Let $W_1=\operatorname{span}\{(1,-1,-3),(3,0,-3),(1,2,3),(4,5,6)\}$ and $W_2=\operatorname{span}\{(6,3,7),(0,0,11),(-4,-2,0)\}$. Find dimension and one basis of $W_1,W_2,W_1+W_2,W_1\cap W_2$.
$$ A= \begin{bmatrix}
1 & -1 & -3 \\
3 & 0 & -3 \\
1 & 2 & 3 \\
4 & 5 & 6 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & 0 & -1 \\
0 & 1 & 2 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}\Rightarrow \operatorname{rank}(A)=2\Rightarrow \dim(W_1)=2$$
How to find one basis for $W_1$?
$$ B= \begin{bmatrix}
6 & 3 & 7 \\
0 & 0 & 11 \\
-4 & -2 & 0 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & 1/2 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
\end{bmatrix}\Rightarrow \operatorname{rank}(B)=2\Rightarrow \dim(W_2)=2$$
How to find one basis for $W_2$?
For $W_1+W_2,$
$$\begin{bmatrix}
1 & -1 & -3 \\
3 & 0 & -3 \\
1 & 2 & 3 \\
4 & 5 & 6 \\
6 & 3 & 7 \\
0 & 0 & 11 \\
-4 & -2 & 0 \\
\end{bmatrix}\sim \begin{bmatrix}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
0 & 0 & 0 \\
\end{bmatrix}\Rightarrow \dim(W_1+W_2)=3$$
How to find one basis of $W_1+W_2$?
From previous, $\dim(W_1\cap W_2)=1$
How to find one basis of $W_1\cap W_2$?
Could someone show the procedure for finding at least one basis for either $W_1,W_2,W_1+W_2$ or $W_1\cap W_2$?
Best Answer
For $W_1$, $W_2$ and $W_1 + W_2$ you can take the non-zero rows as a basis of the vector spaces. To find the elements in the intersection it must be written as a linear combination of the basis of both vector spaces, so it reduces to solving the equation:
$$a(1,0,-1) + b(0,1,2) = c(1,\frac 12,0) + d(0,0,1)$$