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Math Help - Proof TFAE

  1. #1
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    Proof TFAE

    Show that the following properties of an endomorphism d:V-->V of a vector space with a scalar product <.|.> are equivalent:

    (i) d*=d^(-1)
    (ii) <d(u)|d(v)>=<u|v> for all u,v elements of V
    (iii) ||d(u)||= ||u|| for all u element of V
    (iv) ||d(u) - d(v)|| = ||u-v|| for all u,v element of V

    Note: If V is euclidean, a map d with the above properties is orthogonal. If V is unitary, then such a d is likewise unitary.
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  2. #2
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    Quote Originally Posted by TXGirl
    Show that the following properties of an endomorphism d:V-->V of a vector space with a scalar product <.|.> are equivalent:
    Note, sure I understand. How can you have a scalar product involving elements of a vector space!?!
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  3. #3
    Super Member Rebesques's Avatar
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    with a scalar product <.|.> are equivalent:
    I think she means inner product. Doesn't make sence otherwise


    Anyway. Lets deal with the real case.

    i)=>ii) <d(u)|d(v)>=<d*d(u)|v>=<d^{-1}d(u)|v>=<u|v>.

    ii)=>iii) Just let u=v.

    iii)=>iv) ||d(u)-d(v)||=||d(u-v)||=||u-v||.

    iv)=>i) We will show that

    a) <d*d(u)|v>=<u|v> for all u,v in V. This implies d*du=u for all u in V.

    b) <dd*(u)|v>=<u|v> for all u,v in V. This implies dd*u=u for all u in V.

    Now a) and b) imply d*=d^{-1}.

    Lets prove a). We have ||d(u)-d(v)||^2=||u-v||^2 so

    ||d(u)||^2 - 2<du|dv> +||d(v)||^2 = ||u||^2 - 2<u|v> +||v||^2

    and by using ||d(w)||=||w||,

    <du|dv> = <u|v> or <d*d(u)|v>=<u|v>, and a) is proven.


    Now b) is the same.
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  4. #4
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    well it just means that the scalar product operation is defined in the given space
    what is the difference between saying the inner product and the scalar product??
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  5. #5
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    The assumptions here do not seem clearly founded to me:

    i)=>ii) <d(u)|d(v)>=<d*d(u)|v>=<d^{-1}d(u)|v>=<u|v>.
    __________________________________________________ ____

    I said, we have that d* is the adjoint d.

    So, <u|d(v)>=<d*(u)|v>

    So d can then be understood as the identity function thus indicating that
    <d(u)|d(v)>=<u|v> for all u,v elements of V.

    Is that right??
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  6. #6
    Super Member Rebesques's Avatar
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    what is the difference between saying the inner product and the scalar product??
    The inner product is a generalization of the scalar product.


    So, <u|d(v)>=<d*(u)|v>
    Use this identity with v=d(w) for some w to get ii from i.


    So d can then be understood as the identity function thus indicating that
    <d(u)|d(v)>=<u|v> for all u,v elements of V.
    Not true. This is actually the definition of a special operator called isometry. For example, in R^2, consider a rotation by a fixed angle. Then this mapping preserves the length of vectors, and so is an isometry.
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