How to find subspace $W_1\cap W_2$? Where $W_1$ and $W_2$ are subspaces of $\mathbb{R}^5$. Consider $S_1=\{(2,0,3,1,1),(1,0,2,1,1),(2,0,3,1,3)\}$ and $S_2=\{(2,1,1,0,1),(3,2,3,2,3),(1,1,1,1,1)\}$. Suppose $W_1$ and $W_2$ are subspaces of $\mathbb{R}^5$ generated by $S_1$ and $S_2$ respectively. Then what will be $W_1\cap W_2$. I am not able to understand how to start.
 A: Well, there must be an easier method, but here is a solution. Let $\vec z  \in W_1 \cap W_2$. Then it holds:
$$a_1\cdot \begin{bmatrix} 2\\0\\3\\1\\1 \end{bmatrix}+a_2\cdot \begin{bmatrix}1\\0\\2\\1\\1 \end{bmatrix}+a_3 \cdot \begin{bmatrix} 2\\0\\3\\1\\3\end{bmatrix}-\lambda_1\cdot \begin{bmatrix}2\\1\\1\\0\\1\end{bmatrix}-\lambda_2\cdot\begin{bmatrix}3\\2\\3\\2\\3 \end{bmatrix}-\lambda_3\cdot \begin{bmatrix}1\\1\\1\\1\\1\end{bmatrix}=\begin{bmatrix}0\\0\\0\\0\\0\end{bmatrix},$$for some $a_1,a_2,a_3,\lambda_1,\lambda_2,\lambda_3 \in \mathbb R$, which we have to define.
Equivalently, it holds:
$$\begin{bmatrix} 2&1&2&2&-3&-1\\0&0&0&-1&-2&-1\\3&2&3&-1&-3&-1\\1&1&1&0&-2&-1\\1&1&3&-1&-3&-1\end{bmatrix}\cdot\begin{bmatrix}a_1\\a_2\\a_3\\\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}=\begin{bmatrix} 0\\0\\0\\0\\0\end{bmatrix}.$$
We can find the reduced row echelon form of the coefficient matrix, which is (after some work):
$$R=\text{rref }A = \begin{bmatrix}  
1  &  0    &     0   &      0   & 2.5  &   1.5\\
0  &  1    &     0   &      0   & -5   &   -3\\
0  &  0    &     1   &      0   & 0.5  &  0.5\\
0  &  0    &     0   &      1   & 2     &   1\\
0  &  0    &     0   &      0   & 0      &0         
\end{bmatrix}.$$ 
Due to the fact that the system is homogeneous, we can solve the system $$R\cdot\begin{bmatrix}a_1\\a_2\\a_3\\\lambda_1\\\lambda_2\\\lambda_3\end{bmatrix}=\mathbf{0}.$$
Now, we can easily find that $\lambda_3 = y \in \mathbb R, \, \lambda_2 = x \in \mathbb R$ and $\lambda_1 =-2x-y$.
Thus every $\vec z \in W_1\cap W_2$ can be written in the form
$$(-2x-y)*\begin{bmatrix}2\\1\\1\\0\\1\end{bmatrix} + x*\begin{bmatrix}3\\2\\3\\2\\3\end{bmatrix}+ y*\begin{bmatrix}1\\1\\1\\1\\1\end{bmatrix} = x*\begin{bmatrix}-1\\0\\1\\2\\1\end{bmatrix}+y*\begin{bmatrix}-1\\0\\0\\1\\0\end{bmatrix},$$ where $x,y \in \mathbb R$.  
Thus, $$W_1\cap W_2 = \langle(-1,0,1,2,1),\, (-1,0,0,1,0)\rangle.$$

Also, from the rref we can find $a_1 = -2.5x-1.5y$, $a_2 = 5x +3y$, $a_3= -0.5x-0.5y$.
