Spanning Vector Spaces with Trigonometric Functions

[ssb]

Homework Statement

Is there any difference between the vector space spanned by the set cos(t),sin(t) and the vector space spanned by the set cos(t)+sin(t),cos(t)-sin(t)?

Homework Equations

The Attempt at a Solution

Not really a homework question but it will help me answer a problem set given to me by my teacher. I would have to guess that the answer is that there is no difference because of the behavior of sin. if it begins at any place on a number line it doesn't matter if its going in the positive or negative direction, it will still travel the other way in the manor that of which if its sign was reversed...

im sorry if that doesn't make sense. I hope someone who reads this will understand what I mean.

 

[Hurkyl]

I can't figure out what you are thinking -- but there is no linear algebra in your solution, so it's clearly not a solution.

Maybe... before trying to answer the question, you should spend some effort stating the question? Mathematically, what are you given, and what are you trying to show?

 

[ssb]

The exact word for word question given to me by my teacher is

"Is there any difference between the vector space spanned by the set cos(t),sin(t) and the vector space spanned by the set cos(t)+sin(t),cos(t)-sin(t)?"

But he's not asking us to answer that question directly... but for all intensive purposes he might as well be.

What I mean when I try to describe the situation is if you are given sin(x), it doesn't matter where on X you begin graphing sin(x) because it will always cover the same places. You could start at 5 and graph to zero. You could start at zero and graph to 5. The end result would be 2 graphs that look the exact same. That was why I believed that the answer was no because of the properties of Sin.

 

[quantumdude]

But he's not asking us to answer that question directly... but for all intensive purposes he might as well be.

I think you mean "for all intents and purposes".

Here's an example that might help.

Consider the standard basis of $P_2$: $\left\{1,x,x^2\right\}$. Does the set $\left\{1,x+1,x^2-1\right\}$ span $P_2$ as well?

You need to go back to the definition of "span". The span of a set of vectors is the set of all possible linear combinations of those vectors. A general linear combination from my first set would be:

$p(x)=a_2x^2+a_1x+a_0$,

where the coefficients come from some field. Now let's write down a general linear combo from my second set:

$b_2\left(x^2-1\right)+b_1(x+1)+b_0$.

We can rearrange that as follows:

$b_2x^2+b_1x+(-b_2+b_1+b_0)$.

Note that the coefficients are arbitrary elements of the field over which the vector space is defined. So we are free to rename them as follows:

$b_2 \rightarrow a_2$
$b_1 \rightarrow a_1$
$(-b_2+b_1+b_0) \rightarrow a_0$

Do that, and the second linear combo becomes identical to the first. So yes, the second set does span $P_2$.

Understand?

 

[ssb]

I think you mean "for all intents and purposes".

Here's an example that might help.

Consider the standard basis of $P_2$: $\left\{1,x,x^2\right\}$. Does the set $\left\{1,x+1,x^2-1\right\}$ span $P_2$ as well?

You need to go back to the definition of "span". The span of a set of vectors is the set of all possible linear combinations of those vectors. A general linear combination from my first set would be:

$p(x)=a_2x^2+a_1x+a_0$,

where the coefficients come from some field. Now let's write down a general linear combo from my second set:

$b_2\left(x^2-1\right)+b_1(x+1)+b_0$.

We can rearrange that as follows:

$b_2x^2+b_1x+(-b_2+b_1+b_0)$.

Note that the coefficients are arbitrary elements of the field over which the vector space is defined. So we are free to rename them as follows:

$b_2 \rightarrow a_2$
$b_1 \rightarrow a_1$
$(-b_2+b_1+b_0) \rightarrow a_0$

Do that, and the second linear combo becomes identical to the first. So yes, the second set does span $P_2$.

Understand?

WHAHOO! Thanks buddy.