## Theorem of the dimension of the sum

### Task number: 2545

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 \(X_1\cup X_3\).