Isomorphism of the set of $n$ x $n$ matrices with the manifold with dimension $n^2$

Question

I heard the statement that the set of $n \times n$ matrices is isomorphic to $\mathbb{R}^{n^2}$. I see how the set is a submanifold of $\mathbb{R}^{n^2}$, but I am struggling to understand/prove how they are isomorphic. Thank you!

Answer 1

Lets do an example with $2 \times 2$ matrices. Consider the map $f(A) = (a,b,c,d)$ where we let $a = a_{11},b = a_{12}, c = a_{21}, d = a_{22}$ and $A = (a_{ij})$. This is a map into $\mathbb{R}^{2^2}$ which is a bijection. Call its inverse $g$. We can induce a topology on $M=\mathcal{M}(2,2)$ i.e the set of $2 \times 2$ matrices by $X$ is open in $M \iff g^{-1}(X)$ is open in $\mathbb{R}^4$.

This map is clearly linear and so it preserves the vector space structure of both $M$ and $\mathbb{R}^4$. It is also invertible (as mentioned before) i.e $M$ and $\mathbb{R}^4$ are isomorphic as topological vector spaces. We can induce an atlas on $M$ by the charts $\phi_{\alpha} \circ f$ where $\phi_{\alpha}$ are charts of $\mathbb{R}^4$. Hence, $f$ is a homeomorphism of the domain for each $\phi_{\alpha}$ i.e a $C^0$ reparametrization and so $M$ and $\mathbb{R}^4$ are homeomorphic.