Results 1 to 8 of 8

Math Help - Finding the adjoint of a linear operator

  1. #1
    Junior Member
    Joined
    May 2011
    Posts
    59

    Finding the adjoint of a linear operator

    I'm trying to find an explicit expression for the adjoint of a particular linear operator. Are there any very general techniques for doing so when we know very little about the setting we're working in? All I have is that T:V \to V is a linear operator and that it does have an adjoint. I know nothing about V (inner product, dimension (possibly infinite), etc.).

    The book doesn't give any examples like this, and the only one I've found online uses T:\mathbb{R}^3 \to \mathbb{R}^2, so we can work with the inner product formulas explicitly to obtain the adjoint.
    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

    Re: Finding the adjoint of a linear operator

    Quote Originally Posted by AlexP View Post
    I'm trying to find an explicit expression for the adjoint of a particular linear operator. Are there any very general techniques for doing so when we know very little about the setting we're working in? All I have is that T:V \to V is a linear operator and that it does have an adjoint. I know nothing about V (inner product, dimension (possibly infinite), etc.).

    The book doesn't give any examples like this, and the only one I've found online uses T:\mathbb{R}^3 \to \mathbb{R}^2, so we can work with the inner product formulas explicitly to obtain the adjoint.
    There is no general formula that I know of besides the following. If you assume that V is a finite dimensional complex inner product space, and you have some endomorphism T then if \mathscr{B} is an orthonormal basis for V then [T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast where on the left hand side \ast denotes the adjoint of the transformation and on the right it denotes the conjugate transpose. Or, the obvious generlization to non-square matrices.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor FernandoRevilla's Avatar
    Joined
    Nov 2010
    From
    Madrid, Spain
    Posts
    2,162
    Thanks
    45

    Re: Finding the adjoint of a linear operator

    Quote Originally Posted by Drexel28 View Post
    If you assume that V is a finite dimensional complex inner product space, and you have some endomorphism T then if \mathscr{B} is an orthonormal basis for V then [T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast
    More general, if B is a basis of V not necessarily orthonormal, G the Gram's matrix, A the matrix of T and C the matrix of T^* (all these elements with respect to B) then, C=G^{-1}A^*G .
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor Drexel28's Avatar
    Joined
    Nov 2009
    From
    Berkeley, California
    Posts
    4,563
    Thanks
    21

    Re: Finding the adjoint of a linear operator

    Quote Originally Posted by FernandoRevilla View Post
    More general, if B is a basis of V not necessarily orthonormal, G the Gram's matrix, A the matrix of T and C the matrix of T^* (all these elements with respect to B) then, C=G^{-1}A^*G .
    Right, you just change the basis to put it with respect to the orthonormal basis.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    May 2011
    Posts
    59

    Re: Finding the adjoint of a linear operator

    Yes, I do know that [T^*]_\beta=[T]^*_\beta for an orthonormal basis \beta. But since this only works for finite-dimensional vector spaces, I can't use it here. If I work on it for a while longer and don't come up with anything, I'll post here again and give more details. There's a particular aspect that's making it difficult.
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Junior Member
    Joined
    May 2011
    Posts
    59

    Re: Finding the adjoint of a linear operator

    I'm getting nowhere, so here's the actual problem:

    "Let V be an inner product space, and let y, z \in V. Define T: V \to V by T(x)= \langle x,y \rangle z for all x \in V. First prove that T is linear. Then show that T^* exists and find an explicit expression for it."

    It's easy to show that T is linear. It's the second part I'm having difficulty with. T^* is the unique linear operator such that \langle T(x), y \rangle = \langle x, T^*(y) \rangle \quad \forall x,y \in V. In this case T(x)= \langle x,y \rangle z, so we need to show \langle \langle x, y \rangle z , w \rangle = \langle x, T^*(w) \rangle. What's throwing me off is that if we can get from the LHS to the RHS via manipulation by the usual useful properties of inner products, I see no way to get x into/out of the inner product and alone on the RHS. I'm probably over-complicating things, as I often seem to do. I would appreciate hints (and as always, only hints).
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor

    Joined
    Mar 2011
    From
    Tejas
    Posts
    3,401
    Thanks
    762

    Re: Finding the adjoint of a linear operator

    inner-product spaces are usually defined implicitly over a subfield of \mathbb{C}.

    so let T^*(w) = \overline{\langle z,w \rangle}y

    then \langle x,T^*(w) \rangle = \langle x, \overline{\langle z,w \rangle}y \rangle = \langle z,w \rangle\langle x,y \rangle = \langle x,y \rangle\langle z,w \rangle

    = \langle\langle x,y \rangle z,w \rangle = \langle T(x),w \rangle, as desired.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Junior Member
    Joined
    May 2011
    Posts
    59

    Re: Finding the adjoint of a linear operator

    Got it. Thanks.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Adjoint of linear Operator
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 16th 2011, 07:27 AM
  2. self-adjoint linear operator
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: February 23rd 2011, 01:32 AM
  3. Replies: 2
    Last Post: December 5th 2009, 02:30 AM
  4. more Adjoint of Linear operator confusion...
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: December 3rd 2009, 08:01 PM
  5. Normal, self-adjoint, or neither linear operator
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: April 15th 2008, 12:04 PM

Search Tags


/mathhelpforum @mathhelpforum