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?
 A: 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.
A: 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: 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.
A: 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.
A: 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 $\mathbb{R}^3$.
A: 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?
