Isomorphism between $U$ and $\mathbb R^n$

Question

Two vector spaces are called isomorphic if an isomorphism exists between them, and we write $V \cong W$.

A mapping $$f: V\rightarrow W$$ is called an isomorphism if it is linear and bijective.

Show that if $U$ is vector space over $\mathbb{R}$ finitedimensional, then $U\cong \mathbb{R}^n$ for some $n$.

Now I know that a finitedimensional vector space has a basis with finite amount of vectors. And I know what linear and bijective is, but im not sure how to proceed.

Answer 1

Hints:

Choose a basis $\;\{u_1,...,u_n\}\;$ of $\;U\;$ , and then define

$$Tu_i:=e_i\in\Bbb R^n\;\;,\;\;e_i:=i\rightarrow\begin{pmatrix}0\\\ldots\\1\\\ldots\\0\end{pmatrix}$$

and expand definition of $\;T\;$ by linearity. Check it is a linear transformation and bijective

Answer 2

Hint: If a linear transformation takes basis into basis then is an isomorphism.

Take $\{v_1,\ldots, v_n\} \subset U$ you may define a transformation $A : \mathbb R^n \to U$ that takes each $v = (\alpha_1,\ldots,\alpha_n) \in \mathbb R^n$ and

$$Av = \alpha_1v_1 + \ldots + \alpha_n v_n$$

And $Ae_1 = v_1, \ldots , Ae_n = v_n$, where $\{e_1, \ldots, e_n\}\subset \mathbb R^n$ is a canonical basis.