1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    HallsofIvy

    User Avatar
    Staff Emeritus
    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.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Function of determinant
  1. Determinant functions (Replies: 6)

Loading...