Question

Let P(n) be the vector space of polynomials p(t) of degree less than or equal to n. Show that {1, t, t^2, ......, t^n} is a basis for P(n)

Answer 1

@helder_edwin
whenever you get online could you please take a look at this??

also the problem yesterday, when I get my homework back I will let you know what they say about that problem we were having trouble with if you are curious.

Answer 2

glad to see someone. im lost on this

Answer 3

Basically, you need to show two things, that the set {1, t, t^2,..., t^n} spans all polynomials that are of degree n or less, and that the objects {1,t,t^2,...,t^n} are linearly independent.

Answer 4

how do you show that is spans all polynomials?

Answer 5

For the spanning part, you need to show that every polynomial of degree n or less is a linear combination of the kids in the set {1,t,t^2,...,t^n}. That is, if you take:\[c_0(1)+c_1(t)+c_2(t^2)+\ldots+c_n(t^n)\]where the coefficients can be anything, you get all the polynomials of degree n or less.

Answer 6

For the linearly independent part, you need to show that if you have a linear combination that comes out to the zero polynomial for any value of t:\[b_0(1)+b_1(t)+b_2(t^2)+\ldots+b_n(t^n)=0\]then each of the coefficients must be zero.

Answer 7

im not quite seeing the spanning part.

Answer 8

i mean if they are linearly independent doesnt that mean they will span too

Answer 9

Given any polynomial, can you tell me the proper linear combination of the objects in {1,t,t^2,...,t^n}?

If i give you p(x)= 5+6t+4t^2, you would say 5 of 1, 6 of t, and 4 of t^2.

If i give you q(x)= -27+43t^3-5t^5, you would say -27 of 1, 43 of t^3 and -5 of t^5.

So if i give you any polynomial, you would be able to tell me the proper linear combination. This is what it means to span.

Answer 10

The set {1, t, t^2,...,t^n} does span the set of all polynomials of degree n or less because every polynomial can be written as a linear combination of the objects in the set.

Answer 11

Linear independence is something separate. You can have a linearly independent set that doesnt span. For example, the set {1,t,t^2} is linearly independent, but doesnt span the set of polynomials of degree 4. You cant create cubic or quartic equations.

Answer 12

i understand your examples, and i understand about taking the linear combination of the examples you have, i guess what i get hung up on these is that it consists of every polynomial, and i just don't know what to actualy write to show both of those things.

Answer 13

its these general cases that i just am seem to leave me blank

Answer 14

yeah, i understand that frustration. Just make sure you are satisfying the definitions. If you write what I posted a few posts up:

"The set {1, t, t^2,...,t^n} does span the set of all polynomials of degree n or less because every polynomial can be written as a linear combination of the objects in the set."

That can cover the spanning part. If you feel "i should show more", then maybe you can write a random polynomial:\[c_0+c_1t+c_2t^2+\ldots+c_nt^n\]then rewrite it to emphasize the linear combination of the things in the set:\[c_0(1)+c_1(t)+c_2(t^2)+\ldots+c_n(t^n)\]

Answer 15

and with the linear independence all you need to show is that the coeffecients are 0?

Answer 16

that is correct. You need to show that if\[b_0+b_1t+b_2t^2+\ldots+b_nt^n=0\]for any value of t, then all the coefficients must be 0.

Answer 17

this one can be a little tricky without the proper ideas. Theres a long way and short way to do it. Did you have any ideas?

Answer 18

but you can tell by looking at it that all values must be 0 in order for they polynomial to be 0... what is there to prove... ugh.

Answer 19

Think of the polynomial \[t^2-8t+15=0\]There are some non-zero values of t that make this 0 (t=3 and t=5).

Answer 20

So why is it that if\[b_0+b_1t+b_2t^2+\ldots+b_nt^n=0\]for all values of t, that each coefficient must be zero? while you are right that its not a hard question, its also not trivial, something must be said.

Answer 21

If I give you a polynomial of degree n, what is the most number of roots that polynomial can have?

Answer 22

webite kicked me off... but isnt it n

Answer 23

thats correct. There can be at most n roots. So because:\[b_0+b_1t+b_2t^2+\ldots+b_nt^n=0\]for ALL values of t, definitely more than n values of t, tells you that this polynomial breaks that rule. Thats why all the coefficients must be zero.

Answer 24

If you want to be specific, the theorem that is being broken is the fundamental theorem of algebra. Thats the one that say a polynomial can have at most n roots.

Answer 25

so does that say enough to show the set p(t) of polynomials is a basis for P(n)

Answer 26

Yes. If you can show that a set spans and it linearly independent, then you have a basis.

There are plenty of sets that span but arent linearly independent, and plenty that are linearly independent and dont span. Its gotta do both to be a basis.

Answer 27

im going to need to write all of this down and try to wrap my head around this one. i dont know why this one is so difficult. and try to put together what I need to exactly say on my homework. I thank you for your help, and being patient even though I still don't quite understand

Answer 28

its a little rough. this is most peoples first taste of an "abstract" vector space. it doesnt deal with numbers and column vectors any more, now you have to operate purely on the definitions. just keep at it, make sure you know the definitions like the back of your hand.