I tried to solve the following exercise:
In the vector space $\mathbb{R}^4$ , consider the subspace V given by the solutions of the system
$\begin{cases}
x+2y+z=0\\
-x-y+3t = 0
\end{cases}$
and the subspace $W$ generated by the vectors:
$w_1 = \begin{pmatrix} 2\\0\\1\\1 \end{pmatrix}$ and $w_1 = \begin{pmatrix} 3\\-2\\-2\\0 \end{pmatrix}$
Compute $\dim (W\cap W)$ and $\dim (V + W)$.
If I compute the dimension of $V\cap W$ first, by writing $W$ as a system of equations, I find $\dim (V\cap W) = 1$, and by Grassmann's formula I can conclude that $\dim (V + W) = 3$.
If I try to do the opposite, finding a base of $V+W$ first, I can't manage to get to the same result. I always get four linearly independent vectors, which would imply the intersection of the two subspaces is empty.
What do you think I'm doing wrong?
Best Answer
The first subspace $U$ is given by $\begin{cases}x+2y+z=0\\-x-y+3t=0 \end{cases}$, so $x=-2y-z$ and $t=\dfrac{x}{3}+\dfrac{y}{3}=-\dfrac{y}{3}-\dfrac{z}{3}$.
This shows that we have a vector subspace of dimension two given by: $$U=\langle u_1,u_2\rangle=\langle\begin{pmatrix}-6\\3\\0\\-1 \end{pmatrix},\begin{pmatrix}-3\\0\\3\\-1 \end{pmatrix}\rangle.$$ Then we have the second subspace $W$ generated by $w_1=\begin{pmatrix}2\\0\\1\\1 \end{pmatrix}$ and $w_2=\begin{pmatrix}3\\-2\\-2\\0 \end{pmatrix}$. If we want to find the dimension of the subspace $U+W$ we have to study the number of linearly independent vectors: $$\begin{pmatrix}-6&-3&2&3\\3&0&0&-2\\0&3&1&-2\\-1&-1&1&0 \end{pmatrix}\sim \begin{pmatrix}-6&-3&2&3\\0&-2&2&-1\\0&0&1&-1\\0&0&0&0 \end{pmatrix}\implies \text{dim}(U+W)=3$$ hence it's generated by the three vectors $u_1,u_2$ and $w_1$.
At this point we know that $\text{dim}(U\cap W)=2+2-3=1$ and it's easy to see that the vector $(5,-2,-1,1)^T\in U\cap W$ (solving the system $\alpha u_1+\beta u_2=\gamma w_1+\delta w_2$)