V = span{(1,1,0,-1)^T, (0,1,3,1)^T} and W = span{(0,1,2,-1)^T,(1,2,2,-2)^T}
Don't quite know where to start. The coefficient matrix of V and W are inconsistent after gaussian elimination. Also inconsistent with a 4×4 of V and W. We haven't covered basis/dimension yet, or even intersection for that matter, so I assume this problem can be worked out using an algorithm to find constants. Any insight? Thanks.
Span of V:
$$ \left[
\begin{array}{cc|c}
1&0&a\\
1&1&b\\
0&3&c\\
-1&1&d
\end{array}
\right] $$
Row reduced to:
$$ \left[
\begin{array}{cc|c}
1&0&a\\
0&1&b-a\\
0&0&c-3b+3a\\
0&0&a2+d-b
\end{array}
\right] $$
So row 3 and 4 are inconsistent, and these vectors don't span R4.
Similar results for Span of W. Am I approaching this problem correctly?
Best Answer
Let $v \in V \cap W$. We then have $$v=a_1 v_1 + a_2 v_2 = b_1 w_1 + b_2 w_2$$ where $a_i, b_i \in \mathbb{R}$. Thus \begin{align*} 0&=a_1 v_1 + a_2 v_2 - b_1 w_1 - b_2 w_2 \\ &= (a_1, a_1, 0, -a_1)+ (0, a_2, 3a_2, a_2) - (0, b_1, 2b_1, -b_1)-(b_2, 2b_2, 2b_2, -2b_2) \\ &= (a_1-b_2, a_1+a_2-b_1-2b_2, 3a_2 -2b_1-2b_2, -a_1+a_2+b_1+2b_2). \end{align*} Thus we now must solve \begin{cases} a_1-b_2=0 \\ a_1+a_2-b_1-2b_2=0\\ 3a_2 -2b_1-2b_2 =0\\ -a_1+a_2+b_1+2b_2=0. \end{cases} According to Wolfram , this system
has no solutionhas infinitely many solutions.