0
$\begingroup$

I just have to prove that it isn't with O(A+B)=O(A)+O(B) and O(kA)=k(OA) where O is the linear operator (i.e the absolute value), A+B and A would be a complex number, and k is some real constant. I used abs(x+iy). sooo when you break it up it would be abs(x)+abs(iy) which isn't true...right? Or am I completely thinking of this the wrong way?

$\endgroup$
  • $\begingroup$ If it is a linear operator on the complex numbers, then it would be a linear operator on the reals. So, is it a linear operator on the reals? $\endgroup$ – copper.hat Sep 12 '14 at 6:03
  • $\begingroup$ Note that for $z \in \Bbb C$, $\vert z \vert = \vert -z \vert$; but if $\vert \cdot \vert$ were linear, we would have $\vert -z \vert = -\vert z \vert$. If $\vert z \vert \ne 0$, it just won't work. Cheers! $\endgroup$ – Robert Lewis Sep 12 '14 at 6:06
1
$\begingroup$

Yes, you are thinking in a right way. (there are lots of ways to think about this)

To make it more precise, if the absolute value were a linear operator, then

$$ |x + iy| = |x| + |iy|$$

would be an identity. (aside: if this were an identity, that's still not enough to prove the absolute value a linear operator)

So to prove that the absolute value is not a linear operator, it is enough to prove that the equation above is not an identity.

Note it's not enough to just look at them and say they look different: sometimes you can have very different expressions that nonetheless represent the same thing. One simple way to prove that this isn't an identity is to plug in some specific values and see that they are unequal.

$\endgroup$
  • $\begingroup$ Thanks! I was doubting myself more than anything, but this helped me. Thanks a lot! $\endgroup$ – user175470 Sep 12 '14 at 6:18

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.