Re: Why does Transpose[m, {1, 1}] give the diagonal of m?

*To*: mathgroup at christensen.cybernetics.net*Subject*: [mg1888] Re: [mg1859] Why does Transpose[m, {1, 1}] give the diagonal of m?*From*: Richard Mercer <richard at seuss.math.wright.edu>*Date*: Sat, 12 Aug 1995 22:51:22 -0400

David Cabana <drc at gate.net> in [mg1859] Why does Transpose[m, {1, 1}] give the diagonal of m? writes > In a recent article in the thread Re: [mg1741] RealDigits, Roman > Maeder explains one way to test whether a matrix is diagonal: >>even faster and certainly simpler: >>dQ[m_?MatrixQ] /; SameQ @@ Dimensions[m] := >> m == DiagonalMatrix[Transpose[m, {1, 1}]] >>(note: Transpose[m, {1, 1}] returns the list of diagonal elements.) > >An elegant test. My question: why does Transpose[m, {1, 1}] return >the list of diagonal elements? David, Below is a slightly edited version of my previous posting, [mg1131], in which I tried to understand Transpose. Allan Hayes De Montfort University Leicester hay at haystack.demon.co.uk ******** First, let's look at the way we extract the elements of a tensor T. Suppose that its tensor rank is 4 and that its dimensions are {d1,d2,d3,d4} then we need 4 positive integers a1,a2,a3,a4 to get an element : T[[a1,a2,a3,a4]] and, of course ,we must have ai <= di for all i. Transpose[ T, {2,1,3,2}]] is a tensor that given three positive integers a1,a2,a3 will return the value Transpose[ T, {2,1,3,2}] [[a1,a2,a3]] = T[[a2,a1,a3,a2]] We only need three numbers because the repeated 2 uses the second one twice. So the tensor rank is 3 and we must have a1 <= d2, a2 <= d1, a2 <= d4, a3 <= d3. In fact Transpose requires that d1 = d4. The dimensions of the transposed tensor are {d2,d1,d4} Similarly Transpose[ T, {2,1,1,2}] [[a1,a2]] = T[[a2,a1,a1,a2]] and the transposed tensor has rank 2 and dimensions {d2,d1} Check T = Array[SequenceForm,{2,4,4,2}]; Dimensions[T] {2, 4, 4, 2} Transpose[T,{2,1,3,2}][[1,2,3]] 2132 Dimensions[Transpose[T,{2,1,3,2}]] {4, 2, 4} Transpose[T,{2,1,1,2}][[1,2]] 2112 Dimensions[Transpose[T,{2,1,1,2}]] {4, 2} The GENERAL PATTERN (for elements) is (** Transpose[T_, p_List][[a__]] = T[[Sequence@@({a}[[p]])]] **) In the examples so far the length of p has been equal to the tensor rank T (this is needed t o provide the right number of inputs to T. But it would be inconvenient to always have to provide the full sequence so for example, for a tensor T of tensor rank 5 Mathematica interprets Transpose[ T, {2,1,1}] as Transpose[ T, {2,1,1,3,4}] (appending successive integers from the first positive integer not appearing in {2,1,1}). So much for the elements. But how do we construct the transposed tensor itself as an array? Well that's the job of the Array function. Here is my simulation of Transpose, Trans. (Notice that we don't need to allow for extending p -- this is automatically taken care of) Trans[T_, p_List] := Block[{dims, trdims}, dims = Dimensions[T]; trdims = dims[[Position[p,#,{1},1][[1,1]]]]&/@ Range[Length[Union[p]]]; Array[ T[[Sequence@@({##}[[p]])]]&, trdims ] ] The conditions imposed by Transpose are (where the rank of T is r and its dimensions are {d1,d2,...,dr }). 1. Union[ p] is Range[t] for some t <= r; 2. whenever pj = pk then dj = dk. (where p = {p1, p2, . . , pr}) The transposed tensor is of rank r minus the number of repeats in p and, as for dimensions, if s occurs in at position t in p then sth dimension is dt , otherwise the sth dimension can be worked out by extending p as discusssed above. Check Trans[T,{2,1,1,2}] === Transpose[T,{2,1,1,2}] Trans[T,{2,1,1}] === Transpose[T,{2,1,1}] Trans[T,{1,2,2}] === Transpose[T,{1,2,2}] True True True FINDING THE DIAGONAL For a square matrix M, Transpose[M, {1,1}] is the diagonal of M because. Transpose[M, {1,1}][[a]] = M[[a,a]] Thus Transpose[{{a,A},{b,B}}, {1,1}] {a, B} But for non-square matrices Transpose[{{a,A},{b,B},{c,C}}, {1,1}] Transpose::perm: {1, 1} is not a permutation. (*message*) Transpose[{{a, A}, {b, B}, {c, C}}, {1, 1}] The message really means that the condition 2. is not met. This can be allowed for by changing the definition of trdims in the code for Trans Trans2[T_, p_] := Block[{dims}, dims = Dimensions[T]; trdims = Min[dims[[Flatten[Position[p,#,{1}]]]]]&/@ Range[Length[Union[p]]]; Array[ T[[Sequence@@({##}[[p]])]]&, trdims ] ] Now we have Trans2[{{a,A},{b,B},{c,C}}, {1,1}] {a, B} Trans2[Transpose[{{a,A},{b,B},{c,C}}], {1,1}] {a, B} It seems resaonable to call this the diagonal of these matrices