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

>  An elegant test.  My question: why does Transpose[m, {1,
>  1}] return the list of diagonal elements?  Looking under
>  Transpose in Wolfram's Mathematica Reference Guide (an
>  appendix to Mathematica, 2nd ed.), I find:

>  Transpose[list, {n1,n2, ...}] transposes list so that the kth level
>  in list is the nkth level in the result.

>  I expected Depth[Transpose[list, {n1,n2, ...}]] to be
>  the same as Depth[list].  I expected Transpose to swap
>  levels; I did not expect it to throw any away.  Evidently
>  I am laboring under some misconception.  Can someone
>  explain what is going on?

I think so, at the price of a certain amount of abstraction and background  

Consider the matrix 

A = {{16,27},{38,49}}, or
A = [ 16  27 ]
    [ 38  49 ]
C = Transpose[A,{1,1}] gives

*Short Version*
Since {1,1} is not a permutation, Transpose[list,{1,1}] is not defined by the  
Reference Guide. The effect of using {1,1} is to say the two entries must be  
the same, so C[[1]] = A[[1,1]] = 16, and C[[2]] = A[[2,2]] = 49. The effect is  
to compress a two-dimensional list into a one-dimensional list.

If you believe that, I have some Netscape shares I'd like to sell you!
So you'd better read the... 

*Long Version*
(0) The discussion below works just as well for non-square matrices and their  
higher-dimensional analogues, but it's simplest to consider a 2x2 matrix.

(1) The structured list objects on which Transpose operates can be viewed as  
functions on certain integer lattices. Sounds fancy, but an example shows it's  
really a simple idea.

The matrix A can be thought of as a function of two indices with the following  
A(1,1) = 16
A(1,2) = 27
A(2,1) = 38
A(2,2) = 49
(In Mathematica you use double brackets instead of parentheses.)
The domain of the function A is therefore the set of ordered pairs
D = {(1,1), (1,2), (2,1), (2,2)}.
(set notation, not Mathematica notation!)
D is the same as the set (cartesian) product {1,2} X {1,2}.

(2) The ordinary transpose operation can be viewed as the composition of the  
function A and the map t defined from D to D as follows:
t(1,1) = (1,1)
t(1,2) = (2,1)
t(2,1) = (1,2)
t(2,2) = (2,2)

So if B = Transpose[A], then
B(1,1) = A(t(1,1)) = A(1,1) = 16
B(1,2) = A(t(1,2)) = A(2,1) = 38
B(2,1) = B(t(2,1)) = A(1,2) = 27
B(2,2) = B(t(2,2)) = A(2,2) = 49
Therefore B = {{16,38},{27,49}} in Mathematica notation, or
B = [ 16  38 ]
    [ 27  49 ]
as expected.

(3) The key is to understand the map t and how it can be generalized to other  
situations. The map t is defined by a "permutation" or reordering applied to  
the elements of the domain D. In this case the reordering is described by {2,1}  

-- the first entry of the result is the second(2) entry of the input,
-- the second entry of the result is the first(1) entry of the input, i.e.  
t(a,b) = (b,a). Therefore
Transpose[A,{2,1}] means the same as Transpose[A], and
Transpose[A,{1,2}] = A,
because the permutation {1,2} means no reordering at all.

(4) If we now extend this discussion to the non-permutation {1,1}, it should  
define a map s by
-- the first entry of the result is the first(1) entry of the input,
-- the second entry of the result is the first(1) entry of the input, 

i.e. s(a) = (a,a).
Why not s(a,b) = (a,a)? Since only the first entry of the input is used, we  
assume the input has only one entry.
s maps from {1,2} into {1,2} X {1,2} = D.

(5) By analogy, C = Transpose[A,{1,1}] should now represent the composition of  
the function A with the map s. The result is a function on {1,2}, with values
C(1) = A(s(1)) = A(1,1) = 16
C(2) = A(s(2)) = A(2,2) = 49

This discussion could be continued to cases of higher dimensional lists, but  
I'm amazed if you've read this far!

Exercise for the reader:
If you have a three dimension list, e.g.
A = {{{16,27},{38,49}},{{91,82},{73,64}}}
what should be the result of Transpose[A,{1,1,2}] ?
Explain why Mathematica accepts the non-permutation {1,1,2} but not {1,1,3} in  
its place.

Richard Mercer

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