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 $T$.


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 $\det T=\det T^*.$


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 $(\bigwedge^n T)^*.$

  • 1
    $\begingroup$ 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

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.