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If we define a subset $S$ of a vector space $V$ such that whenever

$a_1v_1 + \ldots + a_nv_n = 0$ for $v_i \in V$ and $a_i \in F$,

we have

$a_1 = \ldots = a_n = 0$,

then I am told it can be proven that every vector in $span(S)$ is a unique linear combination of vectors in $S$.

To go about proving this, I have done the following:

Let $x \in span(S)$. Now I want to take two arbitrary linear combinations from $span(S)$ and say they both add up to $x$. Then I want to show that they are in fact identical to one another (that is, they are the same linear combination of vectors in $S$, not just equal to the same resultant vector).

So, let's say:

$\sum a_i v_i = x$ for $a_i \in F$ and $v_i \in S$

and

$\sum b_j w_j = x$ for $b_j \in F$ and $w_j \in S$

Then it must be that $\sum a_i v_i - \sum b_j w_j = 0$, that is:

$a_1v_1 + \ldots + a_nv_n - b_1w_1 - \ldots - b_mw_m = 0$

But by the property defined above for $S \subseteq V$, we must therefore have $a_i = b_i = 0$. Here, it feels a bit strange, because all I am left with is a bunch of zero coefficients. So can I simply wave my hands and say the following?:

Since I defined $\sum a_i v_i = x$ and $\sum b_j w_j = x$ to be arbitrary elements of $span(S)$, with the coefficients taken arbitrarily from $F$, not just $0$, it can't be that all of the coefficients are just $0$? And so the only other possibility is that every vector in $\sum b_j w_j$ is identically equal to those in $\sum a_i v_i$?

I can see the logic in this, but it doesn't feel very "mathematical" or like much of a proof at the end here. Somehow it seems like a circularity or like I am cheating a bit. Am I just overthinking? Or am I missing something?

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  • $\begingroup$ The second linear combination with coefficients $b_j$ must be also a linear combination of the vectors in $S$. In other words, you shuld use $\sum b_j v_j = x$. $\endgroup$
    – Ana S. H.
    Jun 8, 2022 at 19:28
  • $\begingroup$ Yes, the vectors $w_j$ are defined here to be elements of S. However, the purpose of the exercise is to show that they are in fact equal to $v_i$, so I can't simply assert that outright. Although the subscripts change, I also need something to differentiate $v_1$ from $w_1$, for instance. $\endgroup$
    – notadoctor
    Jun 8, 2022 at 19:41
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    $\begingroup$ I don't think that's the purpose of the exercise. The actual purpose is to show that any element of $span(S)$ has a unique linear combination of vectors in $S$. If I understand correctly, $S=\{v_1, v_2, \dots, v_n\}$, i.e. they're giving you a set of specific vectors $v_j$ that span $S$. So the second (distinct) linear combination must be something like $x=\sum b_j v_j$. The only difference between this one and the first one are the coefficients $b_j$. $\endgroup$
    – Ana S. H.
    Jun 8, 2022 at 19:46
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    $\begingroup$ Okay, this is making a bit of sense to me. If that were the case, it would be a much more straightforward conclusion -- the coefficient of each vector must equal zero, so the differences $(a_i - b_i)$ must be zero, so they must be equal to one another. I will have to chew on this a bit though to accept that the set of vectors is not arbitrary... thanks! $\endgroup$
    – notadoctor
    Jun 8, 2022 at 20:02

1 Answer 1

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As Ana S. H. illustrates in the comments above, my approach wasn't quite the right one. Instead, we should let $S = \{v_1, \ldots, v_n\}$ where $v_i \in V$.

Then we can say that any element of $span(S)$ is going to have to be a linear combination of $v_i$. That is, $x = \sum a_iv_i$.

Now, if we let $x = \sum b_jv_j$ as well, we have the simpler result:

$\sum a_iv_i = \sum b_jv_j \Rightarrow \sum a_iv_i - \sum b_jv_j = 0$

This gives us $\sum (a_k - b_k)v_k = 0$, and thus by the above properties of S:

$a_k - b_k = 0 \Rightarrow a_k = b_k$

Therefore, any arbitrary element of $span(S)$ can be uniquely represented as a linear combination of vectors in $S$.

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  • $\begingroup$ Although I'm relatively satisfied with this answer, it leaves two ambiguities to my mind that I haven't cleared up yet. 1) How have we shown that a linear combination of different vectors in $span(S)$ couldn't also equal $x$? Wouldn't that make it not unique? And 2) How can we presume the indices $i$ and $j$ run to the same endpoint? $\endgroup$
    – notadoctor
    Jun 8, 2022 at 22:51
  • $\begingroup$ Nevermind -- to once again answer my own question, a linear combination technically contains every vector in the set, and the coefficient is simply $0$ anywhere we don't want that vector included. So in this case, because every coefficient must be 0, both $a_i$ and $b_j$ must be $0$ if either one is $0$. The indices are the same because they index the same set, and you count every vector in it. $\endgroup$
    – notadoctor
    Jun 9, 2022 at 2:42

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