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The set of injective (surjective) linear transformations is dense in $\mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m})$ if $n\leq m$ (if $n\geq m$).

I don't know how to show this. If $A_{1}$ is the set of injective linear transformations and $A_{2}$ is the set of surjective linear transformations, I shown that $A_{1}$ and $A_{2}$ are open sets in $\mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m})$. This a previous questions so, I think that I should use it, but I don't have any idea. Thanks for any hint.

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    $\begingroup$ What metric are you using? What equivalent definitions of injecitvity do you know? Are you familiar with representing linear mapping as matrix? $\endgroup$
    – user99914
    May 12 '18 at 5:02
  • $\begingroup$ @JohnMa, 1 - Euclidean Metric. 2 - For example: "there is $c > 0$ such that $|Tx| \leq c|x|$", "$Tu \neq 0$ if $u \neq 0$". 3 - Yes! I know too the $||x||_{1}$ norm and $||x||_{\infty}$ norm, but I don't have much of a habit of using them. $\endgroup$
    – Corrêa
    May 12 '18 at 5:10
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First assume $m=n$. Let $\mathcal I(\mathbf R^n, \mathbf R^n)$ be all the injective linear maps taking $\mathbf R^n$ to $\mathbf R^n$. One can identify $\mathcal L(\mathbf R^n, \mathbf R^n)$ with $M_n(\mathbf R)$ after choosing a basis for $\mathbf R^n$. Under this identification, $\mathcal I(\mathbf R^n, \mathbf R^n)$ corresponds to the complement of $\det^{-1}(0)\subseteq M_n(\mathbf R)$, where $\det:M_n(\mathbf R)\to \mathbf R$ is the determinant map. Since $\det$ is a polynomial map, it's zero-set has empty interior (in fact it has measure $0$). Thus $\mathcal I(\mathbf R^n, \mathbf R^n)\cong M_n(\mathbf R)\setminus \det^{-1}(0)$ is dense in $\mathcal L(\mathbf R^n, \mathbf R^n)$.

Now let $m\geq n$. Let $T:\mathbf R^n\to \mathbf R^m$ be an arbitrary linear map. We will show that there are injective linear maps $\mathbf R^n\to \mathbf R^m$ which are arbitrarily close to $T$. Let $V$ be an $n$-dimensional subsapce of $\mathbf R^m$ which contains the image of $T$. Identify $V$ with $\mathbf R^n$. Applying what we have already proved, we see that there are indeed linear map $S:\mathbf R^n\to V\cong \mathbf R^n\subseteq \mathbf R^m$ which are arbitrarily close to $T$ and we are done.

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  • $\begingroup$ Nice! I proved too that all matrice $n \times n$ is limit of a sequence of invertible matrices $n \times n$. Can I use it in case $n = m$? $\endgroup$
    – Corrêa
    May 12 '18 at 17:25
  • $\begingroup$ Yes of course. That says that invertible $n\times n$ matrices are dense in all $n\times n$ matrices. But invertibility of a square matrix is same as the injectivity of the corresponding linear map. $\endgroup$ May 13 '18 at 2:04
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You may consider $m\times n$ matrices instead of abstract linear transformations. An $m\times n$ matrix $A$ is injective iff it has rank $n$, that it to say iff there is a non-zero $n\times n$ minor.

Considering any $n\times n$ submatrix $B$, by density of $GL_n(\mathbb R)$ there exists some invertible $C$ such that $\|B-C\|\leq \epsilon$. Glue the rest of $A$ to $C$ to obtain $C'$, which has rank $n$ (since $C$ is an invertible submatrix) and $\|C'-A\|<\epsilon$.

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It can be shown by induction. Let $\mathcal{I}(\mathbb{R}^{n},\mathbb{R}^{m}) = \{ T \in \mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m}) : T \text{ is injective} \}$.

Start by showing that for all $m \in \mathbb{N}$, $\mathcal{I}(\mathbb{R}^{1},\mathbb{R}^{m})$ is dense in $\mathcal{L}(\mathbb{R}^{1},\mathbb{R}^{m})$.

Now suppose that for some $n \in \mathbb{N}$ and for all $m \in \mathbb{N}_{\ge n}$, $\mathcal{I}(\mathbb{R}^{n},\mathbb{R}^{m})$ is dense in $\mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m})$. Since $\mathcal{I}(\mathbb{R}^{1},\mathbb{R}^{m})$ is dense in $\mathcal{L}(\mathbb{R}^{1},\mathbb{R}^{m})$ and $\mathcal{I}(\mathbb{R}^{n},\mathbb{R}^{m})$ is dense in $\mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m})$, it follows that $\mathcal{I}(\mathbb{R}^{n+1},\mathbb{R}^{m}) \cong \mathcal{I}(\mathbb{R}^{n},\mathbb{R}^{m}) \times \mathcal{I}(\mathbb{R}^{1},\mathbb{R}^{m})$ is dense in $\mathcal{L}(\mathbb{R}^{n},\mathbb{R}^{m}) \oplus \mathcal{L}(\mathbb{R}^{1},\mathbb{R}^{m}) \cong \mathcal{L}(\mathbb{R}^{n+1},\mathbb{R}^{m})$.

The reasoning for the surjective statement is analogous.

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