# Is it acceptable to solve hypothetical statements in Linear Algebra using actual numbers?

I'm taking a Linear Algebra course this semester where we must prove/disprove hypothetical statements. So I'm wondering, is it alright to show that certain theorems hold or not using examples with actual, hard numbers?

For example, one of the problems in our last assignment was this:

Theorem 4.2.1 If W is a set of one or more vectors in a vector space V, then W is a subspace of V if and only if the following conditions hold.

(a) If u and v are vectors in W, then u + v is in W.

(b) If k is any scalar and u is any vector in W, then k$\textbf{u}$ is in W.

Use Theorem 4.2.1 to determine which of the following are subspaces of $R^3$

(a) All vectors of the form $(a, 0, 0)$

So my question is, is it considered bad form to show my answer like so:

If $u = (2, 0, 0)$ and $v = (5, 0, 0)$

$u + v = (7, 0, 0)$ which is on $R^3$ and has the form $(a, 0, 0)$ and holds $\forall a \in R^3$

If $k = -1$ and $u = (1, 0, 0)$

$k\textbf{u} = (-1, 0, 0)$ which is in $R^3$ and has the form $(a, 0, 0)$ and holds $\forall k,a \in R^3$

Thus all vectors of the form $(a, 0, 0)$ are a subspace of $R^3$

Or should I stick to using stand in variables like $\alpha$ and $\beta$ to show my conclusions instead of actual numbers?

-

The general principle is that one example of something can't prove that it is ALWAYS true.

However, one counterexample of something, can prove that it is NOT always true.

In your context, to prove that a subspace is closed, you need to do it for general elements. To prove that a subspace is not closed, you need to find just one example that breaks closure.

-
To this I'd add that trying to find counterexamples to true statements can be a good first step towards understanding how to prove them. –  korrok Feb 4 at 23:46
That makes sense, I take it that it doesn't particularly matter if something is obvious? Like for the first part of my answer, it's pretty clear that the same results will come out of any combination of real numbers. None of my previous courses have really cared about form so I'd like to be as correct as possible. –  TylerFowler Feb 4 at 23:47
@Tyler: if it obvious that the result will come out of any combination, why don't you write it for any combination? Writing for some examples is a good idea to understand how things work. This is what you should do on your scratch paper before writing the actual, formal solution. –  Taladris Feb 4 at 23:58
Prrof by contradiction is actually a subtle variant. You show that the opposite of the original calim is wrong by providing one counterexample. The challenge is coming up with the correct opposite claim. The opposite of "all primes are odd" is NOT "all primes are even". –  MSalters Feb 5 at 12:02

It's not just "considered bad form", it is definitively wrong.

In the conditions of your Theorem 4.2.1 - let's just take condition (a) by way of example - there is an implicit "for all". So (a) actually means $$\hbox{{\bf for\ all} vectors {\bf u} and {\bf v} in W, the vector {\bf u}+{\bf v} is in W.}$$ If you want to say that this sort of thing should be clearly stated and should not be left implicit - I totally agree with you, see if you can get your instructor to change his/her ways ;-)

Since you have to prove an "all" statement, you cannot do it by one example - this only shows that the statement is true in that particular example, not in all cases.

For an analogy, what would you say if someone tried to put this past you: "everyone in your family is under five years old - the reason I know is that your baby brother is only three"? Obviously incorrect reasoning, I hope you agree.

Suggestion. In beginning linear algebra, you will often find that the algebra is pretty easy and the logic is the part that needs care. Don't let your instructor get away with sloppy logic! Good luck.

-

You can disprove statements that "so and so is true for all ..." by showing a counterexample. One (or a few, or many) examples don't prove something is always true.

On the other hand, trying a few examples can help see why the statement is true. Selecting some special/extreme examples might uncover hints (or point at cases that have to be handled separately). Trying to find ways to prove it isn't true can lead to understand why it is true, and to a proof.

-

You want to prove it for all numbers, if you take a number and it doesn't work, then it is false, but if it does work all you know is it works for that particular one, not for others.

Take for example the statement: the sum of 2 prime numbers is always even.

$3+7=10$ is even, so we know it works in that case.

however $2+11=13$ is odd, so it doesn't always work. In fact the numbers 2,11 are what is called a counter-example.

Hope this helps ;)

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 1, "mathjax_display_tex": 1, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0, "katex": 0, "math-container": 0, "wp-katex-eq": 0, "align": 0, "equation": 0, "x-ck12": 0, "texerror": 0, "math_score": 0.7363221645355225, "perplexity": 264.2952094664205}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "plain"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "warc_path": "s3://commoncrawl/crawl-data/CC-MAIN-2014-15/segments/1397609538110.1/warc/CC-MAIN-20140416005218-00121-ip-10-147-4-33.ec2.internal.warc.gz"}