# Finding basis for intersection of vector spaces.

This method of finding a basis of two subspaces is abstracted from How to find basis for intersection of two vector spaces in $\mathbb{R}^n$ where $U$ and $W$ are the subspaces.

1) Construct the matrix $A=(Base(U)|−Base(W))$ and find the basis vectors $\textbf{s}_i=\begin{pmatrix}\textbf{u}_i \\ \textbf{v}_i\end{pmatrix}$ of its nullspace.

2) For each basis vector si construct the vector $\textbf{w}_i=\mathrm{Base}(U)\textbf{u}_i=\mathrm{Base}(W)\textbf{v}_i$.

3) The set $\{ \textbf{w}_1,\ \textbf{w}_2,...,\ \textbf{w}_r \}$ constitute the basis for the intersection space $span(\textbf{w}_1,\ \textbf{w}_2,...,\ \textbf{w}_r)$.

My only question is how do we know that the $w_i$ calculated this way are independent so that we have a basis?

The $u_i$ are all independent because they are part of the basis for Null(A). Since Base(U) is full rank, we get that $w_i = Base(U)u_i$ must also be independent.
More formally, suppose $\sum_{i=0}^n c_iw_i = 0$. Then: $$0 = \sum_{i=0}^n c_iw_i = \sum_{i=0}^n c_iBase(U)u_i = Base(U)\sum_{i=0}^n c_iu_i$$ Since $Base(U)$ is full rank, it follows that $\sum_{i=0}^n c_iu_i = 0$, which is a contradiction.