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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.

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C-star-W-star
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  • What are the $T_n$'s? – Hubble Dec 18 '14 at 16:13
  • You didn't link the previous thread. Paste the link next to the [1]: on the bottom – Ben Grossmann Dec 18 '14 at 16:14
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    What do we know about this norm? Is it a derived norm? Is it multiplicative? Also, what ideas have you come up with here? – Ben Grossmann Dec 18 '14 at 16:16
  • Patience, I'm still editing. ^^ – C-star-W-star Dec 18 '14 at 16:16
  • @Omnomnomnom: Ok so I take any of the equivalent norms on a finite dimensional space, but I don't require it to be multiplicative. – C-star-W-star Dec 18 '14 at 16:25
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    @Omnomnomnom: So far I know the dirty-hands proofs: Norm and determinant are given by polynomials so either way it becomes continuous. – C-star-W-star Dec 18 '14 at 16:27
  • @iHubble: Their linear operators: $T_n\in\mathcal{L}(V)$ – C-star-W-star Dec 18 '14 at 16:29
  • You could use the fact that the determinant is a bounded multilinear map on the columns of the matrix. Then again, that's also somewhat "matrix-dependent". – Ben Grossmann Dec 18 '14 at 16:47
  • I deleted my answer: I'm not sure what you exactly want. – Peter Franek Dec 18 '14 at 16:58
  • How are you defining the determinant? – Santiago Canez Dec 18 '14 at 16:59
  • In most abstract non-sense a linear operator $\det:\mathcal{L}(V)\to\mathbb{C}$. In slightly more concrete sense either the operator that takes the product of eigenvalues or that identifies the induced operator on the exterior algebra. But I left it open as I'm not sure yet which definition fits best for an abstract answer. – C-star-W-star Dec 18 '14 at 17:03
  • @PeterFranek: Just not the definition by polynomials, please. – C-star-W-star Dec 18 '14 at 17:04
  • @SantiagoCanez: You may choose the definition, but please not the polynomial one. – C-star-W-star Dec 18 '14 at 17:05
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    @Freeze_S I can't understand your weird conditions, both in this and in your past post: what is wrong with taking matrices to define determinant of an operator? And what is wrong with pointing out the determinant is a polynomial in $;n;$ unknowns on the entries of a matrix? This is the easiest, smoothest way to check the determinant is a continuous map. Why do you insist so much in not accepting this approach?? – Timbuc Dec 18 '14 at 17:09
  • @Freeze_S If you agree that the volume of a parallelepiped (is it the right word?) depends continuously on the vectors that generate it, then you are done. But it seems to me that if you use any definition, either a geometrical or involving $n$-forms -- if you want to prove continuity to the last $\epsilon$, you sooner or later arrive at the polynomial definition anyway. – Peter Franek Dec 18 '14 at 17:11
  • @Timbuc: Ah because I don't like polynomials =P ...no just a joke. Seriously, because I think there might be a nice answer in the sense of the operator itself not via a representation by matrices. (That message was from yesterday just had no time to right it.) – C-star-W-star Dec 19 '14 at 10:10

1 Answers1

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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 $\det T_m = k_{T_m} \to k_T = \det T$.

PhoemueX
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