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

How do I prove this?

Results 1 to 10 of 10

- Jan 6th 2011, 12:10 PM #1

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Jan 6th 2011, 01:10 PM #2

- Jan 6th 2011, 01:11 PM #3

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Jan 6th 2011, 01:12 PM #4

- Jan 6th 2011, 01:13 PM #5

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Jan 6th 2011, 01:35 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 as your real-valued solution. If not, your problem is a bit more complicated.

- Jan 6th 2011, 01:54 PM #7

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Jan 6th 2011, 03:43 PM #8

- Joined
- Apr 2005
- Posts
- 19,117
- Thanks
- 2803

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, 12:45 PM #9

- Joined
- Mar 2010
- From
- Florida
- Posts
- 3,093
- Thanks
- 9

- Jan 7th 2011, 01:04 PM #10