Vector space isomorphic to R^n+1

**osodud** · October 24th 2010, 05:29 PM

Hello

Let $P_n (R)= \left\{{a_0 + a_1x + a_2x^2 + ... + a_nx^n: a_i \in{R}}\right\}$ 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 $R^n^+^1$ ?

Thanks

**TheEmptySet** · October 24th 2010, 06:08 PM

Originally Posted by osodud

Hello

Let $P_n (R)= \left\{{a_0 + a_1x + a_2x^2 + ... + a_nx^n: a_i \in{R}}\right\}$ 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 $R^n^+^1$ ?

Thanks

Does the map

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

$T(ax^j)=\underbrace{(0,0,...,a,0,...,0))}_{\text{ a is in the j+1 component}}$
Does this define an isomorphism?

**HallsofIvy** · October 25th 2010, 05:16 AM

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

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