3
$\begingroup$

Suppose we have $n$ linearly independent vectors $\mathbf{v}_{1}\ldots\mathbf{v}_{n}$ in $\mathbb{R}^{n}$. I know that they do span $\mathbb{R}^{n}$, because we can easily specify a non-singular map which sends the $\mathbf{v}_{i}$s to the standard basis, and then to whichever vector in $\mathbb{R}^{n}$ we choose.

My question is: do we need all the machinery of linear maps, determinants, etc. or is there a proof which is closer to the definitions? Every time I start writing down a proof I end up wanting to say "and this set of equations can be solved uniquely because this matrix is non-singular". Is this necessary?

$\endgroup$
  • 1
    $\begingroup$ Essentially, you do need this machinery. Since there are many equivalences in linear algebra, you may be able to write such a statement using different language, but the result is that you're saying an equivalent statement. $\endgroup$ – Michael Burr Apr 23 '15 at 21:43
  • $\begingroup$ Just a small typo: I think you mean to say $n$ linearly independent vectors $\endgroup$ – Tyler Hilton Apr 23 '15 at 21:44
  • $\begingroup$ Yep, thanks for spotting that! Just seen a bunch of other typos actually $\endgroup$ – preferred_anon Apr 23 '15 at 21:46
  • $\begingroup$ I would probably say that $n$ linearly dependent spans a $n$-dimensional subspace, but to prove that you've essentially have to go throgh the machinery you're talking about, so essentially yes, but if you're allowed to assume sufficient strong results you can of course say it simpler. $\endgroup$ – Henrik Apr 23 '15 at 21:47
3
$\begingroup$

You can prove as the first thing in Linear Algebra after the primary definitions that:

If there is a linear independent set $\{u_1,...,u_n\}$ and a set $\{v_1,...,v_m\}$ that generates the space, then $n \leq m$.

with an inductive-like argument.

Now, take your $n$ elements from $\mathbb{R}^n$ that are linearly independent. Suppose they don't generate $\mathbb{R}^n$. Take, then, another element of $\mathbb{R}^n$ that cannot be written as linear combination of those you had before. You will have a linearly independent set with $n+1$ elements, which is greater than $n$, and there is a set with $n$ element generating $\mathbb{R}^n$: the canonical vectors. This contradicts the lemma above.

For instance, see the book of Peter Lax.

$\endgroup$
  • $\begingroup$ Ah, perfect! It was actually that lemma I needed, I hadn't thought of using induction! Thank you $\endgroup$ – preferred_anon Apr 23 '15 at 22:13

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.