Intersection of Subspaces

Theorem

If $W_1,W_2,\dots ,W_r$ subspaces of vector space $V$ then intersection of those subspaces is also subspace of $V$.

Proof

Let $W$ be the intersection of the subspaces $W_1,W_2,\dots ,W_r$. This set is not empty because each of these subspaces contains the zero vector of $V$, and hence so does their intersection. Thus, it remains to show that $W$ is closed under addition and scalar multiplication.

To prove closure under addition, let $\mathbf{u}$ and $\mathbf{v}$ be vectors in $W$. Since $W$ is the intersection of $W_1,W_2,\dots ,W_r$, it follows that $\mathbf{u}$ and $\mathbf{v}$ also lie in each of these subspaces. Moreover, since these subspaces are closed under addition and scalar multiplication, they also contain the vectors $\mathbf{u}+\mathbf{v}$ and $k\mathbf{u}$ for every scalar $k$, and hence so does their intersection $W$. This proves that $W$ is closed under addition and scalar multiplication.