Find a vector which is in $\operatorname{span}(e_1,e_2,e_3)$ , but not in the union of [$\operatorname{span}(e_1,e_2)$ and $\operatorname{span}(e_3)$]
So the problem begins by giving an ordered basis of $\mathbb{R}^4$
$$ B = \{e_1 = (1,0,0,0), e_2 = (0,1,0,0), e_3 =(0,0,1,0) , e_4 =(0,0,0,1)\} $$
The actual question is to find a vector $v$ which is in the $\operatorname{span}(e_1,e_2,e_3)$, but not in the union of the $[\operatorname{span}(e_1,e_2)$ and $\operatorname{span}(e_3)]$.
I know this is supposed to be one the easy problems on our assignment, and somehow I am making to difficult. The span of $(e_1, e_2, e_3)$ will be all linear combinations (do I not worry about $e_4$?), so I know I have to find one that will not be including in the linear combinations of $e_1,e_2$ or $e_3$. Can anybody give me a hint, or clue as to how to think about it, since my way isn't working...
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$\begingroup$This exercise is meant to show you the difference between union and span. One way to think about this is that union just sticks things together—it’s an “or” operation—while span also fills in the gaps.
The span of $e_1$ and $e_2$ is two-dimensional—a plane in $\mathbb R^4$. The span of $e_3$ is all of the scalar multiples of $e_3$, so it’s a line. The union of these two spaces consists of vectors that are either in the first space or the second. You can visualize this as a plane with a line sticking out of it. This isn’t even a vector space. The span of the three vectors fills in the gaps that prevent the union from being a vector space. These additional vectors are all of the form $v+w$, where $v\in\operatorname{span}(e_1,e_2)$, $w\in\operatorname{span}(e_3)$ and both are non-zero.
So, to find a vector that is in the span of the three basis vectors but not in their union, take any vector that has nonzero first and second coordinates and zeros for the others (this is $v$) and another that has a nonzero third coordinate, with other zero, (this is $w$) and add them. I.e., any vector with a zero for its fourth coordinate and other coordinates non-zero, as Dhruv Kohli commented.
$\endgroup$ 3 $\begingroup$Take $v=(1,1,1,0)$ then $v\in\operatorname{span}(e_1,e_2,e_3)$, however, $v\notin\operatorname{span}(e_1,e_2)\cup\operatorname{span}(e_3)$. Do you see why?
Here is some geometric understanding of the different spans.
$(1)$ We have that $\operatorname{span}(e_1,e_2,e_3)$ is the entire space of $\Bbb R^3$, so it is the $xyz$-hyperplane.
$(2)$ We have that $\operatorname{span}(e_1,e_2)\cup\operatorname{span}(e_3)$ is the combination of the $xy$-plane and the $z$-axis.
These two spans are not equal. In particular, $(2)$ does not contain points above or below the $xy$-plane that are off the $z$-axis, whereas $(1)$ does.
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