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Let $f: \mathbb R^{n^2} \to \mathbb R$, such that (i) $f$ is a surjection and (ii) $f$ is a homomorphism.

Clearly, $\det$ satisfies those conditions. Is it unique? That is, is there another function, besides the determinant, from matrices to the reals that is a surjective homomorphism?

SRobertJames
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  • What does "homomorphism" mean in this context? What algebraic structure are you trying to preserve? Can't be multiplication since you haven't required invertibility. – lulu May 30 '23 at 23:23
  • How is determinant a homomorphism? – Andrew May 30 '23 at 23:24
  • @lulu But if the determinant is non-zero then it's an invertible matrix, isn't it? If the determinant is zero then it's similar to scalar multiplication. What did I get wrong here? – Suzu Hirose May 30 '23 at 23:25
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    @SuzuHirose The set of square matrices is not a group under multiplication, since inverses don't always exist. Perhaps the OP meant to ask for homomorphisms from $\text {GL}_n(\mathbb R)\to \mathbb R^*$ but that's not what is writtren. – lulu May 30 '23 at 23:28
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    If you intended to specify invertible matrices (and the non-zero reals under multiplication) then this question is a duplicate. – lulu May 30 '23 at 23:30
  • There are exotic group isomorphisms $\Bbb R^\to\Bbb R^$, namely consider that $\Bbb R^\cong C_2\times (0,\infty)\cong C_2\times\Bbb R\cong C_2\times\Bbb Q^{\oplus \Bbb R}$. Now take a $\Bbb Q$-linear bijection $T:\Bbb Q^{\oplus \Bbb R}\to \Bbb Q^{\oplus \Bbb R}$ and transfer the map $(x,v)\to (x,Tv)$ back along the chain of isomorphisms. Once you have $G:\Bbb R^\to\Bbb R^*$ non-trivial isomorphism, extend it to the monoid $\Bbb R$ by sending $0\to 0$ (call it $\hat G$). Finally, $\hat G\circ \det$ is a second surjective whatever. – Sassatelli Giulio May 30 '23 at 23:34
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    Also, obviously, $(\det(x))^3$. – Sassatelli Giulio May 30 '23 at 23:40
  • @lulu I understand your point but $0$ isn't invertible under mulitplication so OP's meaning seems to be the case where the determinant is non-zero, $GL_n({\mathbf R})$ as you note. Also as you note this is then a duplicate. Thank you for responding. – Suzu Hirose May 30 '23 at 23:43

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