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This is part of proving a larger theorem but I suspect my prof has a typo in here (I emailed him about it to be sure)

The lemma is written as follows:

Let $V$ be a vector space. Let {$z, x_1, x_2, ..., x_n$} be a subset of $V$. Show that if $z \in\ span(${$x_1, x_2,..., x_n$}$)$, then $span(${$z, x_1, x_2,..., x_r$}$)=span(${$ x_1, x_2,..., x_r$})

I feel like this should be really simple and I saw a proof that you can take out a vector from a subset and not change the span, but I am unsure of the reverse -- assuming that is what this lemma is about. (To me, the "if" implies something besides just unifying the two sets should follow).

Anyhow, the proof should, I think, start with that we can modify a subset (let's call it S) without affecting the span if we start like this:

$\exists x \in S$ such that $ x \in span(S-${$x$})

then you build up a linearly independent subset somehow. The proof that you can take vectors out of the subset says that since $x\in span(${$x$}$)\ \exists\ \lambda_1, \lambda_2,..., \lambda_n \ \in\ K$ such that $x=\sum_{i=1}^n\lambda_i x_i $

since we know $ span(S) \supset span(S-${$x$}) we just need to show $ span(S) \subseteq span(S-${$x$})

But honestly I am not sure I understand what's happening here well enough to prove the above lemma. (This class moves fast enough that we're essentially memorizing proofs rather than re-deriving them I guess).

I am really starting to hate linear algebra. :-(

(Edited to fix U symbol and make it a "is a member of" symbol)

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  • $\begingroup$ Do you mean "Show that if $z\in span(\{x_1, x_2, ..., x_n\}) ...$ in your third paragraph. I believe you have a typo. $\endgroup$ – Prism Jun 19 '13 at 4:02
  • $\begingroup$ THat's just it: the sheet my prof gave has a union symbol, but it makes more sense if you're showing that it is either in the set or that you're trying to show the span of the union of z and span(x) is the same as the span of x. So I am hoping someone recognizes the lemma at least. $\endgroup$ – Jesse Jun 19 '13 at 4:06
  • $\begingroup$ I think I recognize the lemma: if $z\in span({x_1, x_2, ..., x_r})$, then $span({z, x_1, x_2, ..., x_r})=span({x_1,x_2, ..., x_r})$. $\endgroup$ – Prism Jun 19 '13 at 4:08
  • $\begingroup$ OK, then that makes the whole task easier. So, is it correct to say that since $z\in span (x_1, x_2,.. ,x_r)$ then because a subset of any vector space is in the span of that space (by definition) then the span of the union of x and z would be in $span(x_1, x_2, ... x_r)$? $\endgroup$ – Jesse Jun 19 '13 at 4:12
  • $\begingroup$ Hmmm...that seems a bit tautological, but perhaps my English isn't good enough to recognize the subtlety. In any case, I would prove this explicity: if $z$ is in the $span({x_1, x_2, ..., x_r})$, then $z$ is a linear combination of $x_1, x_2, ..., x_r$. Now, what is $span(z, x_1, x_2, ..., x_r)$? It clearly contains the smaller span $span{(x_1, x_2, ..., x_r})$. But also, since $z$ is a linear combination of $x_1, ..., x_r$, anything that is a linear combination of $z, x_1, ..., x_r$ is automatically linear combination of $x_1, ..., x_r$ (just substitute the expression for $z$). (Cont.) $\endgroup$ – Prism Jun 19 '13 at 4:17
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We want to show $\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots, x_n\}$. In general, to show $X = Y$ where $X, Y$ are sets, we want to show that $X \subseteq Y$ and $Y \subseteq X$.

So suppose $v \in \operatorname{span}\{x_1, \dots, x_n\}$. Then, we can find scalars $c_1, \dots, c_n$ such that $$v = c_1x_1 + \dots + c_nx_n$$ so clearly, $v \in \operatorname{span}\{z, x_1, \dots, x_n\}$. This proves $$\operatorname{span}\{x_1, \dots, x_n\} \subseteq \operatorname{span}\{z, x_1, \dots, x_n\}$$

Now let $v \in \operatorname{span}\{z, x_1, \dots, x_n\}$. Again, by definition, there are scalars $c_1, \dots, c_{n+1}$ such that $v = c_1x_1 + \dots + c_nx_n + c_{n+1}z$. But hold on, $z \in \operatorname{span}\{x_1, \dots, x_n\}$, right? This means there are scalars $a_1, \dots, a_n$ such that $z = a_1x_1 + \dots + a_nx_n$. Hence,

$$v = c_1x_1 + \dots + c_nx_n + c_{n+1}(a_1x_1 + \dots + a_nx_n)$$ $$= (c_1 + c_{n+1}a_1)x_1 + \dots + (c_n+c_{n+1}a_n)x_n$$

and so we conclude that $v \in \operatorname{span}\{x_1, \dots, x_n\}$. Therefore, $$\operatorname{span}\{z, x_1, \dots, x_n\} = \operatorname{span}\{x_1, \dots x_n\}$$

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  • $\begingroup$ thanks a lot. This is the first math class i have been in where memorization was so important. $\endgroup$ – Jesse Jun 19 '13 at 13:13

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