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

How do I prove this?

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- Jan 6th 2011, 11:10 AM #1

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- Jan 6th 2011, 12:10 PM #2

- Jan 6th 2011, 12:11 PM #3

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- Jan 6th 2011, 12:12 PM #4

- Jan 6th 2011, 12:13 PM #5

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- Jan 6th 2011, 12:35 PM #6
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 $\displaystyle \text{Re}(u)$ as your real-valued solution. If not, your problem is a bit more complicated.

- Jan 6th 2011, 12:54 PM #7

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- Jan 6th 2011, 02:43 PM #8

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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.

- Jan 7th 2011, 11:45 AM #9

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- Jan 7th 2011, 12:04 PM #10