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Let $V$ be a vector space over a field $F$ of characteristic $0$. A linear operator $T$ on $V$ induces a linear operator $\Lambda^k T:\Lambda^k V\to \Lambda^k V$ such that $\Lambda^k T(v_1\wedge \cdots\wedge v_k)=Tv_1\wedge\cdots\wedge Tv_k$ for all $v_1, \ldots, v_k\in V$.

If $n=\dim V$, then since $\dim(\Lambda^n V)=1$, we know that there is a unique $c\in F$ such that $\Lambda^n T(v_1 \wedge \cdots\wedge v_n)=c\cdot(v_1\wedge \cdots\wedge v_n)$. We call this constant the determinant of $T$.

From this definition of the determinant, it immediately follows that $\det(TS)=(\det T)(\det S)$ for all linear operators $S$ and $T$ on $V$.

Can we also easily show that $\det T^t=\det T$ for all $T\in \mathcal L(V)$?

Here $T^t$ denotes the transpose of $T$.

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I assume that by transoposition you mean dual mapping $T^*$. That is if $T:V\rightarrow V$ then $T^*:V^*\rightarrow V^*$ in following manner $$\left[T^*\alpha\right](v):=\alpha(T(v)).$$ Let $\alpha_1,\dots,\alpha_n\in V^*.$ Fix arbitrary $v_1,\dots,v_n\in V.$ Then

$$\left[\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)\right](v_1\wedge\dots\wedge v_n)=(T^*\alpha_1\wedge\dots\wedge T^*\alpha_n)(v_1\wedge\dots\wedge v_n)=\\=\det(\left[T^*\alpha_i\right](v_j))=\det(\alpha_i(T(v_j)))=\\=(\alpha_1\wedge\dots\wedge\alpha_n)(T(v_1)\wedge\dots\wedge T(v_n))=(\alpha_1\wedge\dots\wedge\alpha_n)(\bigwedge^nT(v_1\wedge\dots\wedge v_n))=\\=(\alpha_1\wedge\dots\wedge\alpha_n)(\det T\cdot v_1\wedge\dots\wedge v_n)=\det T\cdot(\alpha_1\wedge\dots\wedge\alpha_n)(v_1\wedge\dots\wedge v_n).$$ Since $v_1,\dots,v_n $ were arbitrary we get that $$\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)=\det T\cdot(\alpha_1\wedge\dots\wedge\alpha_n).$$ Hence $\det T$ is determinant of $T^*,$ i.e $\det T=\det T^*.$


Edit

darij grinberg made a good point that we do not know why we can evaluate $$\left[\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)\right](v_1\wedge\dots\wedge v_n).$$ We can do that due to cannonical identification of $\bigwedge^n T^*$ and $(\bigwedge^n T)^*.$

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    $\begingroup$ What does $(T^*\alpha_1\wedge\dots\wedge T^*\alpha_n)(v_1\wedge\dots\wedge v_n)$ mean? There is no such thing as a "wedge product of two linear maps" in general; there are some conventions for how to define such a thing, but they suffer from annoying $n!$ factors. $\endgroup$ – darij grinberg Jun 30 '15 at 15:12
  • $\begingroup$ @darijgrinberg I might write that $$(T^*\alpha_1\wedge\dots\wedge T^*\alpha_n)(v_1\wedge\dots\wedge v_n)=\det(\left[T^*\alpha_i\right](v_j)).$$ The $n!$ does not change much cause we use definition of this map twice. Once to jump in and once to jump out. $\endgroup$ – Fallen Apart Jun 30 '15 at 15:27
  • $\begingroup$ Okay; then, can you explain why $\left[\bigwedge^n T^*(\alpha_1\wedge\dots\wedge\alpha_n)\right](v_1\wedge\dots\wedge v_n)=(T^*\alpha_1\wedge\dots\wedge T^*\alpha_n)(v_1\wedge\dots\wedge v_n)$ ? $\endgroup$ – darij grinberg Jun 30 '15 at 16:41
  • $\begingroup$ @darijgrinberg Look up at the second sentence of the question. It is induced map from $T^*.$ It is the reason why I used square bracets. Mostly to distinguish order of operations (evaluate vectors in the end). $\endgroup$ – Fallen Apart Jun 30 '15 at 16:42
  • $\begingroup$ @darijgrinberg Ok you are right. I forget to mention that there is cannonical identifications $\bigwedge T^*$ and $(\bigwedge T)^*.$ $\endgroup$ – Fallen Apart Jun 30 '15 at 16:50

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