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Math Help - Some inner product proofs help please

  1. #1
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    Some inner product proofs help please

    1) if a,b,c belong to an inner product space X, find all solutions to the equation <a,x>b=c

    I know that I should set c=landa*b but i get stuck from there.

    2) find solutions to equation x + <a,x>c=b

    No idea how to do this one at all

    thanks in advance
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  2. #2
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    I'll start with the first problem...
    <a,x>b=c has solutions only if b and c are parallel, that is, if c = \lambda b for some constant \lambda. So, we set <a,x> = \lambda. We can decompose x into a piece that's parallel to a and a piece that's perpendicular to a, i.e., x = u + v with <u,a> = 0 and v = \sigma a for some constant \sigma.

    Since the dot product is linear, this gives <a,x> = <a,u+v> = <a,u> + <a,v> = <a,v> = \sigma <a,a> = \lambda. Thus, we see that u can be any vector we like as long as it's perpendicular to a, and we must have \sigma = \frac{\lambda}{<a,a>}. The set of solutions, then, is
    x = u + \frac{\lambda}{<a,a>}a for all u orthogonal to a.
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  3. #3
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    wow thanks, that helps alot. Now im guessing the second part isnt as easy as that =s . Any suggestions how to do it ?
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  4. #4
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    Second part is a little trickier, but doable. Think of this as an equation of the form Lx=b, where L is a linear operator. In this case we can write this operator L = I + ca^*. Here I is the identity, and a^* is the conjugate transpose of a, so applying it to a vector gives a^*x = <a,x>.

    So, the equation we want to solve is Lx = (I + ca^*)x = x + <a,x>c = b. In order to solve it, we need to find an inverse of the operator L, because then the solution is just x = L^{-1}b.

    Now I can't remember if there's an easy way to see this, but there happens to be an inverse of the form L^{-1}=I + ua^* for some vector u. Suppose this is true. Then we want (I+ua^*)(I+ca^*) = I. Now expand this out: I + ua^* + ca^* + <a,c>ua^* = I. All that's left is to solve for u, and apply the inverse operator you get to b to get your final result.
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