Linear Algebra Example Problems - A Polynomial Subspace

The vector space P3 is the set of all at most 3rd order polynomials with the "normal" addition and scalar multiplication operators.

In this video we work with a subset of elements from P3 and formally show that it is a subspace. Since P3 is already a vector space, to show this subset is a subspace we only need to check 3 properties, not all 10 normal vector space properties. We work through each of these 3 properties to show the subset is indeed a subspace.

[nightury]

I loved the handwriting more than the explanation....
Well you explained it nicely though.

[danielmoss]

Many thanks for making this video! I am studying this at university and I'm really struggling with some difficult questions (ones that involve determining whether a derivative/integral is a subspace of a vector space) as part of our coursework :(

[cybrok]

Man, thank you very nice simple explanation..

[sobanfarooq]

What if b = 0, then if we do 'b.a(1).x^3' wouldn't it become zero and wouldn't stay a polynomial of degree 3 anymore?

[kunalkheeva]

how will you deal with closing the addition of 1+a^3 and 1-a^3… it gives 2 which has no degree.

[OPOD]

Very nice videos. They are quite helpful. I noticed your last name. Do you happen to be of Greek descent ?

[Learnwithme.]

Is set of upto second degree polynomials where p’(x) = x
a subspace?

[nadimbaraky]

What if a1 = 1 & a2 = -1. It is not closed under addition at all

[lemyul]

but, can't you make a counterexample to show they are not closed under these operations?

[abdullahbsali]

Hi, what if you have a polynomial p(x): p(-1)= 0
How would you prove that it is a subspace of P3(R) as there's no constant a like there was for the example in the video.

[mohammadamananwar]

is a set of polynomials of degree three a vector space?
for vector addition: x^3+(-x^3) is not a third-degree polynomial

[thomasolson]

This seems to be about the scalars being in the polynomial. I was wondering if there was anything on scalars in the solutions of groups of polynomials. Sorry, that's a crappy way of describing it. I can't do better.

[0, (1/10)*sqrt(25-5*sqrt(5)), -(1/10)*sqrt(25-5*sqrt(5)), (1/10)*sqrt(25+5*sqrt(5)), -(1/10)*sqrt(25+5*sqrt(5)), (1/4)*sqrt(2)-(1/20)*sqrt(10), -(1/4)*sqrt(2)+(1/20)*sqrt(10), (1/4)*sqrt(2)+(1/20)*sqrt(10),

[(1/5)*sqrt(5), -(1/5)*sqrt(5), (1/10)*sqrt(20-5*sqrt(10-2*sqrt(5))), -(1/10)*sqrt(20-5*sqrt(10-2*sqrt(5))), (1/10)*sqrt(20-5*sqrt(10+2*sqrt(5))), -(1/10)*sqrt(20-5*sqrt(10+2*sqrt(5))), (1/10)*sqrt(20+5*sqrt(10-2*sqrt(5))), -(1/10)*sqrt(20+5*sqrt(10-2*sqrt(5))), (1/10)*sqrt(20+5*sqrt(10+2*sqrt(5))),
So if I were to take those and compare them to the 10th Chebyshev First Kind and Second kind the scalar would be (1/5)*sqrt(5)*sqrt(2) in each one of the solutions. Maybe... I might have screwed something up.

[JohnSmith-cwhs]

has anyone ever told you how similar you look to the comedian Joe Pera?