Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $V$ be a finite-dimensional vector space over $F$ and let $\tau:V \to V$ be a linear operator. Here's my definition of the determinant:

If $t:U \to U$ is a linear operator and $\dim(U)=n$ then $\det(t)$ is the unique number satisfying $$tu_1 \wedge \cdots \wedge tu_n = \det(t) u_1 \wedge \cdots \wedge u_n$$ for all $u_1,\dots,u_n \in U$.

Define the transpose $\tau^T : V^* \to V^*$ of $\tau$ by $(\tau^Tf)(v)=f \tau v$. Is there a way of proving that $\det(\tau^T)=\det(\tau)$ without choosing a basis? It's clear that

$$\det(\tau) v_1 \wedge\cdots\wedge v_n = \tau v_1 \wedge\cdots\wedge \tau v_n$$ and $$\det(\tau^T) f_1 \wedge\cdots\wedge f_n = \tau^T f_1 \wedge\cdots\wedge \tau^T f_n$$ for all $v_1,\dots,v_n \in V$ and $f_1,\dots,f_n \in V^*$, but how do I "join" them together?

share|improve this question
    
I don't understand $\wedge^k V$ is now one dimensional... –  user38268 Nov 5 '12 at 12:40
    
@BenjaLim: What do you mean? –  wj32 Nov 5 '12 at 12:43
add comment

1 Answer 1

Define the "alternating" multilinear map \begin{align} (V^*)^n \times V^n &\to F \\ (f_1,\dots,f_n,v_1,\dots,v_n) &\mapsto \det(f_i(v_j)) \end{align} where $\det(f_i(v_j))$ refers to the determinant of the matrix with its $(i,j)$ entry equal to $f_i(v_j)$. This gives us (details omitted) a nondegenerate pairing $\langle\cdot,\cdot\rangle:\bigwedge^n V^* \times \bigwedge^n V \to F$ with $$(f_1 \wedge\cdots\wedge f_n,v_1 \wedge\cdots\wedge v_n) \mapsto \det(f_i(v_j)).$$ So for all $f_1,\dots,f_n \in V^*$ and $v_1,\dots,v_n \in V$, \begin{align} \det(\tau)\langle f_1 \wedge\cdots\wedge f_n, v_1 \wedge\cdots\wedge v_n \rangle &= \langle f_1 \wedge\cdots\wedge f_n, \det(\tau)v_1 \wedge\cdots\wedge v_n \rangle \\ &= \langle f_1 \wedge\cdots\wedge f_n, \tau v_1 \wedge\cdots\wedge \tau v_n \rangle \\ &= \det(f_i\tau v_j) \\ &= \det((\tau^T f_i)v_j) \\ &= \langle \tau^T f_1 \wedge\cdots\wedge \tau^T f_n, v_1 \wedge\cdots\wedge v_n \rangle \\ &= \det(\tau^T)\langle f_1 \wedge\cdots\wedge f_n, v_1 \wedge\cdots\wedge v_n \rangle. \end{align}

Therefore $\det(\tau)=\det(\tau^T)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

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