Question

Linear indepedence question:

Would a second degree polynomial be considered in r3 or r2? and can the following theorem apply to polynomials:

If s={v1,v2,...,vr} and it is found in R^n, then if r>n, then it is linear depedent

Answer 1

The vector space consisting of second degree polynomials, having dimension 3, would be isomorphic to \(\mathbb{R}^3\). Any theorems that apply to \(\mathbb{R}^n\) will apply to the vector space of polynomials as well.

Answer 2

Ah I see, how can you visualize a second degree in a dimension of 3? ( I can only picture a parabolla)

Answer 3

A second degree polynomial is of the following form: a_1x^2 + a_2x + a_3

Answer 4

It is easy to see that you can put that into one to one correspondence with ordered triples \((a_1, a_2, a_3)\)

Answer 5

I see that, so you really look at it from the coefficient? (might sounds stupid to say, but thats how I just got it)

Answer 6

Yes you can think of the coefficients as the components of the vector if you like.

Answer 7

The standard basis for the vector space is: \(\{x^2, x, 1\}\)

Answer 8

\(\{1x^2 + 0x + 0, 0x^2 + 1x + 0, 0x^2 + 0x + 1\}\)

Answer 9

Ah, I see what youre getting at. So, those would not be parabollas, just any set of number to a certain degree ( don't know if Im making sense)

Answer 10

When working with it as a vector space, you should ignore that they are parabolas. Just think of \(x\) as a placeholder. These are vectors not functions.

Answer 11

I mean in this context.

Answer 12

I see what you're getting at, they're just a set of numbers defined by x, meaning any numbers possible ( within the given form Im retriceuing). I think I understand, thanks for your patience ! :)