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Thread: two Hermitian problems

  1. #1
    Junior Member
    Oct 2008

    two Hermitian problems

    First problem: Assume T: V -> V is a Hermitian transformation
    Prove T^-1 is Hermitian if T is invertible.

    Here, I can prove that T^n is Hermitian if n > 0 but I'm stuck for n = -1.

    Second Problem: C(0, 1) is linear Space. Inner product is given by:
    (f, g) = integral( f * g, t, 0, 1).

    Let V be the subspace of all f such that integral f, t, 0, 1) = 0.
    Let T: V -> C(0, 1) and T(f(x)) = integral(f, t, 0, x). Prove T is skew - symmetric.

    For this problem, I would apply the transformation, do the inner product, then I would have an integral in an integral. I don't know if I should F and G for their anti derivative or what.
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  2. #2
    A Plied Mathematician
    Jun 2010
    CT, USA
    First Problem: Assume, as the problem indicates, that T:V\to V is Hermitian and invertible. Then we know that

    \langle x|Ty\rangle=\langle Tx|y\rangle for all x,y\in V.

    Let x,y\in V. Then, because the inverse exists, we may define s=T^{-1}x and t=T^{-1}y such that both s,t\in V. Now play around with

    \langle x|T^{-1}y\rangle. You'd like it to equal \langle T^{-1}x|y\rangle.

    Can you get that to happen?

    Second Problem. Let me rephrase your question using more standard notation.

    W=C(0,1) is a linear space with inner product

    \displaystyle \langle f|g\rangle=\int_{0}^{1}\overline{f(t)}\,g(t)\,dt.

    Let V\subseteq W be the subspace consisting of the following:

    \displaystyle V=\left\{f|\int_{0}^{1}f(t)\,dt=0\right\}.

    Let T:V\to W be defined by

    \displaystyle Tf(x)=\int_{0}^{x}f(t)\,dt.

    Prove that T is skew-symmetric.

    Is this a correct re-statement of the problem? If so, I have an extremely strong feeling that integration by parts is going to be the key to solving this problem. The subspace you're in indicates that the typical boundary term of

    \displaystyle \int_{a}^{b}v\,du=\underbrace{(uv)|_{a}^{b}}_{\tex  t{Boundary Term}}-\int_{a}^{b}u\,dv

    will be zero, which means you'll just pick up that minus sign when you try to slap the integral on the other term. Try that and see if it doesn't get you where you need to go.
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