hey i was just wondering how you show this... or well if i am doing or done it right?

 

thanks

 

-Sarah

 

...i'll post my answer in a min, just let me type it up...

ok so for part (a)

 

to show they are linearly independent, you put them in a matrix augmented with a column of 0's (zeros) on the end:

 

[math]

\[ \left( \begin{array}{cccc}

2 & -1 & 3 & 0 \\

-5 & 1 & 0 & 0 \\

0 & 4 & -2 & 0 \end{array} \right)\]

[/math]

 

then row reduce to reduced row echelon form

 

so this goes to this (after a few steps)

[math]

\[ \left( \begin{array}{cccc}

1 & 0 & 4 & 0 \\

0 & 1 & 5 & 0 \\

0 & 0 & -22 & 0 \end{array} \right)\]

[/math]

 

then to this

[math]

\[ \left( \begin{array}{cccc}

1 & 0 & 0 & 0 \\

0 & 1 & 0 & 0 \\

0 & 0 & 1 & 0 \end{array} \right)\]

[/math]

 

and so therefore the subset \beta is a basis for P2

 

hmmm i am not really sure that is right, anyways thats what i did, any help/comments would be welcomed!

aghhhh i can't remember how to do (b), :S

wait, no, i think i do know how to do part (b): like this...

 

and so you have p(t) = (16/27)p1(t) + (1/27)p2(t) + (7/18)p3(t)

 

yep??

I do not really understand why you added the zeroes in an additional collumn. Simply wrinting p1, p2 and p3 in a matrix and taking the determinant would have been enough.

For b: What you are looking for is a set of coefficients a, b, c which satisfy a*p1 + b*p2 + c*p3 = 1 + 1t + 1t². The solution should be straightforward.

 

EDIT: I can only guess what you did to solve b so I cannot comment much on it. I think you solved the equation system I mentioned above so unless you did some error in the computing it should be correct. Just plug your coefficients to the equation above and check if your results are correct.

how do i show that the subset beta spans P_2?

It´s a base for P_2 so it spans P_2.

EDIT: I can only guess what you did to solve b so I cannot comment much on it. I think you solved the equation system I mentioned above so unless you did some error in the computing it should be correct. Just plug your coefficients to the equation above and check if your results are correct.

 

yeah thats what i did, so i think i've got part (b) done, and i sort of showed what i did, with those 2 matrices there (the second one is the rref form of the other)

It´s a base for P_2 so it spans P_2.

 

so how do you show something is a basis then? do you just have to show that it spans P_2 and that the things in the subset are linearly independent?

like how do i show that:

 

P_2 = span{beta} = span{p_1, p_2, p_3}

 

???

I think so, yes. A base always spans the space as being a base implies that each vector can be written as a linear combination of the base vectors. But you could have a set of vectors spanning the space which are not a base (if you were given a 4th vector, for example).

I think so, yes. A base always spans the space as being a base implies that each vector can be written as a linear combination of the base vectors. But you could have a set of vectors spanning the space which are not a base (if you were given a 4th vector, for example).

 

ok, yep, i agree with that

but has what i did in post #2 shown that B is a basis for P_2 ??

or can i just say that well, i have shown they are linearly indepedent, and they obviously span P_2, beause there are non-zero coeffiencts for each bit (t^2, t, constant) etc.

 

but doing this would seem extemely dodget to me :S

what is P^2? obvioulsy it is the space of polynomials of degree at most 2 but ot would be helpful to say so. clearly P^2 is 3 dimensional (the natrual basis being 1,t,t^2), and you have shown you have 3 linearly independent vectors:

 

a general result and in fact the definition of dimension tells us that the set p_1, p_2, and p_3 must be a basis: any linearly independent set iwth n elements in a n n dimensional space must be a basis.

oh ok so i think i was right then

 

thanks matt

 

i just wasnt sure of a formal way of saying it,

"any linearly independent set iwth n elements in a n n dimensional space must be a basis"

i think that must be a thm. somewhere in my book, although i cannot yet find it. but i suppose so long as i say that and include what i did above (showing that p_1, p_2 & p_3 are linearly indepedent) then it sounds complete to me

it may not say it explicitly since it is immediate from the definition of dimension: the maximum cardinality of a linearly independent set, or the minimal cardinality of a spanning set. any linearly independent set can be completed to a basis, so if the cardinality is already maximal (ie no more elements can be added) then it must be a basis.

lol i didnt quite understand that, sorry, but what i said in post #16 was correct?

when you understand it you will see that you are correct; you need to stop being so concerned with unimportant things like this particular question and start to think of the bigger reasons so that you understand why you are correct. if you don't learn why you are correct you will never learn when you are correct and you'll continue to ask us to verify your work. at some point we should not need to do that for a subject.

 

What is a basis? It is a linearly independent spanning set.

1. every basis has the same number of vectors in it, this number is the dimension of the space, let d be the generic letter for dimension.

 

2. all linearly indepedent set have at most d vectors in it.

 

3. any linearly independent set can be completed to a basis.

 

4. d is the maximum number of vectors that can be in a linearly indepedent set.

 

5. put it all together.

i guess i am just a nervous sort of person as well as the stuff you said, so yeah i am never confident, even when i have the answer right the first time

thanks for you help matt

this set is 3-dimensional, right? because there would need to be -1, t and t^{2}, right?

oops sorry yes thats obviously true, i just keep 'typing my mind' as it were, ignore that last post