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I'm a little stuck with this exercise, it is a true or false question. Let V be a finite vectorial space under a field $\mathbb{F}$ and let $T: V \to V$ be a linear operator, then $\det(T)=\det(T^*)$ where $T^*$ is the dual of T.

I think this is a true statement but I do not know how to prove it.

Any suggestion?

Thomas Andrews
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Bayesian guy
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  • Hint: If $A$ is a square matrix, the determinant of the tranpose of $A$ is the same as the one of $A$. – azif00 Oct 13 '21 at 01:49
  • so basically I have to prove that $det[T]=det[T^*]^t$, but $[T]=[T^*]^t$? – Bayesian guy Oct 13 '21 at 01:59
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    Indeed, if $\beta = (e_1,\dots,e_n)$ is an (ordered) basis for $V$, then for each $i$ you can define a linear map $e_i^* : V \to \mathbb F$ by $e_i^(e_i) = 1$ and $e_i^(e_j) = 0$ if $j \neq i$. In this case $\beta^* = (e_1^,\dots,e_n^)$ is an (ordered) basis for $V^$, and $[T^]{\beta^*} = ([T]\beta)^t$, as you can check. – azif00 Oct 13 '21 at 02:08
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    @Bayesianguy See this post for a proof of the fact that $[T^*] = [T]^t$. – Ben Grossmann Oct 13 '21 at 02:32
  • Thanks, finally I could make sense of this exercise – Bayesian guy Oct 13 '21 at 02:44

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