I'm interested in a coordinate-free proof of the statement $\mathrm{det}(A) = \mathrm{det}(A^T).$

Let $V$ be a finite-dimensional vector space over a field $K$, and let $f : V \rightarrow V$ be an endomorphism.

I define $\mathrm{det}(f)$ by $\mathrm{det}(f) \cdot \varphi = \varphi \circ f$, where $\varphi : V \times ... \times V \rightarrow K$ is any alternating function and $(\varphi \circ f)(v_1,...,v_n) := \varphi(f(v_1),...,f(v_n)).$

I define the transpose by $$f^* : V^* \rightarrow V^*, \; \psi \mapsto \psi \circ f.$$

I would like to prove that $\mathrm{det}(f) = \mathrm{det}(f^*)$ using these definitions but haven't been able to find a proof that doesn't involve choosing a basis. Thanks

  • $\begingroup$ I feel like your definition of the transpose is incomplete. What relationship, if any, is there between $f$ and $f^*$? $\endgroup$ – Muphrid May 22 '14 at 4:54
  • 2
    $\begingroup$ @Muphrid $f^*$ takes a linear form $\psi$ and "does $f$ first". If you choose a basis and represent $f$ by a matrix $A$, then $f^*$ is represented by $A^T$ with respect to dual basis, so in that sense it's the 'transpose'. I would rather avoid this argument. $\endgroup$ – user152604 May 22 '14 at 4:57
  • 1
    $\begingroup$ What I'm getting at is that the transpose is the matrix representation of the adjoint of a linear operator only when there is an inner product and that inner product is Euclidean. The inner product helps give a canonical isomorphism between the vector space $V$ and its dual $V^*$. I don't see an overt reference to a choice of inner product here; perhaps you've buried it in some notation I'm unfamiliar with. In any case, I doubt you can prove the statement without some reference to an inner product, for without that canonical isomorphism, the matrix and its transpose have no relationship. $\endgroup$ – Muphrid May 22 '14 at 5:12
  • 1
    $\begingroup$ @Muphrid: No, this has nothing to do with inner products. The preceding comment by user152604 describes the situation perfectly. Inner products only enter if you want to identify elements of $V$ with elements of $V^*$, and this is not done here ($f$ and $f^*$ live in different worlds, so to speak, since $f^*$ acts on $V^*$ and $f$ acts on $V$). $\endgroup$ – Hans Lundmark May 22 '14 at 6:20
  • $\begingroup$ possible duplicate of Determinant of the transpose via exterior products $\endgroup$ – Hans Lundmark May 22 '14 at 6:25

For any simple $k$-vector $v_{(k)}$ and any simple $k$-form $\omega_{(k)}$, the linear operator $f$ and its adjoint $f^*$ should obey

$$\omega_{(k)}[f(v_{(k)})] = [f^*(\omega_{(k)})](v_{(k)})$$

where I have defined the action of a linear operator on $k$-vectors/$k$-forms as follows: if $v_{(k)} = v_1 \wedge v_2 \wedge \ldots \wedge v_k$ for $k$ linearly independent vectors, then

$$f(v_{(k)}) = f(v_1) \wedge f(v_2) \wedge \ldots \wedge f(v_k)$$

and similarly for the adjoint acting on a $k$-form.

It's not obvious to me that the above statements are immediately clear from your definition of the adjoint. It may be you have to prove it starting with the vector/1-form case:

$$\omega[f(v)] = [f^*(\omega)](v)$$

and go from there inductively. Once you have proved this notion, though, then the rest of the proof follows immediately, as $f(v_{(n)}) = (\det f) v_{(n)}$ and $f^*(\omega_{(n)}) = (\det f^*) \omega_{(n)}$.


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.