Results 1 to 4 of 4

Math Help - determinant of T linear operator

  1. #1
    Newbie
    Joined
    Mar 2010
    Posts
    4

    determinant of T linear operator

    Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

    i) 1) Prove that det(T tIv) = det ( [T]β tI ) for any scalar t and any ordered basis β
    ii) 2) Prove that if det (T) ≠ 0 , then T is invertible
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21
    Quote Originally Posted by mystic View Post
    Let T be a linear operator on a finite-dimensional vector space V, and let β be an ordered basis for V, we define det(T) = det ( [T]β )

    i) 1) Prove that det(T tIv) = det ( [T]β tI ) for any scalar t and any ordered basis β
    ii) 2) Prove that if det (T) ≠ 0 , then T is invertible
    I'm not quite sure what you mean for the first one. For the second I assume you're allowed to use that the invertibility of a matrix is equivalent to invertible determinant. If so, then do the following. Let V be an F-space. Then, there exists an isomorphism \phi:V\to F^{\dim_F V} afforded by the choice of the ordered basis \]beta. Suppose then that \det\left([T]_\beta\right)\ne 0 then for every v\in V one has that \phi(v)\in F^{\dim_F V} and since [T]_\beta is invertible one has that [T]_\beta^{-1} exists. Prove then that T\left(\phi^{-1}\left([T]_{\beta}^{-1}(\phi(v))\right)=v. Thus, T is surjective and the conclusion follows since \dim_F V<\infty
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,328
    Thanks
    702
    it seems to me the first problem boils down to showing that [tIv]β = I, the matrix whose entries are the kronecker delta, no matter what β is, since if so, you can subtract the matrices on the RHS. you may need to invoke the isomorphism between Mn(F) and the elements of End(V) (you need β to exhibit this isomorphism, but the fact that they are isomorphic doesn't depend on β).
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45
    The problem defines \det(T)=\det([T_{\beta}]) . In order to see if \det (T) is well defined we need to prove that the right side does not depends on the chosen basis. Indeed, if \beta' is another basis of V then, [T]_{\beta} and [T]_{\beta'} are similar matrices so, \det([T]_{\beta})=\det([T]_{\beta'}) .

    Now, fixed \beta an using the standard isomorphism between \textrm{End}_{\mathbb{K}}(V) and \mathbb{K}^{n\times n} we have [T-tI_V]_{\beta}=[T]_{\beta}-tI_n .
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Linear operator
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: May 30th 2011, 09:43 PM
  2. Linear Operator
    Posted in the Differential Geometry Forum
    Replies: 2
    Last Post: May 25th 2011, 11:59 PM
  3. Is this Linear Operator?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: March 14th 2010, 01:59 PM
  4. linear operator
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: October 16th 2008, 05:12 PM
  5. linear operator??!
    Posted in the Advanced Math Topics Forum
    Replies: 5
    Last Post: June 16th 2007, 07:28 AM

Search Tags


/mathhelpforum @mathhelpforum