Inner Product as a Transformation

[nautolian]

Homework Statement

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

Homework Equations

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!

 

[Dick]

Homework Statement

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

Homework Equations

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!

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.

 

[hedipaldi]

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

 

[Ray Vickson]

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

In complex spaces this is not entirely correct. If we use the *usual* definition of inner product &lt;u,v&gt; \equiv \sum_{i=1}^n \bar{u}_i v_i, where the bar denotes the complex conjugate, then
&lt;u,cv&gt; = c&lt;u,v&gt;, \text{ but } &lt;cu,v&gt; = \bar{c}&lt;u,v&gt;.
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