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Thread: Finding the adjoint of a linear operator

  1. #1
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    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 $\displaystyle T:V \to V$ is a linear operator and that it does have an adjoint. I know nothing about $\displaystyle V$ (inner product, dimension (possibly infinite), etc.).

    The book doesn't give any examples like this, and the only one I've found online uses $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^2$, so we can work with the inner product formulas explicitly to obtain the adjoint.
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    MHF Contributor Drexel28's Avatar
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    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 $\displaystyle T:V \to V$ is a linear operator and that it does have an adjoint. I know nothing about $\displaystyle V$ (inner product, dimension (possibly infinite), etc.).

    The book doesn't give any examples like this, and the only one I've found online uses $\displaystyle T:\mathbb{R}^3 \to \mathbb{R}^2$, 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 $\displaystyle V$ is a finite dimensional complex inner product space, and you have some endomorphism $\displaystyle T$ then if $\displaystyle \mathscr{B}$ is an orthonormal basis for $\displaystyle V$ then $\displaystyle [T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast$ where on the left hand side $\displaystyle \ast$ denotes the adjoint of the transformation and on the right it denotes the conjugate transpose. Or, the obvious generlization to non-square matrices.
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    MHF Contributor FernandoRevilla's Avatar
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    Re: Finding the adjoint of a linear operator

    Quote Originally Posted by Drexel28 View Post
    If you assume that $\displaystyle V$ is a finite dimensional complex inner product space, and you have some endomorphism $\displaystyle T$ then if $\displaystyle \mathscr{B}$ is an orthonormal basis for $\displaystyle V$ then $\displaystyle [T^\ast]_\mathscr{B}=[T]_\mathscr{B}^\ast$
    More general, if $\displaystyle B$ is a basis of $\displaystyle V$ not necessarily orthonormal, $\displaystyle G$ the Gram's matrix, $\displaystyle A$ the matrix of $\displaystyle T$ and $\displaystyle C$ the matrix of $\displaystyle T^*$ (all these elements with respect to $\displaystyle B$) then, $\displaystyle C=G^{-1}A^*G$ .
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    MHF Contributor Drexel28's Avatar
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    Re: Finding the adjoint of a linear operator

    Quote Originally Posted by FernandoRevilla View Post
    More general, if $\displaystyle B$ is a basis of $\displaystyle V$ not necessarily orthonormal, $\displaystyle G$ the Gram's matrix, $\displaystyle A$ the matrix of $\displaystyle T$ and $\displaystyle C$ the matrix of $\displaystyle T^*$ (all these elements with respect to $\displaystyle B$) then, $\displaystyle C=G^{-1}A^*G$ .
    Right, you just change the basis to put it with respect to the orthonormal basis.
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    Re: Finding the adjoint of a linear operator

    Yes, I do know that $\displaystyle [T^*]_\beta=[T]^*_\beta$ for an orthonormal basis $\displaystyle \beta$. 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.
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    Re: Finding the adjoint of a linear operator

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

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

    It's easy to show that $\displaystyle T$ is linear. It's the second part I'm having difficulty with. $\displaystyle T^*$ is the unique linear operator such that $\displaystyle \langle T(x), y \rangle = \langle x, T^*(y) \rangle \quad \forall x,y \in V$. In this case $\displaystyle T(x)= \langle x,y \rangle z$, so we need to show $\displaystyle \langle \langle x, y \rangle z , w \rangle = \langle x, T^*(w) \rangle$. 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).
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  7. #7
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    Re: Finding the adjoint of a linear operator

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

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

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

    $\displaystyle = \langle\langle x,y \rangle z,w \rangle = \langle T(x),w \rangle$, as desired.
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    Re: Finding the adjoint of a linear operator

    Got it. Thanks.
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