# Coordinate-free proof of determinant of transpose

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

• I feel like your definition of the transpose is incomplete. What relationship, if any, is there between $f$ and $f^*$? – Muphrid May 22 '14 at 4:54
• @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. – user152604 May 22 '14 at 4:57
• 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. – Muphrid May 22 '14 at 5:12
• @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 $V$). – Hans Lundmark May 22 '14 at 6:20
• possible duplicate of Determinant of the transpose via exterior products – 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 $f^*(\omega_{(n)}) = (\det f^*) \omega_{(n)}$.