Determinant: Continuity Reference
Build-up on: Determinant: Definition
Problem
Given a vector space $V$.
Consider an endomorphism $T:V\to V$.
Define its determinant $\det:\mathcal{L}(V)\to\mathbb{C}$.
Introduce a norm $\|T\|$.
How to prove the continuity of the determinant:
$$T_n\to T\implies\det T_n\to\det T$$
I'm looking for a non-dirty-hands-matrix-polynomial-answer. ;)
I assume basic knowledge of Differential Geometry and Functional Analysis.
 A: Do you like the following?
If we take the definition proposed in the linked question, we can do the following:
Take $v_1, \dots, v_n \in V$ with $v_1 \wedge \dots \wedge v_n \neq 0$ (with $n = \dim V$). Such $v_1, \dots, v_n$ exist, because otherwise $\Lambda^n V = \{0\}$.
By the definition in the other post, we have
$$
(Tv_1) \wedge \dots \wedge (T v_n) = k_T \cdot v_1 \wedge \dots \wedge v_n
$$
for each $T \in B(V)$, where $k_T \in \Bbb{C}$ is a suitable (unique!) constant (which we call the determinant $\det T = k_T$).
The existence of such a constant follows because of $\dim ( \Lambda^n V) = 1$.
The map
$$
\Phi : V^n \to \Lambda^n V, (w_1, \dots, w_n) \mapsto w_1 \wedge \dots \wedge w_n
$$
is multilinear. Now every multilinear map on a finite dimensional normed vector space is bounded/continuous (to see this, we have to choose bases and so on, which maybe makes the construction less appealing to you, but we are also using bases, ... in arguments like $\dim (\Lambda^n V) = 1$).
Hence, if $T_m \to T$, then $T_m v_j \to T v_j$ for all $j = 1, \dots, n$, which yields
$$
k_{T_m} \cdot v_1 \wedge \dots \wedge v_n =(T_m v_1) \wedge \dots \wedge (T_m v_n) \to (Tv_1 ) \wedge \dots \wedge (Tv_n) = k_T \cdot v_1 \wedge \dots \wedge v_n.
$$
But this implies $\det T_m = k_{T_m} \to k_T = \det T$.
