I'm trying to find a way to prove that the determinant of the transpose of an endomorphism is the determinant of the original linear map (i.e. det(A) = det(Aᵀ) in matrix language) using Dieudonne's definition of the determinant expressed in terms of an alternating bilinear form but am having problems with it. Let me set the stage a little:

Start with a Bilinear form ψ : V × V → ℝ | (u,v) ↦ ψ(u,v) = ψ(∑ᵢ₌₁²λᵢeᵢ,∑ᵢ₌₁²μᵢeᵢ).

It is alternating if ψ(v,v) = 0 & anti-symmetry, ψ(u,v) = - ψ(v,u), follows by expanding ψ(u + v,u + v) = 0

Consider the endomorphism f : V → V where V is finite-dimensional (in this case, 2-D). If ψ(f(u),f(v)) = γψ(u,v) for some γ ∈ ℝ (a unique γ which follows by expanding the alternating form & showing γ is determined by the action of ψ on it's basis) & we intuitively interpret the alternating bilinear form ψ(u,v) as the area spanned by u & v the geometric interpretation of γ is inescapably obvious. Here γ is known as the determinant of the map f, i.e. ψ(f(u),f(v)) = det(f)ψ(u,v).

Some quick properties that follow immediately, without any stress, from the definition:

1: ψ(u,v) = det(I)ψ(u,v) → det(I) = 1

2: ψ((fg)(u),(fg)(v)) = det(fg)ψ(u,v) → ψ(f(g(u)),f(g(v))) = det(f)ψ(g(u),g(v)) = det(f)det(g)ψ(u,v) → det(fg) = det(f)det(g)

3: f Bijective → ff⁻¹ = I → 1 = det(I) = det(ff⁻¹) = det(f)det(f⁻¹) → det(f) ≠ 0

4: ψ(f(u + w),f(v)) = det(f)ψ(u + w,v) = ψ(f(u + w),f(v)) = det(f)ψ(u,v) + det(f)ψ(w,v)

5: ψ(f(u + av),f(v)) = = det(f)ψ(u + av,v) = = det(f)ψ(u,v)

6: ψ(f(b),f(v)) = ψ(f(αu + βv),f(v)) = ψ(αf(u) + βf(v),f(v)) = αψ(f(u),f(v)) → α = ψ(f(b),f(v))/ψ(f(u),f(v)), β = ψ(f(u),f(b))/ψ(f(u),f(v))

7: ψ((f + g)(u),(f + g)(v)) = det((f + g))ψ(u,v) = [det(f) + det(g)]ψ(u,v) + ψ(f(u),g(v)) + ψ(g(u),f(v))

(I can't think of what to do right now that doesn't result in a dead end but at least we have a nice geometric interpretation of why det(A + B) ≠ det(A) + det(B)).

All of the above extends easily to multilinear forms, now here are the questions:

[1] Using this approach to show det(f) = det(fᵀ) I get stuck. There seem to be a few approaches. One, taken by Dieudonne, is to just say that on expanding the determinant we see that the expansion equals the expansion of the transpose matrix. Another, taken by Bourbaki, is to prove using permutations that the determinant of the matrices are equal then using the fact that the dual basis representation of the map is the transpose of the matrix representation in the original basis to prove this for the maps... However these approaches seem disconnected with the simplicity the above picture so I give it a shot myself, basically if det(f) = det(fᵀ) holds when we do it with matrices, it should be that:

ψ(f(u),f(v)) = ψ(u,fᵀ(f(v))) = ψ(fᵀ(f(u)),v) = det(f)ψ(u,v)


ψ(fᵀ(u),fᵀ(v)) = ψ(u,f(fᵀ(v))) = ψ(f(fᵀ(u)),v) = det(fᵀ)ψ(u,v)

give the same result, though I can find no way of showing this. However, if the above equalities are taken seriously then it amounts to showing ψ(u,fᵀ(f(v))) = ψ(u,f(fᵀ(v))), but quite honestly I'm not sure any this is even valid in the first place. For instance:

If we define:

f : V → U | v ↦ f(v)

Φ : U → ℝ | w ↦ Φ(w)

(Φf) : V → ℝ | (ΦT) : v ↦ (Φf)(v) = Φ(f(v))

fᵀ: L(U,ℝ) → L(V,ℝ) | Φ ↦ fᵀ(Φ) = Φ(f)

where fᵀ(Φ)(v) = (Φf)(v) = Φ(f(v))

then given a concrete example:

f : V → U | (x,y) ↦ f(x,y) = (y,x + y)

Φ : U → ℝ | (u,v) ↦ Φ(u,v) = u - 2v

(Φf) : V → ℝ | (Φf) : (x,y) ↦ (Φf)(x,y) = Φ(f(x,y)) = Φ(y,x + y) = y - 2(x + y)

fᵀ: L(U,ℝ) → L(V,ℝ) | Φ ↦ fᵀ(Φ) = Φ(f)

where fᵀ(Φ)(x,y) = (Φf)(x,y) = Φ(f(x,y)) = Φ(y,x + y) = y - 2(x + y)

even considering the map ψ(u,fᵀ(f(v))) = ψ(u,f(fᵀ(v))) makes no sense. How in the world can you write fᵀ(f(v))? fᵀf : V → L(V,ℝ) isn't a valid map, fᵀ has to be defined on the image of f whereas it's not defined on the vector space U it's defined on the vector space of linear functionals from U to ℝ. It implies that f(x,y) = (y,x + y) is an element of L(U,ℝ). If you want to compose f with fᵀ then fᵀ has to be defined on L(U,ℝ), you're talking about entirely separate concepts in doing this so I don't see why ψ(u,fᵀ(f(v))) = ψ(u,f(fᵀ(v))) makes any sense, it seems fast and loose to me... Even pretending this composition makes sense we get fᵀf : V → L(V,ℝ) | (x,y) ↦ (fᵀf)(x,y), which - now an element of L(V,ℝ) - gives (fᵀf)(x,y) : V → ℝ | (z,w) ↦ (fᵀf)(x,y)(z,w) which again makes no sense to me.

I can't even make up an example to test whether ψ(u,fᵀ(f(v))) = ψ(u,f(fᵀ(v))) holds inside an alternating bilinear form so hopefully you see my problem.

Basically I don't know

a) why fᵀ(f(v)) is a valid definition

b) whether ψ(u,fᵀ(f(v))) = ψ(u,f(fᵀ(v))) holds inside of an alternating bilinear form

c) an alternative way to show ψ(f(u),f(v)) = ψ(fᵀ(u),fᵀ(v)) = det(f)ψ(u,v) = det(fᵀ)ψ(u,v) without giving in, losing the flow, & resorting to permutations or explicit calculations.

[2] If it works out that we can prove things this way & I were to try to adapt this process to multilinear forms, I don't know how one can talk about the transpose of a linear map since in most sources, e.g. the wiki page on the transpose (can't link it), it's defined in terms of bilinear forms - but we're working inside of a multilinear form. Would I have to create a valid definition of the transpose inside of a multilinear form? I can't find anything on this & I don't know if one can just write ψ(f(u),f(v),f(w)) = ψ(u,fᵀ(f(v)),fᵀ(f(w))) etc...?

[3] Dieudonne goes on to define the adjoint f* of a map f, one way using a bilinear form ψ(f(x),y) = ψ(x,f*(y)), & an alternative way I'll describe below.

If f is an endomorphism of V then fᵀ is an endomorphism of L(V,ℝ), & so using the linear maps:

s_ψ : V → L(V,ℝ) |x ↦ s_ψ(x) w/ s_ψ(x) : V → ℝ | y ↦ s_ψ(x) = ψ(x,y)

d_ψ : V → L(V,ℝ) |y ↦ d_ψ(x) w/ d_ψ(y): V → ℝ | y ↦ d_ψ(y) = ψ(x,y)

it "readily follows" that the map f* defined as either

f* = (s_ψ)⁻¹(fᵀ)(s_ψ) : V → V


f* = (d_ψ)⁻¹(fᵀ)(d_ψ) : V → V

known as the Adjoint of f Relative to ψ is an endomorphism of V.

I have no idea in the world how the two definitions are equivalent (this was included in a book aimed at French high school students!!!), but assuming we've proven that det(f) = det(fᵀ) we can show that det(f) = det(f*) since det(f*) = det[(s_ψ)⁻¹(fᵀ)(s_ψ)] = det(fᵀ) = det(f).

If there is an alternative way to prove det(f) = det(f*) we could invert f* to show det(f) = det(fᵀ), so I offer this as an extra possible approach to this problem, though moreso I post it to see if anybody understands this & can justify why both definitions are equivalent because I certainly can't. However this approach seems to me to also be constrained by the fact that fᵀ doesn't seem to be defined for multilinear forms so I'm not sure if this is flawed or not.

Thanks for taking the time with this!

  • 4
    $\begingroup$ This is an amazing post, both by virtue of its sheer length and because you managed to typeset almost all of it legibly without using $\TeX$. Still I think it would be preferable to use the latter in the future. To do so, use dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. In case you don't know how to use $\TeX$, you can look at the source for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". $\endgroup$ – joriki Sep 2 '12 at 14:09
  • $\begingroup$ Sure thing, but apart from the subscript _ψ terms I love the way these symbols look & would have used tex if fractions or uglyness came into the picture. $\endgroup$ – Feng Shuei Sep 2 '12 at 14:33
  • $\begingroup$ @FengShuei, proofwiki.org/wiki/Determinant_of_Transpose $\endgroup$ – alancalvitti Sep 2 '12 at 14:57

Your Answer

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

Browse other questions tagged or ask your own question.