Theorem

If ${latex.inlineV_{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})}.

Proof

Let ${latex.inlinev_{1}, ..., v_{m}} be a basis of ${latex.inlineV_{1} \cap V_{2}}; thus ${latex.inlinedim(V_{1} \cap V_{2}) = m}. Because ${latex.inlinev_{1}, ..., v_{m}} is a basis of ${latex.inlineV_{1} \cap V_{2}}, it is linearly independent in ${latex.inlineV_{1}}. Hence this list can be extended to a basis ${latex.inlinev_{1}, ..., v_{m}, u_{1}, ..., u_{j}} of ${latex.inlineV_{1}} per 2.32. Thus ${latex.inlinedim\ V_{1} = m + j}. Also extend ${latex.inlinev_{1}, ..., v_{m}} to be a basis of ${latex.inlineV_{2}} like ${latex.inlinev_{1}, ..., v_{m}, w_{1}, ..., w_{k}}. Thus ${latex.inlinedim\ 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.inlineV_{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.inlineV_{1} \cup V_{2}} and thus contained in ${latex.inlineV_{1} + V_{2}}. The span of the list contains ${latex.inlineV_{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.inlinec_{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.inlineV_{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.inlinev_{1}, ..., v_{m}} is a basis of ${latex.inlineV_{1} \cap V_{2}}, we have that:

${latex.inlinec_{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.inlinea_{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.