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Math Help - Au=1

  1. #1
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    Au=1

    Let A be a linear operator. If Au=1 has a complex-valued solution, then it also has a real-valued solution.

    How do I prove this?
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  2. #2
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    What kind of space is u in? Is A a differential operator? Do you know anything about A other than it's a linear operator?
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    What kind of space is u in? Is A a differential operator? Do you know anything about A other than it's a linear operator?
    That is everything I know.
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  4. #4
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    Is that the real number 1 on the RHS? Or is that a vector?
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  5. #5
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    Real number 1.
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  6. #6
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    What is the overall context for this problem? Is it from a textbook? If so, how did the book define a linear operator?

    In particular, I'm interested in knowing if A always maps real-valued vectors/functions to real-valued vectors/functions. If it does, then you can choose \text{Re}(u) as your real-valued solution. If not, your problem is a bit more complicated.
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  7. #7
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    It is from the particular solutions section of PDE.

    We say an operator A is defined if there is a rule which assigns to each function u of one given class another function v = Au of a second class.
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  8. #8
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    Quote Originally Posted by dwsmith View Post
    Let A be a linear operator. If Au=1 has a complex-valued solution, then it also has a real-valued solution.

    How do I prove this?
    This is all you have? Okay, if Au= 1 has a "complex-valued solution" then u must be a complex number and A is a linear operator that maps numbers to numbers. In particular, A could be multiplication by -i: (-i)(i)= 1. That problem has only i as solution, there is no real-valued solution.

    Note, that, since the real numbers are a subset of the complex numbers, saying there is a complex solution could mean that there is a real number solution. However, the problem says "it also has a real solution" and that is not true.
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  9. #9
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    Here is the solution the professor furnished.

    Let u=v+iw

    Au=A(v+iw)=Av+Aiw=1\Rightarrow Av=1 \ \mbox{and} \ Aw=0i\Rightarrow w=0 and v is a real-valued solution.
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  10. #10
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    That is the solution I hinted at in post # 6. However, this solution assumes that A maps real-valued solutions to real-valued solutions. That may or may not be a good assumption.
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