# Au=1

• Jan 6th 2011, 11:10 AM
dwsmith
Au=1
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?
• Jan 6th 2011, 12:10 PM
Ackbeet
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?
• Jan 6th 2011, 12:11 PM
dwsmith
Quote:

Originally Posted by Ackbeet
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.
• Jan 6th 2011, 12:12 PM
Ackbeet
Is that the real number 1 on the RHS? Or is that a vector?
• Jan 6th 2011, 12:13 PM
dwsmith
Real number 1.
• Jan 6th 2011, 12:35 PM
Ackbeet
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
dwsmith
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 $\displaystyle v = Au$ of a second class.
• Jan 6th 2011, 02:43 PM
HallsofIvy
Quote:

Originally Posted by dwsmith
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?

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
dwsmith
Here is the solution the professor furnished.

Let $\displaystyle u=v+iw$

$\displaystyle Au=A(v+iw)=Av+Aiw=1\Rightarrow Av=1 \ \mbox{and} \ Aw=0i\Rightarrow w=0$ and v is a real-valued solution.
• Jan 7th 2011, 12:04 PM
Ackbeet
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.