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Homework Help: Function of determinant

  1. Sep 24, 2010 #1
    1. The problem statement, all variables and given/known data

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

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

    2. Relevant equations

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

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

    3. 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.
  2. jcsd
  3. Sep 24, 2010 #2


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    Science Advisor

    A linear functional on a vector space is completely determined by its action on a basis.
    What would [itex]det(\psi_{ij})[/itex] and [itex]det(\phi_j)[/itex] be? If F obeys exactly the same laws as the determinant, it is the determinant!
  4. Sep 24, 2010 #3
    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 necessery 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.
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