Question

Show that the set \[\left\{ 1,x,x^2,x^3,... \right\}\]is a basis of the vector space \[\mathbb{R}[x]\]

Answer 1

\(\mathbb{R}[x]\) is the vector space of n'th degree polynomials involving one variable?

Answer 2

Just not sure about the notation I'm afraid..

Answer 3

im trying to think of how to word this lol <.<

Answer 4

um... ok. you want to show that if you have a linear combination of the basis "vectors", that:
\[c_0(1)+c_1(x)+c_2(x^2)+\ldots c_n(x^n) = 0 \iff c_0 =c_1 = \ldots = c_n = 0 \]
Then you claim that this is true for any n.

Answer 5

now that first statement:
\[c_0(1)+c_1(x)+c_2(x^2)+…+c_n(x^n)=0\] has to hold for all x.
By the Fundamental Theorem of Algebra, we know that a polynomial of degree n has at most n roots. So for the above to hold for all x, that does imply that each coefficient is 0.

So the set:
\[\left\{ 1,x,x^2,\ldots,x^n \right\}\]is linearly independent, and because this holds for all n, the set:
\[\left\{ 1,x,x^2,\ldots \right\}\]is also linearly independent.

Answer 6

You also have to satisfy the other requirements of a basis.

Answer 7

Closure under vector addition, and closure under scalar multiplication.

Answer 8

those are the requirements for a space to be a vector space. I dont think that has anything to do with the basis itself.
\[\mathbb R[x]\]
does follow those 2 properties (along with 0 being in the space as well.)
It is the set of all polynomials in one variable x, whose coefficients are real numbers.