Different choice of bases leads to a homeomorphic metric on Hom(V,W)

Question

Let Hom(V,W) denote all linear transformations from V to W. Choosing a basis for V and W, we can identify Hom(V,M) with the m*n matrices, and consequently give it the metric of R^nm. Show that a different choice of bases leads to a homeomorphic metric on Hom(V,W)

Answer 1

1. The metric on $X = \mathbb{R}^{nm}$ arises as a norm. $$ \|(x_j)\| = \sqrt{\sum_{j=1}^{nm} |x_j|^2} $$
2. Any two norms on a finite dimensional space are equivalent : Define a fixed norm on $X$ as $$ \|(x_j)\|_1 = \sum_{j=1}^n |x_j| $$ Now, if $\|\cdot \|$ is any other norm on $X$, then $$ \|x\| = \|\sum_{j=1}^n x_j e_j\| \leq M\|x\|_1 $$ where $$ M = \max\{\|e_j\| : 1\leq j\leq n\} $$ Hence, the identity map $$ \iota : (X,\|\cdot\|_1) \to (X,\|\cdot\|) $$ is continuous.
3. The unit ball $$ B = \{x\in X : \|x\|_1 \leq 1\} $$ is compact by Bolzano-Weierstrass Hence the map $\iota$ restricted to the unit ball maps a compact set to another compact set and is hence a homeomorphism. In other words, $\iota^{-1}$ is a bounded linear map, and so there exists a $M'\in \mathbb{R}$ such that $$ \|x\|_1 \leq M' \|x\| $$ Thus, $\|\cdot \|$ and $\|\cdot \|_1$ are equivalent norms, and so generate the same topology.