$f(A)=0 \iff A$ is not invertible

Question

Let $f: M_n(\mathbb{K}) \rightarrow \mathbb{K}$ such that $\forall A,B\in M_n(\mathbb{K}) : f(AB)=f(A)f(B)$ and $f$ not constant.

Prove that: $f(A)=0 \iff A$ is not invertible.

I can also show that $f$ is injective.

Answer 1

The claim is not true as stated. The function $f(X)=0 \forall X$ is a counterexample.

With the extra assumption that $f$ is not constant function, the claim holds. Here are the hints:

Hint 1: Show $f(0)=0$.

Hint 2: If $f(Y) \neq 0$ for some $Y$ set $A=Y, B=I$. Deduce that $f(I) \neq 0$.

Hint 3: If $A$ is invertible set $B=A^{-1}$.

Hint 4: If $A$ is a diagonal matrix with $n-1$ ones and a zero on the diagonal, show that $f(A)=0$.

To do this, consider all such matrices $A_1,..., A_n$. Show that for each $j$ there exists some $P$ such that $A_j=PA_1P^{-1}$, and deduce that $f(A_j)=f(A_1)$.

What is $f(A_1...A_n)=?$

Hint 5 Use hint $4$ to deduce that if $A$ is a diagonal matrix with at least a zero on diagonal then $f(A)=0$.

Hint 6: If $A$ is not invertible show that there exists some $D,B$ with $D$ diagonal and $B$ invertible so that $A=DB$.