Question

Show that the set P(n) of all polynomials of degree less than or equal to n is a subspace of C(1). What is dim P(2)? What is dim P(n)?

Answer 1

@helder_edwin would you by chance have any ideas on this. I need to show closed under addition, and scalar multiplication. \[p _{1}(t)+...+p _{n}(t)=?????\]
and then \[rp _{1}(t)+....+rp _{n}(t)=?????\]
I really am not sure what to do here?

Answer 2

@joemath314159 if you have a moment, could you maybe look at this problem?

Answer 3

Let \(p(x)=a_0+a_1x+\dots+a_nx^n\) and \(q(x)=b_0+b_1x+\dots+b_nx^n\) in \(P(n)\) then
\[ \large p(x)+q(x)=(a_0+a_1x+\dots+a_nx^n)+(b_0+b_1x+\dots+b_nx^n) \]
\[ \large =(a_0+b_0)+(a_1+b_1)x+\dots+(a_n+b_n)x^n \]
is in P(n).

Also, if \(\alpha\in\mathbb{R}\) then
\[ \large \alpha\cdot p(x)=\alpha(a_0+a_1x+\dots+a_nx^n) \]
\[ \large =(\alpha a_0)+(\alpha a_1)x+\dots+(\alpha a_n)x^n \]
is also in P(n).

Answer 4

how would you find out the dim p(2) and also dim p(n)
is the dim the n value?

Answer 5

no. u know that 1 and x and x^2 are polynomials. right?

Answer 6

yes

Answer 7

I guess im confused on what they mean by p(2) does the mean the first three , 1, x, and x^2? and then it would be dim =3

Answer 8

exactly. because
\[ \large a+bx+cx^2=a\cdot\color{red}{1}+b\cdot\color{red}{x}+c\cdot
\color{red}{x^2} \]
so dim P(2)=3.

and in general dim P(n)=n+1

Answer 9

thank you, well explained. i have one more problem i am having trouble with, i hope you can help.

Answer 10

u r welcome

Answer 11

i'll be logged in for a while.