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Is it possible to have a one-to-one (injective) linear transformation:

$$f: \mathbb R^4 \to \mathbb R^3$$

If so, is it possible to prove that using dimension theorem?

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    $\begingroup$ Have you heard of the Rank-Nullity Theorem? $\endgroup$ – TomGrubb Jul 13 '15 at 20:13
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    $\begingroup$ As vector spaces over $\mathbb{R}$, the answer is no. As vector spaces over $\mathbb{Q}$, the answer is yes. $\endgroup$ – Zev Chonoles Jul 13 '15 at 20:16
  • $\begingroup$ @ZevChonoles Could you post it as an answer please? Excelent remark. $\endgroup$ – Peter Franek Jul 13 '15 at 20:17
  • $\begingroup$ @ZevChonoles : Will this require the axiom of choice? $\endgroup$ – Michael Hardy Jul 13 '15 at 20:18
  • $\begingroup$ @MichaelHardy: I believe so, I seem to remember that $|X|=|X\times X|$ is not necessarily true in just $\mathsf{ZF}$, and that's the result I'd want to use on $\mathbb{Q}$-bases of these spaces. $\endgroup$ – Zev Chonoles Jul 13 '15 at 20:27
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No. This follows from $$\dim \mathbb R^4 = \dim (\operatorname{Im} f) + \dim(\operatorname{Ker} f).$$ Since $\dim\mathbb R^3=3$, it is clear that $\dim(\mathrm{Im} f)\leqslant 3$, so that $\dim(\operatorname{Ker}f)\geqslant 1$ and hence $f$ is not injective.

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  • $\begingroup$ Thanks @Michael Hardy, forgot about \dim. As for \mathrm vs \operatorname, for some reason Wordpress's LaTeX support doesn't support \operatorname so I got in the habit of writing \mathrm instead. $\endgroup$ – Math1000 Jul 13 '15 at 20:56
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As vector spaces over $\mathbb{R}$, the answer is no, as the other answers have amply described.

However, we can consider $\mathbb{R}$ (and indeed any $\mathbb{R}^n$) as a vector space over $\mathbb{Q}$. We know that the dimension of $\mathbb{R}$ over $\mathbb{Q}$ is $\mathfrak{c}$, the continuum. Suppose that $S\subset\mathbb{R}$ is a $\mathbb{Q}$-basis for $\mathbb{R}$. Then assuming the axiom of choice, we can find a bijection $S^4\to S^3$, i.e. a bijection of a $\mathbb{Q}$-basis for $\mathbb{R}^4$ to a $\mathbb{Q}$-basis for $\mathbb{R}^3$, thus obtaining a $\mathbb{Q}$-linear isomorphism from $\mathbb{R}^4$ to $\mathbb{R}^3$.

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Use the rank–nullity theorem:

The dimension of the domain (in this case $4$) is the sum of the dimension of the image (in this case $\le 3$) and the dimension of the null space (which must in this case therefore be $\ge 1$).

If the dimension of the null space is more than $0$, then more than one point in the domain is mapped to $0$ in the image, so the function is not one-to-one.

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No. Every linear map $T : \Bbb R^4 \to \Bbb R^3$ must satisfy: $$4 = \dim \ker T + \dim \operatorname{Im}\,T,$$ and $\ker T = \{0\}$ will imply that $4 \leq 3$.

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  • $\begingroup$ A minor gripe: the right word is "satisfy". (Even though it is nevertheless common to misuse "verify" this way in mathematics.) See, for example, Milne's page on "Mathlish". $\endgroup$ – Zev Chonoles Jul 13 '15 at 20:43
  • $\begingroup$ Thanks, I'll look it! More culture is always good (also, English is not my first language) $\endgroup$ – Ivo Terek Jul 13 '15 at 20:50
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Suppose that there exists such a transformation $f$. Let $\{b_1,b_2,b_3,b_4\}$ be a base of $\Bbb R^4$. Then $$\{f(b_1), f(b_2), f(b_3), f(b_4)\}$$ is a l.i. subset with four elements of $\Bbb R^3$, which can not exist.

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Never from $\mathbb{R}^{n}\to \mathbb{R}^{m}$ for $n>m$ because the matrix of this transformation must have dimension $m \times n$, so what is its maximum rank? minimum dimension of its null space?

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