Can anyone tell me if my proof is correct or if there is something lacking in my argument? I am not too confident about it. I used some hints to construct my argument but I took a different route to that of the book. If anyone has some advice I would really appreciate it. The problem states the following:
Prove that if $W_1$ and $W_2$ are finite-dimensional subspaces of a vector space $V$, then the subspace $W_1+W_2$ is finite-dimensional, and $\dim{(W_1+W_2)}=\dim{(W_1)}+\dim{(W_2)}-\dim{(W_1\cap W_2)}$.
Proof: Let $\alpha=\{u_1, u_2, \dots,u_k\}$ be a basis for $W_1\cap W_2$. Then $\alpha$ is a linearly independent set in both $W_1$ and $W_2$. So we can extend $\alpha$ to $\beta=\{u_1, \dots, u_k, v_1,\dots, v_n\}$, a basis for $W_1$ and we can also extend $\alpha$ to $\gamma=\{u_1, \dots, u_k, w_1,\dots, w_m\}$, a basis for $W_2$. We first show that $W_1+W_2$ is finite-dimesional. Any vector $w$ in $W_1+W_2$, by definition, can be expressed as $w=x+y$, where $x$ belongs to $W_1$ and $y$ belongs to $W_2$. Then, we can find scalars in $F$ such that $x=\sum_{i=1}^{n}{a_i v_i}+\sum_{i=1}^{k}{b_i u_i}$ and $y=\sum_{i=1}^{m}{c_i w_i}+\sum_{i=1}^{k}{d_i u_i}$. So $w$ is a linear combination of the vectors in the set $\delta=\beta\cup\gamma=\{u_1, \dots, u_k, v_1,\dots, v_n, w_1,\dots, w_m\}$, a finite set, which means it is also a generating set. Consequently, we can reduce $\delta$ to a basis for $W_1+W_2$ which yields the result. We claim now that $\delta$ is linearly independent. To show it we express the linear combination: $$\sum_{i=1}^{k}{x_i u_i}+\sum_{i=1}^{n}{y_i v_i}+\sum_{i=1}^{m}{z_i w_i}=0$$ From which $$v=\sum_{i=1}^{n}{y_i v_i}=-\sum_{i=1}^{k}{x_i u_i}-\sum_{i=1}^{m}{z_i w_i}$$ Then, $v$ is a linear combination of vectors in $W_1\cap W_2$ and vectors in $W_2$. We also have that $v$ is a linear combination of the vectors in $W_1$, so $v$ belongs to $W_1\cap W_2$. Then, there exist $e_i$ in $F$ such that $$v=\sum_{i=1}^{k}{e_i u_i}$$ Thus, $$\sum_{i=1}^{k}{e_i u_i}+\sum_{i=1}^{k}{x_i u_i}+\sum_{i=1}^{m}{z_i w_i}=0$$ so $e_1=\dots=e_k=x_1=\dots=x_k=z_1=\dots=z_m=0$ since the set $\gamma$ is linearly independent. This also implies $v=0=\sum_{i=1}^{n}{y_i v_i}$, so $y_1=\dots=y_n=0$ since the set $\{v_1, \dots, v_n\}$ is a subset of a linearly independent set $\beta$. Combining our results we find that $\delta$ is a generating and linearly independent set, so it is a basis for $W_1+W_2$ and $$\dim{(W_1+W_2)}=k+n+m=k+n+k+m-k=\dim{(W_1)}+\dim{(W_2)}-\dim{(W_1\cap W_2)}$$