# Does a one-to-one linear transformation from $\mathbb R^4$ to $\mathbb R^3$ exist?

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?

• Have you heard of the Rank-Nullity Theorem? Commented Jul 13, 2015 at 20:13
• As vector spaces over $\mathbb{R}$, the answer is no. As vector spaces over $\mathbb{Q}$, the answer is yes. Commented Jul 13, 2015 at 20:16
• @ZevChonoles Could you post it as an answer please? Excelent remark. Commented Jul 13, 2015 at 20:17
• @ZevChonoles : Will this require the axiom of choice? Commented Jul 13, 2015 at 20:18
• @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. Commented Jul 13, 2015 at 20:27

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.

• 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. Commented Jul 13, 2015 at 20:56

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 $\mathbb{R}^3$.

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.

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 $4 \leq 3$.

• 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". Commented Jul 13, 2015 at 20:43
• Thanks, I'll look it! More culture is always good (also, English is not my first language) Commented Jul 13, 2015 at 20:50

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.

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?