# Prove that each vector in span(S) is a unique linear combination, where S is a particular subset of a vector space V

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?

• 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 $\sum b_j v_j = x$. Jun 8, 2022 at 19:28
• 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. Jun 8, 2022 at 19:41
• 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 $b_j$. Jun 8, 2022 at 19:46
• 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! Jun 8, 2022 at 20:02

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 $$S$$.

• 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? Jun 8, 2022 at 22:51
• 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. Jun 9, 2022 at 2:42