Take both of the diagonals through an element in a matrix

Question

I have a matrix like this:

Reverse@Array[List, {8, 8}] // MatrixForm

I need a function that given an element of this matrix will return all the elements corresponding to the two diagonals that run through it.

The first diagonal can be found like this:

Diagonal[matrix, j-i]

where {i,j} is the element. How can I get the diagonal that runs orthogonally to this through the same given element {i,j}?

(It is somewhat related to this question but my problem is more that I can't figure out which diagonal I want once I've reversed the matrix.)

Answer 1

As you said, to get the main diagonal through $(i,j)$, you use

Diagonal[mat, j - i]

To get the other one, you can use

Diagonal[Reverse[mat, 2], (Dimensions[mat][[2]] - j + 1) - i]

Check correctness visually:

{i, j} = {3, 5}
MatrixForm[mat /. ((# -> Style[#, Red]) & /@ Diagonal[mat, j - i])]
MatrixForm[
 mat /. ((# -> Style[#, Red]) & /@ 
    Diagonal[Reverse[mat, 2], (Dimensions[mat][[2]] - j + 1) - i])]

Answer 2

The relation for the indices of matrix diagonals is easy to infer, and you can observe this from the following image:

If $A_{i_0,j_0}=A_{5,3}$ is the chosen element, then the diagonal elements have indices satisfying $i-j = i_0-j_0$ and the skew diagonal elements have indices satisfying $i+j=i_0+j_0$.

The problem is now almost solved! We just feed these exact rules to SparseArray:

Clear@orthoDiags
orthoDiags[{p_, q_}, n_] := 
    SparseArray[{
        {i_, j_} /; j + i == p + q :> 1, 
        {i_, j_} /; i - j == p - q :> 1}, 
        {n, n}
]

and use Pick to extract the elements:

Flatten@Pick[mat, orthoDiags[{5, 3}, 8], 1]

Answer 3

There are many solutions. I like one based on FrobeniusSolve (see e.g. this answer How to find lattice points on a line segment?):

d[i_Integer, j_Integer] /; i > 0 && j > 0 := 
  DeleteCases[ FrobeniusSolve[{1, 1}, i + j], {___, 0, ___}]

Then

matrix[[#1, #2]]& @@@ d[3, 4]

{{8, 6}, {7, 5}, {6, 4}, {5, 3}, {4, 2}, {3, 1}}

yields that second diagonal orthogonal to Diagonal[matrix, 4 - 3].
But as far as you work with matrices like matrix you needn't use neiter Apply nor Part, one has only to put adequate initial values (equations) to FrobeniusSolve.

As an alternative to d function one can use diag based on Array:

diag[i_, j_] := Array[{# + i, j - #} &, i + j - 1, -i + 1]

Answer 4

May late entry (called away). Short and sweet I think.

Edit: Ooops, did not read OP closely enough - this works for nXm shapes, not nXmXo... I'll leave answer for searchers with the former unless OP wants removal. See update below...

xDiag[mat_, {er_, ec_}] := 
 SparseArray[{r_, c_} :> mat[[r, c]] /; Abs[r - er] == Abs[c - ec], 
   Dimensions[mat]]["NonzeroValues"]

Row[{(test = Table[10 x + y, {x, 1, 5}, {y, 1, 5}]) // MatrixForm
  , xDiag[test, {2, 3}]},"  "]

After discovering the flaw in the above (shape limited to nXm, and fails with zero elements in diagonals) came up with one I like even better. It is shape agnostic, that is, it works with any array shape treating "diagonals" at the top level:

(* cross diagonals *)
Diag2[mat_, {er_, ec_}] := 
 Join @@ Cases[
   ReplacePart[mat, {r_, c_, ___} /; Abs[r - er] != Abs[c - ec] -> Sequence[]], {__}, {1}]

(* tests *)
Column[{Column[{(test1 = Table[10 x + y, {x, 1, 5}, {y, 1, 5}]) // 
     MatrixForm, xDiag2[test1, {3, 2}]}, Left, 1],
  Column[{(test2 = Array[List, {5, 5}]) // MatrixForm, 
     xDiag2[test2, {3, 2}]}, Left, 1]
   Column[{(test3 = Array[List, {5, 5, 2}]) // MatrixForm, 
     xDiag2[test3, {3, 2}]}, Left, 1]}, Left, 2]

(* results *)

Answer 5

As far as I can see, using Diagonal it should be something like this:

extract[m_, {i_, j_}] := 
 {Diagonal[m, j - i], Diagonal[Reverse[m], (i + j) - (Length[m] + 1)]}

And then:

m = Array[List, {8, 8}];
m // MatrixForm

you have:

extract[m,{6,2}]//Column
(* {{5,1},{6,2},{7,3},{8,4}}
   {{7,1},{6,2},{5,3},{4,4},{3,5},{2,6},{1,7}} *)