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Math Help - Proof with Transformations and Subspaces

  1. #1
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    Proof with Transformations and Subspaces

    I have this proof that I cannot figure out. If anyone could give me a little headstart that would be great. I am still struggling to fully grasp the concepts so maximum explanation is useful please.

    Also I dont know how to make the perp symbol that looks like an upside down T so ill use an L instead.
    V^L=V perp
    W c V means W is a subspace of V

    Suppose that A is a real symmetric n x n matrix. Show that if V is
    a subspace of Rn and that A(V) c V, then A(V^L) c V^L.

    Proof with Transformations and Subspaces-screen-shot-2010-11-29-11.32.53-pm.png

    Any help??!?!
    Thanks
    Last edited by BrianMath; November 29th 2010 at 07:34 PM. Reason: more clarity
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  2. #2
    MHF Contributor FernandoRevilla's Avatar
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    If y\in V^{\perp} then, for all x\in V we have:

    <Ay,x>=(Ay)^tx=y^tA^tx=y^tAx=<y,Ax>

    But by hypothesis Ax\in V, so:

    <y,Ax>=0=<Ay,x>\;\forall x\in V

    That is, Ay\in V^{\perp} .

    Regards.

    Fernando Revilla
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  3. #3
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    Quote Originally Posted by BrianMath View Post
    I have this proof that I cannot figure out. If anyone could give me a little headstart that would be great. I am still struggling to fully grasp the concepts so maximum explanation is useful please.

    Also I dont know how to make the perp symbol that looks like an upside down T so ill use an L instead.
    V^L=V perp
    W c V means W is a subspace of V

    Suppose that A is a real symmetric n x n matrix. Show that if V is
    a subspace of Rn and that A(V) c V, then A(V^L) c V^L.

    Click image for larger version. 

Name:	Screen shot 2010-11-29 at 11.32.53 PM.png 
Views:	11 
Size:	18.5 KB 
ID:	19900

    Any help??!?!
    Thanks


    Denote by \langle,\rangle the usual inner product in \mathbb{R}^n , so that y\in L^\bot\Longleftrightarrow \langle x,y\rangle=0\,\,\,\forall x\in L

    Since A is symmetric this means it is self-adjoint , so \langle Av,u\rangle=\langle v,Au\rangle,\,\,\,\forall u,v\in\mathbb{R}^n

    Let now x\in L^\bot\Longrightarrow \langle x,v\rangle=0\,\,\,\forall v\in L , and thus for any v\in L we get:

    \langle Ax,v\rangle=\langle x,Av\rangle = 0 , since Av\in L by assumption.

    Tonio
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  4. #4
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    Thanks

    Damn, this is a lot simpler than I thought. I should have thought of this myself. I was trying things with the basis of V and some other things that were just unnecessary haha.
    Thanks a bunch.
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