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Denote by $\mathcal L'(\mathrm R^n,\mathrm R^m)$ and $\mathcal L_\prime (\mathrm R^n,\mathrm R^m)$ the subsets formed by the surjective and the injective mappings, respectively, of the normed space $\mathcal L(\mathrm R^n,\mathrm R^m)$ (whose vectors are the linear mappings from $\mathrm R^n$ to $\mathrm R^m$).

I'm trying to prove the following:

  • If $m \leq n$, then $\mathcal L'(\mathrm R^n,\mathrm R^m)$ is dense in $\mathcal L(\mathrm R^n,\mathrm R^m)$;

  • If $n \leq m$, then $\mathcal L_\prime (\mathrm R^n,\mathrm R^m)$ is dense in $\mathcal L(\mathrm R^n,\mathrm R^m)$.

I have already proved that $\mathcal L'(\mathrm R^n,\mathrm R^m)$ and $\mathcal L_\prime (\mathrm R^n,\mathrm R^m)$ are open subsets (regardless of whether $m \leq n$ or $n \leq m$), using continuity of determinant and adequate restrictions/extensions of domain. However, the trick does not seem to work in the present case.

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I think you should add a "analysis" tag since you need the concept of "limit"? –  Jack Aug 5 '11 at 1:13
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Let $\lambda \in \mathcal L(\mathrm R^n,\mathrm R^m)$ be not surjective, $m\le n$ and $(e'_1,\dots,e'_m)$ a base of $\mathrm R^m$ such that $(e'_1, \dots , e'_k)$ is a base of $\mathrm {Im}(\lambda)$.
Now let's choose a base $(e_1,\dots,e_n)$ of $\mathrm R^n$ such that $$ \lambda(e_i) = e'_i\quad \forall 1\le i \le k $$ and define a linear map $\delta:\mathrm R^n \to \mathrm R^m$ such that $$ \delta(e_i) := \begin{cases} e'_i & \text{iff $k < i \le m$} \\ 0 & \text{otherwise} \end{cases} $$ For each $\epsilon > 0$, $\lambda_\epsilon := \lambda + \epsilon \delta$ is surjective.
For each $1\le i \le k$ $$ \lambda_\epsilon(e_i) = \lambda(e_i) = e'_i $$ If $k+1 \le i \le m$, let $x$ be a linear combination of $(e_1, \dots, e_k)$ such that $\lambda(x) = \lambda(e_i)$ we have $$ \lambda_\epsilon(e_i - x) = \lambda(e_i) - \lambda(x) + \epsilon \delta(e_i) - \epsilon \delta(x) = \epsilon e'_i $$ Moreover $$ \lim_{\epsilon\to 0} \lambda_\epsilon = \lambda $$

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