven even without knowing $\text{det}$ exists? – Jagerber48 Jan 05 '22 at 21:04
  • @Jagerber48 Probably yes – Alvaro Martinez Jan 05 '22 at 22:19