0
$\begingroup$

Let $S$ be a subspace of an n-dimensional vector space $V$. Prove that $dim(S)\leq\ n$.

Let $B = \{v_1, v_2, v_3,...v_n\}$ be the basis set for vector space $V$.

Let $B_s = \{v_1, v_2, v_3,...v_n, v_{n+1}\}$ be the basis set for the subspace $S$.

Then $span(S) > V$ because $B_s$ has more vectors than $B$ and $span(S)$ can potentially reach a vector that is outside of the vector space $V$. But this cannot be the case because $S$ is a subset of $V$. Hence it must be the case that $B_s \leq\ B$ and therefore, $dim(S) \leq\ n$.

Note: If the proof is incorrect can you point out where I went wrong?

$\endgroup$
  • 1
    $\begingroup$ The subspace $S$ will have a basis, but you have no prior information about the size of basis for $S$, so you are not correct when you start with a basis for $S$ with more than $n$ elements. $\endgroup$ – Geoff Robinson Mar 21 '17 at 18:01
2
$\begingroup$

There's a few things, mostly stemming from improper use of notation.

Let $B = \{v_1, v_2, v_3,...v_n\}$ be the basis set for vector space $V$.

Let $B_s = \{v_1, v_2, v_3,...v_n, v_{n+1}\}$ be the basis set for the subspace $S$.

When you select $v_1$ thru $v_n$, you're defining them as part of a basis for $V$. There's no reason they need to lie in $S$, much less be a basis for it. For example, let $V = \Bbb R^3$ and $S$ be the space spanned by $(1, 1, 1)$. If $B$ is $\{ (1, 0, 0), (0, 1, 0), (0, 0, 1) \}$, what would you say $B_s$ was?

Then $span(S) > V$

I'm not sure what you mean by this. The span of a subspace is simply itself. Also, what's $>$ mean in this context?

$span(S)$ can potentially reach a vector that is outside of the vector space $V$

How did you reach this conclusion? Every element in $S$ lies in $V$, and so every element in $span(S) = S$ must as well.

it must be the case that $B_s \le B$

Again, I'm not sure what $\le$ means in this context.


Overall, I think these issues could be avoided if your sentences are more precise (not to be confused with "use more symbols"). You don't have to write out every step of every lemma, but you should write your proof in such a way that if someone asked you to elaborate on a particular sentence, you could do so rigorously.

Here's a cleaner intro to your proof, if that helps you get started:

Let $B_s = \{ v_1, \ldots, v_k \}$ be a basis for $S$. We would like to show that $k \le n$. (proceed however you like from here. direct proof, contradiction, etc)

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Thank you for guiding me in the write direction. This is exactly what I was looking for! $\endgroup$ – Iamlearningmath Mar 21 '17 at 19:24

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.