# Vector space isomorphic to R^n+1

• Oct 24th 2010, 05:29 PM
osodud
Vector space isomorphic to R^n+1
Hello

Let $\displaystyle P_n (R)= \left\{{a_0 + a_1x + a_2x^2 + ... + a_nx^n: a_i \in{R}}\right\}$ be our set.

The sum and the scalar product are defined normally.

How can i prove that this set is with those operations is isomorphic to$\displaystyle R^n^+^1$ ?

Thanks
• Oct 24th 2010, 06:08 PM
TheEmptySet
Quote:

Originally Posted by osodud
Hello

Let $\displaystyle P_n (R)= \left\{{a_0 + a_1x + a_2x^2 + ... + a_nx^n: a_i \in{R}}\right\}$ be our set.

The sum and the scalar product are defined normally.

How can i prove that this set is with those operations is isomorphic to$\displaystyle R^n^+^1$ ?

Thanks

Does the map

$\displaystyle T:\mathbb{P}_n \to \mathnn{R^{n+1}}$ defined by

$\displaystyle T(ax^j)=\underbrace{(0,0,...,a,0,...,0))}_{\text{ a is in the j+1 component}}$
Does this define an isomorphism?
• Oct 25th 2010, 05:16 AM
HallsofIvy
A very general and not very difficult theorem- if two finite dimensional vector spaces have the same dimension, then they are isomorphic.

That is, if U has dimension n, then it has a basis $\displaystyle \{u_1, u_2, \cdot\cdot\cdot, u_n\}$. If V also has dimension n, then it has a basis $\displaystyle \{v_1, v_2, \cdot\cdot\cdot, \v_n\}$. The function, $\displaystyle f:U\to V$, defined by $\displaystyle f(u_i)= v_i$ and extended "by linearity" is an isomorphism.