# Inner Product as a Transformation

1. Oct 24, 2012

### nautolian

1. The problem statement, all variables and given/known data

Let V be an inner product space. For v ∈ V fixed, show
that T(u) =< v, u > is a linear operator on V .

2. Relevant equations

3. The attempt at a solution

First to show it is a linear operator, you show that T(u+g)=T(u)+T(g) and T(ku)=kT(u)
So,
T(u+g)=<v, u+g>=<v,u>+<v,g>=T(u)+T(g)
Then T(ku)=<v, ku>=k<v,u>
And since both the results are in the inner product space it is a linear operator on V? However, I don't know if this is right, because can't you not split up the second value like I did? Only the first? A little clarification would be appreciated! Thanks!

2. Oct 24, 2012

### Dick

If your inner product is over the real numbers then both are fine. If it's over the complex numbers then you'll have to say exactly how your inner product is defined.

3. Oct 25, 2012

### hedipaldi

By definition the inner product is bilinear,that is,linear in eahh variable while the other variable is held constant.

4. Oct 25, 2012

### Ray Vickson

In complex spaces this is not entirely correct. If we use the *usual* definition of inner product $$<u,v> \equiv \sum_{i=1}^n \bar{u}_i v_i,$$ where the bar denotes the complex conjugate, then
$$<u,cv> = c<u,v>, \text{ but } <cu,v> = \bar{c}<u,v>.$$
Note: this definition of inner product gives <u,u> ≥ 0 and real; the other type
$$(u,v) \equiv \sum_{i=1}^n u_i v_i$$ gives (u,u) = complex number, in general.

RGV