Let $W_1=\{(1,1,2,1), (3,1,0,0)\}$ and $W_2=\{(-1,-2,0,1), (-4,-2,-2,-1)\}$
Apparently $\dim W_1=\dim W_2=2$.
For $\dim W_1\cap W_2$, since $(-4,-2,-2,-1)$ can be expressed as $-(1,1,2,1)-(3,1,0,0)$, $\dim W_1\cap W_2=1$. But what if the spans are complicated, how do you find $\dim W_1\cap W_2$. Do you try matrix on each vector and see which ones has a nontrivial solution?
For $\dim W_1+W_2$, how do you know it without calculating $\dim W_1\cap W_2$?
Best Answer
Put the given vectors into the columns of matrices \begin{align*} w_1 &= \left[\begin{array}{rr} 1 & 3 \\ 1 & 1 \\ 2 & 0 \\ 1 & 0 \end{array}\right] & w_2 &= \left[\begin{array}{rr} -1 & -4 \\ -2 & -2 \\ 0 & -2 \\ 1 & -1 \end{array}\right] \end{align*} The dimensions of $W_1$ and $W_2$ are the ranks of the matrices $w_1$ and $w_2$ respectively. Row reducing gives $\DeclareMathOperator{rref}{rref}$ \begin{align*} \rref(w_1) &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right] & \rref(w_2) &= \left[\begin{array}{rr} 1 & 0 \\ 0 & 1 \\ 0 & 0 \\ 0 & 0 \end{array}\right] \end{align*} These reduced row-echelon forms show that both $w_1$ and $w_2$ have rank two. This proves that $\dim W_1=\dim W_2=2$, as you have observed.
Now, the dimension of $W_1+W_2$ is the rank of the matrix $$ [w_1,w_2] = \left[\begin{array}{rrrr} 1 & 3 & -1 & -4 \\ 1 & 1 & -2 & -2 \\ 2 & 0 & 0 & -2 \\ 1 & 0 & 1 & -1 \end{array}\right] $$ Row reducing gives $$ \rref[w_1,w_2]= \left[\begin{array}{rrrr} 1 & 0 & 0 & -1 \\ 0 & 1 & 0 & -1 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0 \end{array}\right] $$ This proves that $\dim (W_1+W_2)=3$. The dimension formula then implies that $\dim(W_1\cap W_2)=1$.
This method also provides bases for the spaces in question. The pivot columns of $\rref[w_1,w_2]$ correspond to the columns in $[w_1,w_2]$ that form a basis for $W_1+W_2$. The "free" columns of $\rref[w_1,w_2]$ correspond to the columns of $[w_1,w_2]$ that span $W_1\cap W_2$. Here, we have $\DeclareMathOperator{Span}{Span}$ \begin{align*} W_1+W_2 &= \Span \left\{ \left[\begin{array}{r} 1 \\ 1 \\ 2 \\ 1 \end{array}\right], \left[\begin{array}{r} 3 \\ 1 \\ 0 \\ 0 \end{array}\right], \left[\begin{array}{r} -1 \\ -2 \\ 0 \\ 1 \end{array}\right] \right\} & W_1\cap W_1 &= \Span\left\{ \left[\begin{array}{r} -4 \\ -2 \\ -2 \\ -1 \end{array}\right] \right\} \end{align*}