Euclidean metric for set of linear transformations. I was going through Analysis and Topology Problem sheets from Cambridge University (I do not study there and I do not have access to their lectures.) and I stumbled upon the following.

Let $(T_k)_{k∈ \mathbb{N}}$ be a sequence in $L(\mathbb{R}^m, \mathbb{R}^n)$. Show that if $T_n → 0$ in the euclidean metric, then $T_n → 0$ pointwise. Is the converse true? Do your answers change if $(T_k)$ is a sequence in $Bil(\mathbb{R}^m × \mathbb{R}^n, \mathbb{R}^p)$?

It had me guessing what Euclidean Metric could possibly mean here. I can think of such a metric; write a matrix with respect to a certain basis (most likely the standard basis). Treat these matrices as elements of $R^{mn}$ and use Euclidean Metric there. The only problem is that this metric depends on the choice of basis. And if the question were referring to this metric, I doubt they would have used $L(\mathbb{R}^n, \mathbb{R}^m) $ as it would make more sense to use $Mat(\mathbb{R}, m \times n)$.
The other metric I could think of was the supremum (operator) norm on the difference of matrices. But it feels there is nothing Euclidean about it.
So, what other metric would befit the name "Euclidean Metric for the set of Linear transformations"?
Thank you in advance!
 A: You don't actually need to mention matrices at all to define a metric over here. You can use them as inspiration but you don't need them.
The Euclidean metric on $\mathbb{R}^n$ is really just induced by the standard norm on $\mathbb{R}^n$, which is induced by the standard inner product:
$$\langle x,y \rangle = \sum_{i=1}^{n} x_i y_i$$
$$\left| \left|x \right|\right| = \sqrt{\langle x,x \rangle} = \sqrt{\sum_{i=1}^{n} x_i^2}$$
$$d(x,y) = \left|\left|x-y\right|\right| = \sqrt{\sum_{i=1}^{n} (x_i-y_i)^2}$$
This does depend on the standard basis of $\mathbb{R}^n$ (in the sense that if we used a different basis, the coordinates of $x$ relative to that basis would be different) but why would this be a problem? The standard metric for $\mathbb{R}^n$ mimics what we observe in Pythagoras's Theorem. It gives us a reasonable way to measure distances that conforms with how we would normally go about doing this in $\mathbb{R}^2$ and $\mathbb{R}^3$. It is as 'Euclidean' as it gets. Of course, you can introduce other norms on $\mathbb{R}^2$ and $\mathbb{R}^3$ if you wish.
In any event, your intuition is correct. Observe that it really doesn't matter if we talk about $Hom(\mathbb{R}^n,\mathbb{R}^m)$ or $M(m \times n,\mathbb{R})$. Both vector spaces are isomorphic to each other (can you see why?) so it is perfectly reasonable to try and define a norm on $Hom(\mathbb{R}^n,\mathbb{R}^m)$ by taking pointers from $M(m\times n,\mathbb{R})$.
I promised that you don't need matrices over here. Let me make good on that promise now. Let $T \in Hom(\mathbb{R}^n,\mathbb{R}^m)$. Let $\{e_1,\ldots,e_n\}$ be the standard basis of $\mathbb{R}^n$. We define
$$\left| \left| T \right| \right|_{Eucl} :=\sqrt{\sum_{i=1}^{n} \left| \left| T(e_i) \right|\right|^2}$$
Notice that this is precisely the strategy that you proposed but we didn't need to think about matrices to make this definition work.
Now, it is an entirely different thing to show that that all norms $\mathbb{R}^n$ are equivalent, in the sense that there are simple inequalities relating one norm to another. So, if $p_1$ and $p_2$ are norms on $\mathbb{R}^n$, you can show that:
$$\exists c_1,c_2 > 0: c_1 p_1 \leq \left| \left| \cdot \right| \right|_{standard} \leq c_2 p_2$$
So, the specific norm you use to define convergence will not matter at all. I'll leave you to try and prove this on your own.
