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Let $f:\mathscr M_n(\mathbb K) \to \mathbb K$ be a non constant function such as $f(AB) = f(A)f(B)$ for all $A,B$ in $\mathscr M_n(\mathbb K)$. The question is to show that $M\in GL_n(\mathbb K)$ iff $f(M)\neq 0$.

I've already shown that $f(Id) = 1$ and $M\in GL_n(\mathbb K) \Rightarrow f(M)\neq 0$. I'm stuck on the converse.

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    $\begingroup$ Have you proven that all singular matrices (over a field) are zero divisors? $\endgroup$ – Henry Swanson Jan 7 '15 at 19:57
  • $\begingroup$ Ok, I got it! It's easy with this indication. $\endgroup$ – Sebastien Jan 7 '15 at 20:10
  • $\begingroup$ Indeed.... No, I don't succeed: if $M\notin GL_n(K), \exists N\in\mathscr M_n(K)$ such as $MN = 0$, so $f(M)f(N) = f(0) = 0$. Thus $f(M) = 0$ or $f(N) = 0$... so what? $\endgroup$ – Sebastien Jan 7 '15 at 21:21
  • $\begingroup$ maybe too complicated, but if $M$ is singular then $MM_1M_2\dots M_k=0$ where all $M_i=A_iMA_i^{-1}$ for some non-singular $A_i$ $\endgroup$ – user8268 Jan 7 '15 at 22:24
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If $M$ is invertible, $f(M)$ must be nonzero, otherwise $f(A)=f(MM^{-1}A)=f(M)f(M^{-1}A)=0$ for every matrix $A$, which is a contradiction to the assumption that $f$ is nonconstant.

Every (singular or invertible) square matrix $M$ of rank $k$ can be written in the form of $M=P(I_k\oplus 0_{n-k})Q$ for some invertible matrices $P$ and $Q$ that are products of elementary matrices (here $0_{n-k}$ denotes a zero square matrix of size $n-k$). In the previous paragraph we have established that $f(P)$ and $f(Q)$ are nonzero. So, we see that $f(M)$ is nonzero for some $M$ of rank $k$ if and only if $f$ is nonzero for every matrix of rank $k$.

Now, suppose $f$ is nonzero at some singular matrix of rank $k\ (<n)$. Then $f(A)$ is nonzero for every matrix $A$ of rank $k$. By mathematical induction, $f(B)$ is nonzero for every matrix $B$ of rank $m=k-1$ down to $0$, because $$ f(I_m\oplus 0_{n-m}) = f\left(I_m\oplus \operatorname{diag}(1,0,\ldots,0)\right)\, f\left(I_m\oplus\operatorname{diag}(0,0,\ldots,1)\right). $$ In particular, $f(0)$ is nonzero. Yet, $f(0)=f(0\,C)=f(0)\,f(C)$ for every matrix $C$. Hence $f=1$, which is a contradiction to the given assumption on $f$.

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    $\begingroup$ The same with a different viewpoint: If $A$ is singular, then $BA$ is nilpotent for some invertible $B$. From $f((BA)^n)=f(0)=0$ we get $f(BA)=0$ and finally $f(A)=0$ $\endgroup$ – Hagen von Eitzen Jan 8 '15 at 11:30
  • $\begingroup$ @HagenvonEitzen Your answer is much more concise! $\endgroup$ – user1551 Jan 8 '15 at 11:37

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