What is the Function of Determinant in Expressing a Linear Map?

[losiu99]

Homework Statement

Let E be a vector space of finite dimension over $$\Gamma$$ (char 0), and $$F\colon L(E,E)\rightarrow \Gamma$$ satisfies
(1) $$F(\phi \circ \psi)=F(\phi)F(\psi)$$
(2) $$F(\hbox{id})=1$$

Prove F can be expressed as a function of determinant, $$F(\phi)=f(\hbox{det}\phi)$$.

Homework Equations

Hint: Let $$e_\nu$$ be a basis and define
$$\psi_{ij}e_{\nu}=\begin{cases}e_\nu & \text{if } \nu\neq i\\e_i+\lambda e_j & \text{if }\nu = i\end{cases}$$
$$\phi_i e_\nu =\begin{cases}e_\nu & \text{if } \nu \neq i \\ \lambda e_\nu & \text{if }\nu = i\end{cases}$$

Show that $$F(\psi_{ij})=1$$ and that $$F(\phi_j)$$ doesn't depend on i.

The Attempt at a Solution

I have proved facts in the hint, but I cannot clearly see how to carry on. I guess it's about using this two families of functions to transform functions having the same determinant without changing the value of F to the point we can conclude F's values are equal. Unfortunetely, I fail miserably trying to accomplish this.
I appreciate any help.

 

[HallsofIvy]

A linear functional on a vector space is completely determined by its action on a basis.
What would $det(\psi_{ij})$ and $det(\phi_j)$ be? If F obeys exactly the same laws as the determinant, it is the determinant!

 

[losiu99]

Well, for the record, F doesn't obey the same laws as a determinant. Nor is it necessarily a linear functional. Any multiplicative function defined in terms of value of determinant with f(id)=1 will satisfy this conditions. Anyway, I believe I have roughly proved all the necessary facts, I think it's just the matter of carefully writing down the proof, making sure I haven't overlooked something. Will do it tomorrow and let you know if I succeed. Thanks for your response.