1) A determinant function on $n$-dim vector space $V$ is map $\Delta: V*V*V*\dots*V \to K$ with two properties:

  • determinant function is $n$-multilinear;

  • if determinant function has in input two equal vectors, than determinant function is 0.

2) Suppose $V$ is $n$-dim $K$-vector space and $f \in \mathrm{End}_k(V)$. Choose $0 \neq A \in \Lambda^n V^*$ and define $\Delta f$ as

$$ \Delta f = \Delta(f(a_1),\dots,f(a_n)) \implies \exists \,\det(f) \in K: \Delta f = (\det(f))\Delta. $$

The scalar $\det(f)$ is determinant of $f$.

Maybe it's a weird question, but I cannot connect and intuitive understand definitions of determinant function and determinant. As I understood, determinant function takes set of vectors(matrix) and returns scalar.

But what does the formula $(\det(f))\Delta$ mean? what is $\Delta f$ and why does the relationship $\Delta f = (\det(f))\Delta$ hold?

Can you provide also the intuition?

  • $\begingroup$ Looking at this question, it looks like most of your confusion will be cleared up when you reach the end of the chapter. $\Delta f$ is defined here. There's a lot I don't understand here (what does the lambda mean, and what function does $A$ have, etc) But there exists only one determinant function. It seems like you're making preparations to prove that, so maybe that will give some context? $\endgroup$ – Mitchell Faas Apr 12 '17 at 11:10
  • $\begingroup$ You must have made a mistake in transcribing 2), as it makes no sense. First, you have not explained what the $a_i$ and their relation to $A$ are. I assume $a_i \in V$, to make sense of $\Delta((f(a_1),\ldots,f(a_n))$, which is a scalar; but then there is no way for this scalar to be equal to $\det(f) \Delta$, which is a multilinear map. It would be best if you included a link to the original reference. $\endgroup$ – Alex Provost Apr 12 '17 at 17:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.