Prove that if $W_1$ is any subspace of a finite-dimensional vector space $V$, then there exists a subspace $W_2$ of $V$ such that $V = W_1 \oplus W_2$
What I have done so far is to note that since $V$ is finite and $W_1$ is a subspace of $V$, we have $\dim(W_1) \leq \dim(V)$. If we have equality, then let $W_2 = \{0\}$, so that we have $V= W_1 \oplus W_2$.
So now I need to look at the case when $\dim(W_1) \lt \dim(V)$.
What I have tried for this case is to let $$\beta = \{v_1,v_2,..,v_n\}$$ be a basis for $V$ and $$\gamma=\{u_1,u_2,..,u_m\}$$ a basis for $W_1$. My idea was to extend $\gamma$ to a basis for $V$, so let $$\alpha=\{u_1,u_2,..,u_m,w_1,w_2,..,w_{n-m}\}$$ be the extension of $\gamma$ to $V$, where $w_1,w_2,..,w_{n-m}$ are basis vectors for $W_2$. If I'm doing this right then I would just have to show that $W_1 \cap W_2 = \{0\}$ and $W_1 + W_2= V$
Am I heading in the right direction? Any hints would be greatly appreciated.
Best Answer
If you already know that you can compete a basis of a subspace to a basis for the whole space then you are practically done.
Hint: Note that $\alpha$ is a basis for $V$ (this should give you $W_{1}+W_{2}=V$, why ?) and that the $w_{i}$ are linearly independent of the $u_{i}$ (this should show that $W_{1}\cap W_{2}=\{0\}$, why ?)
Note: The way I see it, there is no use for $\beta$ or of the $v_{i}$ in the proof