If \({latex.inline[V_{1}, V_{2}](V_{1}, V_{2})} are subspaces of a finite dimensional vector space, then \){latex.inlinedim(V{1} + V{2}) = dim\ V{1} + dim\ V{2} - dim(V{1} \cap V{2})}.
Let \({latex.inline[v_{1}, ..., v_{m}](v_{1}, ..., v_{m})} be a basis of \){latex.inlineV{1} \cap V{2}}; thus \({latex.inline[dim(V_{1} \cap V_{2}) = m](dim(V_{1} \cap V_{2}) = m)}. Because \){latex.inlinev{1}, ..., v{m}} is a basis of \({latex.inline[V_{1} \cap V_{2}](V_{1} \cap V_{2})}, it is linearly independent in \){latex.inlineV_{1}}. Hence this list can be extended to a basis \({latex.inline[v_{1}, ..., v_{m}, u_{1}, ..., u_{j}](v_{1}, ..., v_{m}, u_{1}, ..., u_{j})} of \){latex.inlineV_{1}} per 1753318250 - Axler 2.32 Every linearly independent list extends to a basis|2.32. Thus \({latex.inline[dim\ V_{1} = m + j](dim\ V_{1} = m + j)}. Also extend \){latex.inlinev{1}, ..., v{m}} to be a basis of \({latex.inline[V_{2}](V_{2})} like \){latex.inlinev{1}, ..., v{m}, w{1}, ..., w{k}}. Thus \({latex.inline[dim\ V_{2} = m + k](dim\ V_{2} = m + k)}. We will show that \){latex.inlinev{1}, ..., v{m}, u{1}, ..., u{j}, w{1}, ..., w{k}} is a basis of \({latex.inline[V_{1} + V_{2}](V_{1} + V_{2})}. This will complete the proof because it shows that \){latex.inlinedim(V{1} + V{2}) = m + j + k = (m + j) + (m + k) - m = dim\ V{1} + dim\ V{2} - dim(V{1} \cap V{2})}.
The combined list of v’s, u’s, and w’s is contained in \({latex.inline[V_{1} \cup V_{2}](V_{1} \cup V_{2})} and thus contained in \){latex.inlineV{1} + V{2}}. The span of the list contains \({latex.inline[V_{1}\ and\ V_{2}](V_{1}\ and\ V_{2})} and hence is equal to \){latex.inlineV{1} + V{2}} (why???). Thus, we only need to show that this list is linearly independent. To do so, suppose:
${latex.inlinea{1}v{1} + ... + a{m}v{m} + b{1}u{1} + ... + b{j}u{j} + c{1}w{1} + ... + c{k}w{k} = 0}
where all a’s, b’s, and c’s are scalars. We need to prove that all a’s, b’s, and c’s are equal to 0. Well, that equation can be rewritten as:
\({latex.inline[c_{1}w_{1} + ... + c_{k}w_{k} = -a_{1}v_{1} - ... - a_{m}v_{m} - b_{1}u_{1} - ... - b_{j}u_{j}](c_{1}w_{1} + ... + c_{k}w_{k} = -a_{1}v_{1} - ... - a_{m}v_{m} - b_{1}u_{1} - ... - b_{j}u_{j})} which shows that \){latex.inlinec{1}w{1} + ... + c{k}w{k} \in V_{1}}. All the w’s are in \({latex.inline[V_{2}](V_{2})} so this implies that the linear combinations of w’s with the c’s as scalar coefficients are in \){latex.inlineV{1} \cap V{2}}. Because \({latex.inline[v_{1}, ..., v_{m}](v_{1}, ..., v_{m})} is a basis of \){latex.inlineV{1} \cap V{2}}, we have that:
\({latex.inline[c_{1}w_{1} + ... c_{k}w_{k} = d_{1}v_{1} + ... + d_{m}v_{m}](c_{1}w_{1} + ... c_{k}w_{k} = d_{1}v_{1} + ... + d_{m}v_{m})} but \){latex.inlinev{1}, ..., v{m}, w{1}, ..., w{k}} is linearly independent so the last equation implies that all c’s and d’s equal 0. Thus we have:
\({latex.inline[a_{1}v_{1} + ... a_{m}v_{m} + b_{1}u_{1} + ... + b_{j}u_{j} = 0](a_{1}v_{1} + ... a_{m}v_{m} + b_{1}u_{1} + ... + b_{j}u_{j} = 0)}. But the list \){latex.inlinev{1}, ..., v{m}, u{1}, ..., u{j}} is linearly independent, so this means all a’s and b’s are 0 which completes the proof.