Direct Sum Decomposition

Question

Let $V_1, V_2 $ be two subspaces of a real vector space $V$.

Define $ V_1 + V_2 = \{v_1 + v_2 : v_1 ∈ V_1, v_2 ∈ V_2\} $.

Consider the case $V = R^{n+m}$ Provide a natural direct sum decomposition of $R^{n+m}$ as $R^n⊕ R^m $

I really am not sure how to begin. My thought is to show that since a basis for $R^{n+m}$ has $n+m$ elements, I can extract two bases from the set: one with n elements and another with m elements. Then I somehow prove they are bases for two subspaces $(V_1, V_2)$ of $V$ and $R^n⊕ R^m = R^{n+m}$ I don't think I am on the right track or understanding this question well at all. Any help would be much appreciated.

Answer 1

Let$$V_1=\{(x_1,x_2,\ldots,x_{n+m})\,|\,x_{n+1}=x_{n+2}=\cdots=x_{n+m}=0\}$$and let$$V_2=\{(x_1,x_2,\ldots,x_{n+m})\,|\,x_1=x_2=\cdots=x_n=0\}.$$Then $V_1\simeq\mathbb{R}^n$, $V_2\simeq\mathbb{R}^m$, and$$\begin{array}{ccc}V_1\oplus V_2&\longrightarrow&\mathbb{R}^{m+n}\\(v_1,v_2)&\mapsto&v_1+v_2\end{array}$$is an isomorphism.