Prove or Disprove:
Vector spaces $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$ are isomorphic.
It is generally known that two real vector spaces are isomorphic to one another if and only if they have the same dimensions (see also this question).
This if and only if relationship holds even when their dimensions are infinite.
Vector spaces $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$ both appear to have dim $\mathbb{R^{N}}$ = $\infty$, and dim $\mathbb{R^{N \times N}}$ = $\infty$, which appears to suggest that both vector spaces are isomorphic to one another – that is, there exists an invertible, linear map between $\mathbb{R^{N}}$ and $\mathbb{R^{N \times N}}$.
But specific elements from both vector spaces, such as $\mathbb{R^{2}}$ and $\mathbb{R^{2 \times 2}}$, have different dimensions – dim $\mathbb{R^{2}}$ = 2 and dim $\mathbb{R^{2 \times 2}}$ = 4. This means $\nexists \;\; T : \mathbb{R^{2}} \longrightarrow \mathbb{R^{2 \times 2}}$ such that T is both linear and invertible.
Is the above line of reasoning correct?
Best Answer
Your reasoning is almost correct. Indeed, vector spaces are isomorphic when they are characterized by the same dimension. In other words, their respective bases possess the same cardinality, which itself translates as the existence of bijection between those bases. The spaces $\Bbb{R}^\Bbb{N}$ and $\Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ are isomorphic not because their dimensions are both infinite, but because the sets $\Bbb{N}$ and $\Bbb{N}\times\Bbb{N}$ are related by a bijection.
Concretely, let's call $\phi : \Bbb{N} \to \Bbb{N}\times\Bbb{N}$ the said bijection, whatever its form. Let's also name the basis of $\Bbb{R}^\Bbb{N}$ as $\{e_i\}_{i\in\Bbb{N}}$ and the one of $\Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ as $\{\varepsilon_{(n,m)}\}_{(n,m)\in\Bbb{N}\times\Bbb{N}}$. Then, the linear map $\psi : \Bbb{R}^\Bbb{N} \to \Bbb{R}^{\Bbb{N}\times\Bbb{N}}$ such that $\psi(e_i) = \varepsilon_{\phi(i)}$ is an isomorphism.