# Prove a property of inner product spaces

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. – 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

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