# Kenneth Hoffman | Ray Kunze: An Inquiry into Symbolic Meaning

When those authors state the following

$\bf{Theorem 6.}$ If $W_1$ and $W_2$ are finite-dimensional subspaces of a vector space $V$, then $W_1+W_2$ is finite-dimensional and \begin{eqnarray} \dim W_1 + \dim W_2 = \dim (W_1 \cap W_2) + \dim (W_1+W_2). \end{eqnarray} Proof. By Theorem 5 and its corollaries, $W_1 \cap W_2$ has a finite basis $\{\alpha_1,\cdots,\alpha_k\}$ which is part of a basis \begin{eqnarray} \{\alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_m\} \hspace{.3cm} \text{for $W_1$} \end{eqnarray} and part of a basis \begin{eqnarray} \{\alpha_1,\dots,\alpha_k,\gamma_1,\dots,\gamma_m\} \hspace{.3cm} \text{for $W_2$}. \end{eqnarray} Then subspace $W_1+W_2$ is spanned by the vectors \begin{eqnarray} \alpha_1,\dots,\alpha_k,\beta_1,\dots,\beta_m,\gamma_1,\dots,\gamma_n \end{eqnarray}

How is this last part so?

Anything in $W_1$ is a sum of the $\alpha$'s and $\beta$'s, and everything in $W_2$ is a sum of the $\alpha$'s and the $\gamma$'s. Therefore, the span of the $\alpha$'s, $\beta$'s and $\gamma$'s contains both $W_1$ and $W_2$... but since the span is closed under addition it contains $W_1+W_2$.
On the other hand, all those basis vectors are in $W_1+W_2$, so the span can't "escape" $W_1+W_2$. So, you have equality.
I'm not sure exactly what you mean by "do these data suffice," but it's certain that they are relevant steps for proving the proposition. The follow-up is to consider the mapping from $W_1\times W_2\to W_1+W_2$ given by $(x,y)\mapsto x+y$. You can check that the kernel of the map is $W_1\cap W_2$, and so the rank-nullity theorem gives you the equation you seek. (The dimension of $W_1\times W_2$ is $dim(W_1)+dim(W_2)$, and the image of the map is $W_1+W_2$.)