I'm just curious as to why the following procedure for finding the intersection of two vector spaces doesn't work:
Say I'm asked to find the intersection of the vector spaces $V=\langle(1, 1, 1), (1, 0, 1)\rangle$ & $W=\langle (1, 1, 0), (0, 1, 1)\rangle$.
A vector $v$ belongs to $V\cap W$ if and only if it can be expressed as both
$\alpha \begin{pmatrix} 1\\ 1\\ 1\end{pmatrix}+\beta\begin{pmatrix} 1\\ 0\\ 1\end{pmatrix}$ and $\mu\begin{pmatrix} 1\\ 1\\ 0\end{pmatrix}+\nu\begin{pmatrix} 0\\ 1\\ 1\end{pmatrix}$ for some $\alpha, \beta, \mu, \nu\in \mathbb{R}$. Thus we try to solve
$$\alpha \begin{pmatrix} 1\\ 1\\ 1\end{pmatrix}+\beta\begin{pmatrix} 1\\ 0\\ 1\end{pmatrix}=\mu\begin{pmatrix} 1\\ 1\\ 0\end{pmatrix}+\nu\begin{pmatrix} 0\\ 1\\ 1\end{pmatrix}$$, or
$$\begin{pmatrix} 1 & 1 & -1 & 0 \\ 1 & 0 & -1 & -1 \\ 1 & 1 & 0 & -1 \end{pmatrix}\begin{pmatrix}\alpha \\ \beta \\ \mu \\ \nu \end{pmatrix}=0$$
Since applying elementary row operations corresponds with multiplying on the left by elementary matrices, such operations preserve the solutions $(\alpha, \beta, \mu, \nu)$ of the equation above. Using elementary row operations I end up getting
$$\begin{pmatrix} 1 & 0 & 0 & -1 \\ 0 & -1 & 0 & -1 \\ 0 & 0 & 1 & 0 \end{pmatrix}\begin{pmatrix}\alpha \\ \beta \\ \mu \\ \nu \end{pmatrix}=0$$
for which $(\alpha, \beta, \mu, \nu)$ is a solution precisely when $\alpha = \nu, \beta = -\nu, \mu=0$. But letting $\mu=0$ & $\nu=1$ we get
$$\begin{pmatrix} 1\\ 1\\ 1\end{pmatrix}-\begin{pmatrix} 1\\ 0\\ 1\end{pmatrix}=\begin{pmatrix} 0\\ 1\\ 1\end{pmatrix}$$
which is false.