1
$\begingroup$

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...

$\endgroup$
1
  • $\begingroup$ Any vector with first three components non-zero. $\endgroup$ – Dhruv Kohli - expiTTp1z0 Sep 12 '17 at 20:09
2
$\begingroup$

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
  • $\begingroup$ So the span(e1,e2,e3) would be any three numbers, but the union of span (e1,e2) with span of (e3) would be just like zero, zero and a number and those that are number, number, zero......in simply terms since typing on my phone....which makes sense because I was thinking wrong. So basically 2,1,1 would work. Any thought on whether I need to include the fourth term as zero for my final answer? $\endgroup$ – andemw01 Sep 12 '17 at 22:42
  • $\begingroup$ @andemw01 The enclosing space is $\mathbb R^4$, so you do need that final zero. Otherwise, the vectors would all be elements of $\mathbb R^3$, which is not a subspace of $\mathbb R^4$. $\endgroup$ – amd Sep 13 '17 at 0:12
  • $\begingroup$ Thanks everybody $\endgroup$ – andemw01 Sep 13 '17 at 0:15
0
$\begingroup$

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.

$\endgroup$
2
  • $\begingroup$ The OP is working in $\mathbb R^4$. $\endgroup$ – amd Sep 12 '17 at 21:37
  • $\begingroup$ @amd Yes, but since his/her question doesn't need a 4th component, I got rid of it in my geometric explanation to make visualization easier. $\endgroup$ – Dave Sep 12 '17 at 21:38

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.