Defining matrices Lie groups I have noticed that we use the clear identification of Mat($n,m$) and $\mathbb{R}^{nm}$ given by $(a_{i,j})_{i,j} \to (a_{1,1},\cdots a_{n,m})$.
This seems quite important for computation of tangent space for example, everytime we use a limit to compute a differential in term of the coefficients of the matrix. What I was asking was, if we put on Mat($n,m$) the operator norm $||A|| := \text{max}\left\lbrace \frac{||Ax||}{||x||} : x \in \mathbb{R}^{n} -\textbf{0}\right\rbrace$, the induced topology should turn the identification above between Mat($n,m$) and $\mathbb{R}^{nm}$ an homeomorphism.
There is a nice way to see it explicitly ? How could I prove it ? My difficulty relies on the fact that I'm not confortable using the topology induced by the operator norm to prove the continuity of the map.
On a finite dimensional real vector space, all norms are equivalent. So, since that map is a homeomorphism if we consider the norm$$\bigl\|(a_{ij})_{1\leqslant i,j\leqslant n}\bigr\|=\sqrt{\sum_{i=1}^n\sum_{j=1}^ma_{ij}^{\,2}}$$on $\operatorname{Mat}(n,m)$ (it's actually an isometry then), it will still be a homeomorphism if we use the operator norm there.