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Math Help - Inverse of a self adjoint linear transformation

  1. #1
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    Inverse of a self adjoint linear transformation

    How would I go about showing that the inverse of a self-adjoint linear transformation on a finite dimensional real vector space is also self-adjoint?
    Last edited by laura9010; January 8th 2010 at 03:04 PM.
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  2. #2
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    Quote Originally Posted by laura9010 View Post
    How would I go about showing that the inverse of a self-adjoint linear transformation on a finite dimensional real vector space is also self-adjoint?
    By definition, A^*:H\to H is the adjoint of A:H\to H if  \langle Ax , y \rangle = \langle x , A^* y \rangle \forall x,y\in H.

    Now, (A^{-1})^*:H\to H is the transformation for which  \langle A^{-1}x , y \rangle = \langle x , (A^{-1})^* y \rangle \forall x,y\in H, which means

     \langle AA^{-1}x , y \rangle = \langle x , A^*(A^{-1})^* y \rangle \forall x,y\in H, and therefore

     \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H.

    The trick is to use this result to prove that A(A^{-1})^*=I_{|H|}. It follows that A^{-1}=(A^{-1})^*, which is to say that the inverse of A is self-adjoint.
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  3. #3
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    Quote Originally Posted by hatsoff View Post
    By definition, A^*:H\to H is the adjoint of A:H\to H if  \langle Ax , y \rangle = \langle x , A^* y \rangle \forall x,y\in H.

    Now, (A^{-1})^*:H\to H is the transformation for which  \langle A^{-1}x , y \rangle = \langle x , (A^{-1})^* y \rangle \forall x,y\in H, which means

     \langle AA^{-1}x , y \rangle = \langle x , A^*(A^{-1})^* y \rangle \forall x,y\in H, and therefore

     \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H.

    The trick is to use this result to prove that A(A^{-1})^*=I_{|H|}. It follows that A^{-1}=(A^{-1})^*, which is to say that the inverse of A is self-adjoint.
    Possible alternative:

    If M is a hermitian matrix of A, then

    B = {B^t} \Rightarrow {B^t}{B^{ - 1}} = I \Rightarrow {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} \Rightarrow {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}, and so the inverse of B is hermitian, and so the inverse of the linear transformation is self-adjoint.

    Is that correct?
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  4. #4
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    Quote Originally Posted by laura9010 View Post
    Possible alternative:

    If M is a hermitian matrix of A, then

    B = {B^t} \Rightarrow {B^t}{B^{ - 1}} = I \Rightarrow {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} \Rightarrow {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}, and so the inverse of B is hermitian, and so the inverse of the linear transformation is self-adjoint.

    Is that correct?
    I don't see how {B^{ - 1}} = {\left( {{B^t}} \right)^{ - 1}} implies {B^{ - 1}} = {\left( {{B^{ - 1}}} \right)^t}, but then again I'm very rusty with this sort of thing.

    The missing step from my proof is as follows:

    \langle x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H

    \implies \langle I_{|H|}x , y \rangle = \langle x , A(A^{-1})^* y \rangle \forall x,y\in H

    \implies A(A^{-1})^*=I_{|H|}^*=I_{|H|}.
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