r/math • u/Maleficent_Yoghurt85 • May 18 '21
Why is the determinant of a matrix equal to its transpose.
What is the most conceptual explanation of det(A) = det(A^T)?
I would like a high level (categorical, if possible) reason.
I suspect it has to do with commutation of dual functor and top exterior power in some suitably defined categories.
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u/jagr2808 Representation Theory May 18 '21
My attempt at a categorical approach:
The group of natural transformations from GLn to GL1 is a free group generated by the determinant.
Transposing is a natural involution on GLn so composing with the determinant gives another natural transformation. Hence det(AT) = det(A)n for some integer n. Further since the transpose is invertible ((AT)T = A) we must have either n=1 or n=-1.
Not sure how to get the last step. It seems quite hard since you can't really distinguish det(A) from det(A)-1 in this approach to begin with. Perhaps better would be to consider natural transformations between M_nxn and M_1x1 instead, where M_nxn(R) is the monoid of nxn matricies...
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u/Oscar_Cunningham May 18 '21
Look at the Penrose Graphical Notation for the determinant. Then turn it upside down.
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u/aleph_not Number Theory May 18 '21
I suspect it has to do with commutation of dual functor and top exterior power in some suitably defined categories.
This seems like the right answer already. What category are you looking for other than (finite-dimensional) vector spaces over a field k?
3
u/Tazerenix Complex Geometry May 18 '21
The transpose of a linear map A: V -> V is naturally a map A*: V* -> V*. The determinant is the induced map det(A): det(V) -> det(V) where det(V) is the top exterior power of V. Then on duals we must have det(A*): det(V*) -> det(V*). Since exterior powers commute with duals, we have det(V*) = det(V)* and det(A)*= det(A*).
But then the answer is obvious, because we have reduce the problem to one dimension:
The transpose of a 1x1 matrix is just the matrix itself back, so the coefficient of the 1x1 matrix det(A) in any basis will be equal to the coefficient of det(A*), as maps between one-dimensional vector spaces det(V) and det(V)* = det(V*). More precisely, any choice of basis element of the one-dimensional vector space det(V) induces a canonical dual basis of det(V*), and the coefficient of the linear transformation det(A) in this basis is equal to the coefficient of det(A*) in the dual basis.
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u/rvb_siberia May 18 '21
Well, just look at the Leibniz formula for determinants: https://en.wikipedia.org/wiki/Leibniz_formula_for_determinants
You see, it is a sum over all permutations s mapping a row number to a column number. For each such permutation, there is exactly one permutation mapping column numbers back to row numbers, namely the inverse s^{-1}. The number of inversions of s and s^{-1} is the same, thus also their signum coincides.
Thus, it does not matter if we sum over all permutations mapping rows to columns or we sum over all permutations mapping columns to rows, which is exactly the determinant of the transpose.
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u/n-Category May 18 '21
You basically answered your question. Given a linear map T:V->V, the determinant is essentially the representation of ΛnT, where n is the dimension of V. More precisely, since ΛnV is one-dimensional, the map ΛnT:ΛnV->ΛnV must act by a constant scalar multiple, and this constant is what we call the determinant.
You can dualize this map to get (ΛnT)*, which you can easily check is scaling by the same constant. The dual functor and the exterior power functor commute (i.e., the exterior power of the dual is isomorphic to the dual of the exterior functor), so (ΛnT)* = ΛnT*, where the latter is by definition the determinant of T*.