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Prove: if $u$ and $v$ are vectors in an inner product space and $c$ is a scalar, then $\langle u,cv\rangle =c\langle u,v\rangle$.

I am a little confused since in my textbook it shows 2 contradictory properties in two different theorems. In one it shows $c\langle u,v\rangle = \langle cu,v\rangle $ and here it wants the opposite. Please explain

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what do you mean by " then =c" –  zapkm Feb 27 '12 at 16:49
    
Welcome to MathSE. I see that you are relatively new here. So I wanted to let you know a few things about MathSE. We like to know where the problem is from what you've tried on a problem; this prevents people from wasting their time telling you thinks you already know, and helps make sure the answers are at an appropriate level. If this is homework, please consider adding the [homework] tag; people will still help, so don't worry. –  Arturo Magidin Feb 27 '12 at 16:52
    
sorry i don't know how yo use latex and like because of this it isnt appearing. Does anyone know of a good site to teach me latex? –  Sarah Feb 27 '12 at 16:53
    
Thanks @ArturoMagidin –  Sarah Feb 27 '12 at 16:56

1 Answer 1

up vote 1 down vote accepted

The two properties are not "contradictory", they are complementary. Both of them are true.

(To say that they are contradictory would be like saying that "$30 = 2\times 15$" is contradictory with "$30 = 3\times 10$". They aren't contradictory, they can both hold at the same time).

For inner products over the real numbers, both equalities hold: $$\langle c\mathbf{u},\mathbf{v}\rangle = c\langle\mathbf{u},\mathbf{v}\rangle = \langle\mathbf{u},c\mathbf{v}\rangle$$ for all vectors $\mathbf{u}$ and $\mathbf{v}$ and all scalars $c$.

In order to prove it, however, one needs to know exactly what properties of the inner product you are assuming. I'm guessing that they are the following:

  1. $\langle \mathbf{u},\mathbf{u}\rangle\geq 0$ for all $\mathbf{u}$; $\langle \mathbf{u},\mathbf{u}\rangle = 0$ if and only if $\mathbf{u}=\mathbf{0}$;
  2. $\langle \mathbf{u}+\mathbf{w},\mathbf{v}\rangle = \langle \mathbf{u},\mathbf{v}\rangle + \langle\mathbf{w},\mathbf{v}\rangle$ for all $\mathbf{u},\mathbf{v},\mathbf{w}$.
  3. $c\langle \mathbf{u},\mathbf{v}\rangle = \langle c\mathbf{u},\mathbf{v}\rangle$ for all $\mathbf{u}, \mathbf{v}$ and all $c$.
  4. $\langle\mathbf{u},\mathbf{v}\rangle = \langle\mathbf{v},\mathbf{u}\rangle$ for all $\mathbf{u},\mathbf{v}$.

If this ist he case, start with $\langle \mathbf{u},c\mathbf{v}\rangle$, and then use 4, 3, and 4 again to get the desired result.

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Thanks @Arturo Magidin –  Sarah Feb 27 '12 at 17:03

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