Here is the question which I received in a test a few days ago: Prove that if $W$ is a subspace of a vector space $V$ and $w_1, w_2,\ldots,w_n$ are in $W$, then $a_1 w_1 + a_2 w_2 + \cdots + a_n w_n$ belongs to $W$ for any scalars $a_1,a_2,a_3,\ldots,a_n$.

My professor asked us to prove it by using the axiom of induction. However, I have a question before that.

How are we going to be able to prove this question if the statement did not even mention whether $a_1,a_2,a_3,\ldots,a_n$ is in the field which $V$ is over?

Everyone around me says that I'm stupid to think that much, it's simply too much overanalyzing. But how can we assume it is the field which $V$ is over while there exist not only a field? Or is it a "goes without saying" stuff?

Since I'm a newbie to this forum, my question may be very very stupid. As a result, I'm going to apologize for that in advance.


  • $\begingroup$ I would take that saying "$a_i \in \mathbb{F} ~\forall i$, where $V$ is a vector space of $\mathbb{F}$" falls into the "goes without saying" category. $\endgroup$ – erfink Oct 15 '16 at 3:38
  • $\begingroup$ You should ask your teacher first. $\endgroup$ – Ana S. H. Oct 15 '16 at 3:44
  • $\begingroup$ I asked, and he as well thought I'm thinking too over. $\endgroup$ – 李智修 Oct 15 '16 at 3:45
  • $\begingroup$ Alexis Olson, thanks for you edit!! It seems much better now. It was so ugly...... $\endgroup$ – 李智修 Oct 15 '16 at 3:50
  • $\begingroup$ @erfink yeah I think so too $\endgroup$ – 李智修 Oct 15 '16 at 3:55

As an addition to the other two answers, I would like to give a little exposition on the "goes without saying" types of statements that often accompany questions. Note that the question could have been written as

Given that $V$ is a vector space over field $\mathbb{F}$ with operations of scalar multiplication and vector addition ($\cdot, +$ respectively) defined by $\dots$ such that they satisfy the following axioms: $\dots$; and that $W$ is a subspace of $V$ and that vectors $\{\vec{w}_1 , \vec{v}_2, \dots, \vec{w}_n \} \in W$, show that $\sum_{i=0}^n a_i \vec{w}_i = a_1 \vec{w}_1 + a_2 \vec{w}_2 + \cdots + a_n \vec{w}_n$ is also in $W$, given that the scalars $a_i \in \mathbb{F}$ for all $i$.

Do we really gain anything with this phrasing? Certainly it removes some ambiguities (now we know that $+$ is well defined!), but at a great cost: there are an awful lot of extra words and symbols for the same question. Having written my share of math exams, this rephrased version would be a terrible exam question: there are so many pieces of math touched upon that the actual question becomes very lost. In any writing (including proofs!), you have to direct your writing to the appropriate audience. As such, you strike a balance between clarity and precision. If you demand absolute precision, your writing becomes nigh impossible to comprehend and reads like the instruction manual for an Ikea table; if you sacrifice precision for clarity, there will necessarily be ambiguities introduced. The art of writing is in striking a balance, wherein the audience will safely be able to understand and gloss over the ambiguities.

Given that this was a question to see if you understand that a vector subspace must be closed under vector addition and scalar multiplication, as well as seeing how well you can use the axiom of induction in a proof, bickering about which field the scalars belong to is well outside the scope of the question.

General Advice

Since this won't be the last time you'll ever come across ambiguous phrasing, here's some suggestions that have served me well.

  1. Ask for clarification! Most professors I've met wouldn't mind a quick question about clarifying a nebulous statement, even during an exam. Granted, phrasing is important here: "Can I assume that the scalars $a_i$ come from the field $\mathbb{F}$ of which $V$ is a vector space over?" versus "What's a scalar? / What is $a_i$?"
  2. If in doubt, you can always be more verbose yourself. Particularly on an exam, wherein the purpose is to show your mathematical understanding, starting your proof with

    It should be noted that if the scalars $a_i$ are not from the appropriate field, then the statement is false by [simple counter-example]. As such, we proceed under the assumption that $a_i \in \mathbb{F} \forall i$. Let $\dots$

This second method is basically just a way to "cover all your bases," as it were, allowing you to answer the question with what you feel are reasonable assumptions. It would be hard to take points off for such a response, unless the assumption in question is way below the content of the exam or proceeds directly from a definition. Answering every question in this manner can also be seen as being a bit of a "smart-a**" or insufferable, or that there might be a lack of mathematical comprehension. Additionally, don't use it as an excuse to not solve the intended problem!

As a final note, it is a good thing that you are bringing this level of scrutiny to the phrasing of question--make sure you're applying the same level of rigor to your own writing and proofs.

  • 1
    $\begingroup$ Thank you @erfink, I now have a full understanding now. $\endgroup$ – 李智修 Oct 16 '16 at 4:02

A field is part of the definition of a vector space and any subspace will have the same field as the original space.

If you are talking about a vector space, you can nearly always assume that any scalars mentioned belong to the associated field as anything else doesn't really make sense.

  • $\begingroup$ I agree your viewpoints....but I'm just a bit surprised that we have to assume something even if it is not unique..... $\endgroup$ – 李智修 Oct 15 '16 at 3:53
  • $\begingroup$ @李智修 If you spell everything out every time, things end up very long, repetitive, and harder to read. When there is very little chance of misunderstanding, including extra information is more obfuscating than helpful. $\endgroup$ – Alexis Olson Oct 15 '16 at 3:59

The subspace inherits both scalars and vectors from the parent space. This means that $a_s$ is an element of the field the space is over, and that the set of vectors are a subset of the set of vectors of the space.

  • $\begingroup$ Yes, that's right. But the statement didn't mention the dependency of the element "a" and the v.s operation is only defined on a given field. Never mind, I think I'm going to simply ignore this contradiction.....it's a bit trivial in this learning stage. $\endgroup$ – 李智修 Oct 15 '16 at 4:04

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