Verifying dot product and finding h

[dirk_mec1]

Homework Statement

http://img152.imageshack.us/img152/3851/33495448dh9.png

Homework Equations

http://img146.imageshack.us/img146/4655/37276835io7.png

The Attempt at a Solution

Well I found:

<br /> ||f|| = \frac{1}{ \sqrt{3}}<br />

<br /> ||g||=\frac{i}{ \sqrt{3}}<br />


<br /> &lt;f,g&gt; = \frac{1-4i}{17}<br />

Can someone verify the above?



The last question really bothers me. I know that the inproduct must be zero but what does the span explicitly mean? Is it the term below?

<br /> \alpha \cdot \left( \frac{-1}{2} \right) ^n + \beta \left( \frac{i}{2} \right) ^n <br />

But if that's the case how can I find h?

 

[gabbagabbahey]

I'm not seeing any questions. I can't check your solutions if I don't know the question.

 

[dirk_mec1]

I'm not seeing any questions. I can't check your solutions if I don't know the question.

Oops, sorry! I've edited the post.

 

[HallsofIvy]

Each of these is correct except ||g||. The norm is MUST be a real number. That is true even for a space over the complex numbers. Each of <f, f>, <f, g>, and <g, g> is a geometric series so you can use
\sum_{n=0}^\infty ar^n= \frac{a}{1- r}

 

[gabbagabbahey]

Your ||g|| and <f,g> are a little off, perhaps you should show me your work for those two.

 

[dirk_mec1]

||g|| = \sum_{k=0}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k = \sum_{k=0}^{\infty} \left( \frac{-1}{2} \right) ^k = |-1/3| =1/3

 

[gabbagabbahey]

shouldn't you have (-1)^k/4 ? ;0) ...(and the summation should start at k=1, since n is a natural number)

 

[dirk_mec1]

Yes, you're right! I'll try again:

<br /> ||g|| = | \sum_{k=1}^{\infty} \left( \frac{i}{2} \right) ^k \left( \frac{i}{2} \right)^k | =| \sum_{k=1}^{\infty} \left( \frac{-1}{4} \right) ^k |= |\frac{-1}{5}| = \frac{1}{5} <br />

 

[Office_Shredder]

(and the summation should start at k=1, since n is a natural number)

That's a bold assertion to make in a math forum

For the last part, the span of f and g is all linear combinations of f and g (so what you put up in the OP describes the nth term of an arbitrary element in the span)

 

[gabbagabbahey]

||g|| looks good now, but I think you are still missing a neg sign in your <f,g>

 

[dirk_mec1]

&lt;f,g&gt; = | \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m | =| \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m |= | \frac{-i}{i+4} |= | \frac{-1-4i}{17} | = <br />

 

[gabbagabbahey]

Why do you have modulus signs in that equation? ...And (4-i)(-i)= (-1)-4i not (+1)-4i ;0)

 

[dirk_mec1]

So the modulus sign is only inserted in the norm calculation?

<br /> = \sum_{m=1}^{\infty} \left( \frac{-1}{2}\ \right) ^m \left( \frac{i}{2} \right) ^m = \sum_{m=1}^{\infty} \left( \frac{-i}{4} \right) ^m = \frac{-i}{i+4} = \frac{-1-4i}{17} = <br /> <br />

 

[gabbagabbahey]

Yes, if you were asked to find ||<f,g>|| then you would use them, but <f,g> doesn't even need to be real, let alone positive.

 

[dirk_mec1]

That's a bold assertion to make in a math forum
For the last part, the span of f and g is all linear combinations of f and g (so what you put up in the OP describes the nth term of an arbitrary element in the span)

So it is: <br /> <br /> \alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ \left( \frac{i}{2} \right) ^n \right] _{n \in \mathbb{N} }<br /> <br />

But how are you then suppose to find h?

 

[gabbagabbahey]

If you knew what h was, what would span{f,h} be?

 

[dirk_mec1]

I think this:

\alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ h(n)] _ {n \in \mathbb{N} }<br /> <br /> <br />

 

[gabbagabbahey]

right, now use that along with the fact that <f,h>=0 (and span{f,g}=span{f,h}) to find h(n).

 

[dirk_mec1]

Wait something is wrong...

There holds: &lt;g,g&gt; = -1/5 but <...> is always larger than 0...so what's wrong?

Okay, I thought of something for h(n)

h(n) = 2^n + (-1)^{n+1}

 

[dirk_mec1]

@gabbagabbahey is

<br /> h(n) = 2^n + (-1)^{n+1}<br />

a good choice?

 

[gabbagabbahey]

hmmm... let's see, that mean that:

\left&lt;f,h \right&gt; = \sum_{n=1}^{\infty} \left( \frac{-1}{2} \right) ^n (2^n + (-1)^{n+1})=\sum_{n=1}^{\infty}\left( (-1)^n- \left( \frac{1}{2} \right) ^n \right)=\frac{-3}{2} \neq 0

...so, no it doesn't work...any other ideas?

 

[gabbagabbahey]

Wait something is wrong...

There holds: &lt;g,g&gt; = -1/5 but <...> is always larger than 0...so what's wrong?

||g||=||\left&lt; g,g \right&gt; || is always real and greater than or equal to zero, but \left&lt; g,g \right&gt; can be any complex number.

 

[gabbagabbahey]

So it is: <br /> <br /> \alpha \cdot \left[ \left( \frac{-1}{2} \right) ^n \right] _{n \in \mathbb{N} } + \beta \left[ \left( \frac{i}{2} \right) ^n \right] _{n \in \mathbb{N} }<br /> <br />

But how are you then suppose to find h?

Hmmm...I'm not too familiar with the notation here. What exactly do the []'s mean?

Here is how I would write the span:

\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} \}

 

[HallsofIvy]

||g||=||\left&lt; g,g \right&gt; || is always real and greater than or equal to zero, but \left&lt; g,g \right&gt; can be any complex number.

No. One of the conditions for an inner product is that &lt;g, g&gt; \ge 0. Another is that <g,g>= 0 if and only if g= 0. The inner product <f, g>, for two different vectors f and g, over the complex numbers can be any complex number.

 

[gabbagabbahey]

No. One of the conditions for an inner product is that &lt;g, g&gt; \ge 0. Another is that <g,g>= 0 if and only if g= 0. The inner product <f, g>, for two different vectors f and g, over the complex numbers can be any complex number.

Doesn't that depend on which definition of inner product you use?

The one given in his original post is just:

(a,b)=\sum_i a_ib_i

Although, it isn't clear if that applies to complex elements or just reals.

Usually, for complex valued elements, the definition is:

(a,b)=\sum_i a_i\overline{b_i} (where the overbar denote complex conjugation)I admit, I've never seen the inner product defined without the complex conjugation for complex valued elements, but I decided to go with the definition given in his post.

 

[dirk_mec1]

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

So then the answer is ||g|| = \sqrt{ \frac{1}{3} }

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

 

[gabbagabbahey]

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.

 

[dirk_mec1]

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 &lt;f,g&gt; = \frac{4i-1}{17}

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

 

[morphism]

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.

 

[HallsofIvy]

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> \ge 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.

 

[HallsofIvy]

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

So then the answer is ||g|| = \sqrt{ \frac{1}{3} }

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:
&lt;f, g&gt;= \overline{&lt;g, f&gt;}

 

[dirk_mec1]

Hmmm...I'm not too familiar with the notation here. What exactly do the []'s mean?

Here is how I would write the span:

\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} \}

I talked to the instructor and the expression above is incorrect.

It should be like this:

\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} \}

Notice that the constants are independent of n.

 

[dirk_mec1]

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> \ge 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.

 

[dirk_mec1]

Yes, you can. The innerproduct, over the complex numbers, is NOT symmetric:
&lt;f, g&gt;= \overline{&lt;g, f&gt;}

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.