Theorem of the dimension of the sum

Prove that if $$V$$ and $$W$$ are finitely generated, then $\dim(V)+\dim(W)=\dim(V\cap W)+\dim({\mathcal L}(V\cup W)).$

• Hint

Use suitable bases.

• Resolution

Take any basis $$X_1$$ of $$V\cap W$$. Extend $$X_1$$ by $$X_2$$ to a basis of $$V$$, and similarly extend $$X_1$$ by $$X_3$$ to a basis of $$W$$.

It suffices to show that $$X_1\cup X_2\cup X_3$$ is a basis of $${\mathcal L}(V\cup W)$$ since all dimensions are determined by expressions

$$\dim(V)+\dim(W)=(|X_1|+|X_2|)+(|X_1|+|X_3|)=\\|X_1|+(|X_1|+|X_2|+|X_3|)=\dim(V\cap W)+\dim({\mathcal L}(V\cup W)).$$

Any vector of $${\mathcal L}(V\cup W)$$ is a linear combination of some vectors from $$V$$ and $$W$$. These auxiliary vectors could be expressed with respect to the basis $$X_1\cup X_2$$ for vectors from $$V$$, and analogously w.r.t. the basis $$X_1\cup X_3$$ for vectors from $$W$$. The overall linear combination can be rearranged w.r.t. $$X_1\cup X_2\cup X_3$$, i.e., this set generates $${\mathcal L}(V\cup W)$$.

Also holds that $$X_1\cup X_2\cup X_3$$ is linearly independent – $$X_1\cup X_2$$ is linearly independent, since it has been chosen as a basis. If some combination of vectors from $$X_3$$ belongs $$V$$, it would have to belong to $$V\cap W$$, that would violate the linear independence of the set $$X_1\cup X_3$$.