1
$\begingroup$

Show that a finite-dimensional vector space $V$ is complete.


I'm not sure on how to start with this problem. I kind of think that one should use the fact that any two norms on the finite vector space are equivalent, but I don't know how to proceed forward. What would be a way to solve the problem?

Thanks for the help.

$\endgroup$
6
  • $\begingroup$ Can you show that $\mathbb R^n$ is complete? $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '15 at 3:54
  • $\begingroup$ Yes. Under the Euclidean norm. It follows that it is complete under any other norm on $\mathbb{R}^{n}$ since we have equivalence of norms? Not so sure about the last statement. $\endgroup$ – Guy Nov 27 '15 at 3:57
  • 2
    $\begingroup$ Well, you should try to make sure it that last statement is true or not! $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '15 at 3:59
  • $\begingroup$ First of all, you'd want to give a metric on the vector space - there isn't a single metric that can be found just knowing that a space is an abstract vector space over the reals. Secondly, you'll need your field to be a complete field, first. $\endgroup$ – Thomas Andrews Nov 27 '15 at 7:27
  • 2
    $\begingroup$ The question should mention what (kind of) field this vector space is supposed to be over (I don't think it makes any sense for vector spaces over a finite field, for instance). And also what sense of "complete" is intended. $\endgroup$ – Marc van Leeuwen Nov 27 '15 at 9:24
2
$\begingroup$

Pick a basis $\{v_1,\dots,v_n\}$ of $V$. If $a=\sum a_iv_i$ and $b=\sum b_iv_i$, then $\Vert a-b\Vert\leq\sum\vert a_i-b_i\vert\Vert v_i\Vert$ using the triangular inequality, and if $M$ is larger than the norm of all the $v_i$s, then clearly $\Vert a-b\Vert\leq M\sum|a_i-b_i|$.

It follows that the map $T:(a_1,\dots,a_n)\in\mathbb R^n\mapsto \sum a_iv_i\in V$, which is a bijective linear map, is continuous if we endow the domain with the $1$-norm. It follows that the function $a=(a_1,\dots,a_n)\in\mathbb R^n\mapsto\Vert\sum a_iv_i\Vert$ is continuous and non-zero on the unit sphere of $\mathbb R^N$,so its minimum there is a positive number. This implies that the map $T$ has a continuous inverse.

Since $T$ and its inverse map Cauchy sequences to Cauchy sequences and limits to limits, one can use this to show that $V$ is complete.

$\endgroup$
3
$\begingroup$

Hint:

  1. Show on finite dimensional spaces, any two norms are equivalent.
  2. Show if two norms are equivalent, complete with respect to one norm is equivalent to complete wrt another norm.
  3. Hence it suffices to show it's complete wrt to $\|\cdot\|_1$, which is defined as $\|x\|_1=\sum_{i=1}^n|\alpha_i|$ with $x=\sum_{i=1}^n\alpha_i e_i$ where $e_i$ are basis for the finite dimensional space. Remember to use the completeness of $\mathbb{R}$ or $\mathbb{C}$
$\endgroup$
3
$\begingroup$

To start, let's pick a particularly convenient norm. We know that we can find a finite basis for our space so that we may write any vector in the space as its coordinates with respect to this basis. Now equip the space with the standard Euclidean norm (if $\textbf{v}=(a_1,\ldots,a_n)$ then $\vert \vert \textbf{v}\vert\vert=\sqrt{a_1^2+\ldots+a_n^2}$). Now show that because the sequence of vectors is Cauchy with respect to this norm, the $n$ sequences defined by the coordinates of this norm must each be Cauchy. This means that each sequence of coordinates will converge in the field of scalars (this requires that the field of scalars of your vector space is itself complete). It follows that the sequence of vectors will converge to the point defined by the the limit of these $n$ scalar sequences.
Finally, apply the equivalence of norms to show that the sequence of vectors will converge with respect to any norm.

$\endgroup$
1
  • 3
    $\begingroup$ How do you prove that all norms are equivalent before knowing that the space is complete? $\endgroup$ – Mariano Suárez-Álvarez Nov 27 '15 at 4:27
0
$\begingroup$

I would advise using the norm $\Vert x \Vert_\infty = \mathrm{max} \{ \vert x_1 \vert, \ldots, \vert x_n \vert \}$.

That way the notions of convergent and Cauchy sequence are equivalent to every component being convergent or Cauchy respectively. Then the claim follows immediately by using the completeness of $\mathbb{R}$.

$\endgroup$

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