Linear Algebra

[willmoore21]

Ok so here is the question I'm having problems with. We are only just learning this side of linear algebra and it's thrown me off, we don't really have any examples in lectures that I can go through and sort of work it out.



I don't think I need to show spanning because I think it's obvious. The first thing I want to go about is showing linear independance.
I have done an example before where I show independance when I have given an R2 example and it's just numbers, but am not sure how to do it for functions. How do I write these as a linear combination of the other vectors?

EDIT: I know that to show linear dependance you add the vectors and set them to zero with unknown coeffiecients and solve, but I'm not sure whether to replace the x or not.

What I mean:

Asin(x) + Bcos(x) =0
sin(A) + cos(B) = 0

I don't know which one to use, and if I did I don't know how to solve the top one because x could be between 0 and pi.


SECOND EDIT: NEW QUESTION

I have found the definate integral of sin(x)cos(x)dx between 0 and pi, it is 0. Does this make it orthonormal? Do I have to prove something else?

 

[daon2]

Suppose there are constants a,b such that \(\displaystyle a\sin x + b\cos x \equiv \textbf{0}\) for all x. It is important to realize this is a vector space of functions, and by \(\displaystyle \textbf{0}\) here, I do not mean the real number 0. I mean the zero function, which is the function 0(x)=0

Then show that \(\displaystyle a=b=0\)

Hint: plug in some "nice" values.

 

[willmoore21]

Ok, every value I plug in, 0,1,pi,pi/2 always make one positive and one zero. So a and b have to be 0. But how do I write this in a proof type form. Surely I cant say for 0<x<pi this makes one positive and one zero so a and b have to equal zero?

 

[willmoore21]

I have worked out the inner product and found it to be zero, but does the integral of sin^2(x)=1, and cos^2(x)=1 for it be to be an othornormal basis? They equal each other but not 1.

 

[willmoore21]

Update

Ok, so now I am still at the same place with linear independance.

I know that when sin(x) is present cos(x) is 0 and visa versa for 0<x<pi.

However I do not know how to write this out properly? Can anyone help with that?

and Secondly,

I know that the inner product = 1.

However <f,f> = <g,g> does not equal 1. It equals pi

I think this means if I take the f(x) say, and redefine a new function say h(x), where h(x)=(pi^-0.5)sin(x), now <h,h>=1,

so is that part correct, is that problem solved of orthonomality?

 

[pka]

Ok, so now I am still at the same place with linear independance.

Suppose that \(\displaystyle a\cos(x)+b\sin(x)=0\) for all \(\displaystyle x\).
If \(\displaystyle x=0\text{ then }a=~?\)

If \(\displaystyle x=\frac{\pi}{2}\text{ then }b=~?\)

What does that tell you

 

[willmoore21]

Suppose that \(\displaystyle a\cos(x)+b\sin(x)=0\) for all \(\displaystyle x\).
If \(\displaystyle x=0\text{ then }a=~?\)

If \(\displaystyle x=\frac{\pi}{2}\text{ then }b=~?\)

What does that tell you

that a=b=0? For all x? <<< Well for some x, but how do I write it for all x?

 

[pka]

that a=b=0? For all x? <<< Well for some x, but how do I write it for all x?

I don't think you really understand linear independence of functions. Go back and review Daon2's reply.

 

[daon2]

You are assuming the equality holds for all x. If the equality must hold for all x, then it must hold for x=0 and x=pi/2.

 

[willmoore21]

You are assuming the equality holds for all x. If the equality must hold for all x, then it must hold for x=0 and x=pi/2.

ahhhhh ok. Thanks.

So

acos(0)+bsin(0)=0
So a must be 0,

acos(pi/2)+bsin(pi/2)=0
So b must be 0,

So this proves that a=b=0.




EDIT: Thanks guys, have the LI sorted and I have the orthonormal part all figured out. Thanks.