Inner product

  • Thread starter dirk_mec1
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  • #26
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The instructor has put a pdf file on the internet (which I didn't see), apparently the inner product is defined as above:

http://img221.imageshack.us/img221/2899/58958904pd7.png [Broken]

So then the answer is [tex]||g|| = \sqrt{ \frac{1}{3} } [/tex]

But for <f,g> the inner product can have two values depending which function you use for the overbar, right?
 
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  • #27
gabbagabbahey
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That's better, but <f,g> only has one value...its is the one where g is complex conjugated...<g,f> would have f conjugated.

Is there a definition like this from your course notes that you can post for span{}?....different authors use different notation sometimes.
 
  • #28
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That's better, but <f,g> only has one value...its is the one where g is complex conjugated...<g,f> would have f conjugated.
Yes, you're right I get then [tex]<f,g> = \frac{4i-1}{17} [/tex]

Is there a definition like this from your course notes that you can post for span{}?....different authors use different notation sometimes.

I found this in the lecture notes:

http://img135.imageshack.us/img135/3959/11448859oi8.png [Broken]
 
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  • #29
morphism
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Doesn't that depend on which definition of inner product you use?
Halls is saying that (f,f) (both terms are f) is always real and nonnegative. This is certainly always true. Of course (f,g) can be any complex number if g!=f.
 
  • #30
HallsofIvy
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Doesn't that depend on which definition of inner product you use?
How many definitions of "inner product" are there?

The only one I know is
"An inner product on a vector space V is a mapping from VxV to C (or R is the vector space is over R) such that
1) <au+ bv,w>= a<u,v>+ b<v,w>

2) <u, v>= complex conjugate of <v, u> (or just <v,u> if the vector space is over R)

3) <u, u> [itex]\ge [/itex] 0

4) <u, u>= 0 if and only if u= 0."

If by "defining" the inner product, you mean defining such a function on a specific vector space, if it does not satisfy those, it is NOT an inner product.
 
  • #31
HallsofIvy
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The instructor has put a pdf file on the internet (which I didn't see), apparently the inner product is defined as above:

http://img221.imageshack.us/img221/2899/58958904pd7.png [Broken]

So then the answer is [tex]||g|| = \sqrt{ \frac{1}{3} } [/tex]

But for <f,g> the inner product can have two values depending which function you use for the overbar, right?

Yes, you can. The innerproduct, over the complex numbers, is NOT symmetric:
[tex]<f, g>= \overline{<g, f>}[/tex]
 
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  • #32
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Hmmm...I'm not too familiar with the notation here. What exactly do the []'s mean?

Here is how I would write the span:

[tex]\text{span} \{ f,g \} =\{ \alpha_n \left( \frac{-1}{2} \right) ^n+ \beta_n \left( \frac{i}{2} \right) ^n|\alpha_n,\beta_n \in \mathbb{C} \quad \forall \; n \in \mathbb{N} \}[/tex]
I talked to the instructor and the expression above is incorrect.

It should be like this:

[tex]\text{span} \{ f,g \} =\{ \alpha \left( \frac{-1}{2} \right) ^n+ \beta \left( \frac{i}{2} \right) ^n|\alpha,\beta \in \mathbb{C} \quad \forall \; n \in \mathbb{N} \}[/tex]

Notice that the constants are independent of n.
 
  • #33
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How many definitions of "inner product" are there?

The only one I know is
"An inner product on a vector space V is a mapping from VxV to C (or R is the vector space is over R) such that
1) <au+ bv,w>= a<u,v>+ b<v,w>

2) <u, v>= complex conjugate of <v, u> (or just <v,u> if the vector space is over R)

3) <u, u> [itex]\ge [/itex] 0

4) <u, u>= 0 if and only if u= 0."

If by "defining" the inner product, you mean defining such a function on a specific vector space, if it does not satisfy those, it is NOT an inner product.

Yes, this definition is used in my lecture notes.
 
  • #34
761
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Yes, you can. The innerproduct, over the complex numbers, is NOT symmetric:
[tex]<f, g>= \overline{<g, f>}[/tex]

What do you mean by "yes, you can"? I presume you mean that only one value corresponds to <f,g> with f and g complex.
 

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