Inner Product as a Transformation

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Homework Help Overview

The problem involves demonstrating that a specific transformation, defined as T(u) = for a fixed vector v in an inner product space V, is a linear operator. The context is rooted in the properties of inner product spaces and linear operators.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the linearity of the transformation by examining the properties of the inner product, specifically questioning the ability to split terms in the inner product for different cases (real vs. complex). There is also a focus on the bilinearity of the inner product and its implications for the transformation.

Discussion Status

The discussion is exploring the nuances of the inner product's properties, particularly in relation to real and complex numbers. Some participants have provided clarifications regarding the definitions and implications of bilinearity, but there is no explicit consensus on the correctness of the original poster's reasoning.

Contextual Notes

Participants note the distinction between real and complex inner products, indicating that the definitions and properties may vary based on the field over which the inner product is defined. This suggests that assumptions about the inner product's behavior may need to be clarified further.

nautolian
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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!
 
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nautolian said:

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

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