# Forming an upper triangular matrix from the elements of a list

I asked this question before, but i was closed because it the question was not comprehensible enough. So i reworded it, hope that's ok.

I have a list with 6 elements.

{a1, a2, a3, a4, a5, a6}


Now, I want to generate a upper triangle matrix with the elements above as matrix elements, so that the matrix reads

{{a1, a4, a6}, {0, a2, a5}, {0, 0, a3}}


That is, the triangle upper matrix matrix shall be filled up with the elements of the list. It is not important where each element is placed.

I want to do this also with larger matrices, so for example a list with 36 elements shall form a 8 x 8 upper triangle matrix.

Any ideas?

• Related: (55659) Commented Aug 2, 2014 at 14:12

For the filling pattern you showed:

x = {a1, a2, a3, a4, a5, a6};

n = 3;

x ~InternalPartitionRagged~ Range[n, 1, -1] ~Flatten~ {2} // PadLeft

{{a1, a4, a6}, {0, a2, a5}, {0, 0, a3}}


To find n given a complete input list x you can use:

n = Sqrt[1 + 8 Length@x]/2 - 1/2

• I get the same result without the Flatten which would make your +1 answer even shorter.
– eldo
Commented Aug 2, 2014 at 15:00
• @eldo Thanks for vote. However Flatten is needed for the filling order shown in the question. I'll leave the "any order" filling to others, unless I think of something particularly clean. Commented Aug 2, 2014 at 15:05
• @Mr.Wizard you many need to generalize n. +1 for the Partition trick. Commented Aug 2, 2014 at 15:09
• @Algohi Updated. Commented Aug 3, 2014 at 0:02

Since "It is not important where each element is placed":

a = {a1, a2, a3, a4, a5, a6}
mat = SparseArray[
Rule[#, #2] & @@@ Thread@{Flatten[Table[{i, j}, {i, 3}, {j, i, 3}], 1], a}];
mat // MatrixForm


$\left( \begin{array}{ccc} \text{a1} & \text{a2} & \text{a3} \\ 0 & \text{a4} & \text{a5} \\ 0 & 0 & \text{a6} \\ \end{array} \right)$

If you have a list of 36 elements that you want to turn into a 8x8 upper triangle matrix:

l = Range@36;
mat = SparseArray[
Rule[#, #2] & @@@ Thread@{Flatten[Table[{i, j}, {i, 8}, {j, i, 8}], 1], l}];
mat // MatrixForm


$\left( \begin{array}{cccccccc} 1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ 0 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ 0 & 0 & 16 & 17 & 18 & 19 & 20 & 21 \\ 0 & 0 & 0 & 22 & 23 & 24 & 25 & 26 \\ 0 & 0 & 0 & 0 & 27 & 28 & 29 & 30 \\ 0 & 0 & 0 & 0 & 0 & 31 & 32 & 33 \\ 0 & 0 & 0 & 0 & 0 & 0 & 34 & 35 \\ 0 & 0 & 0 & 0 & 0 & 0 & 0 & 36 \\ \end{array} \right)$

Here is an attempt to make it more robust:

mat[l_] := Module[{n = Abs[1/2 (1 - Sqrt[1 + 8*Length@l])]},
If[IntegerQ@n,
SparseArray[Rule[#, #2] & @@@ Thread@{Flatten[Table[{i, j}, {i, n}, {j, i, n}], 1], l}],
"Wrong length. The length should be " <>
ToString[Or @@ ((#[n]*(#[n] + 1)/2) & /@ {Ceiling, Floor})] <> "."]]


Usage:

l = Range@11;
mat@l


Wrong length. The length should be 15 || 10.

l = Range@10;
mat@l


$\left( \begin{array}{cccc} 1 & 2 & 3 & 4 \\ 0 & 5 & 6 & 7 \\ 0 & 0 & 8 & 9 \\ 0 & 0 & 0 & 10 \\ \end{array} \right)$

• Your answer only functions when there are exactly 6 elements. But the OP wants a more general solution: "... so for example a list with 36 elements shall form a 8x8 upper triangle matrix."
– eldo
Commented Aug 2, 2014 at 13:39
• @eldo I assumed that you were capable of replacing 3 by 8 ;o) But there you go.
– Öskå
Commented Aug 2, 2014 at 13:46
• Now I can see +1
– eldo
Commented Aug 2, 2014 at 13:56

You should try to avoid procedural language in mathematica. I think this should work, but maybe it is not the most efficient:

Code edited for clarity

utMatrix[list_List] := With[{matrixsize = -1/2 + 1/2 Sqrt[1 + 8*Length@list]},
PadLeft[Take[list, #], matrixsize, 0] & /@ (Transpose@{Most@#, Rest@# - 1} &
[Accumulate@Join[{1}, Range[matrixsize, 1, -1]]])
/;IntegerQ[matrixsize]]


Basically this function partitions the list in sublists of appropriate lengths and then pads them with 0s on the left. The condition at the end checks for lists of wrong lengths.

Maybe it's not the most efficient solution, but it's quite simple. I had to fill in the vectors in the specific order.

First, I define a vector, for example:

MatrixDim = 5;
VecLength = (MatrixDim + 1) MatrixDim/2;
InputVector = Table[i, {i, VecLength}];


(1,2,3,4,5,6,7,8,9,10,11,12,13,14,15)

And declare a matrix:

MatrixLUp = Table[0 i + 0 j, {i, 1, MatrixDim}, {j, 1, MatrixDim}];


The actual formula that feeds in the matrix is like that:

OldX = 0;
j = 1;
Do[
i = x - OldX;
MatrixLUp[[i, j]] = InputVector[[x]];
If[x == (j + 1) j/2, OldX = x; j++;]
, {x, 1, VecLength}
];


Finally, the elements in the matrix are shown here:

MatrixLUp//MatrixForm//TeXForm


$$\begin{array}{ccccc} 1 & 2 & 4 & 7 & 11 \\ 0 & 3 & 5 & 8 & 12 \\ 0 & 0 & 6 & 9 & 13 \\ 0 & 0 & 0 & 10 & 14 \\ 0 & 0 & 0 & 0 & 15 \\ \end{array}$$

toUpperMatrix[l_List] := Module[{i, n, m, k, t},
n = Length[l];
m = {};
k = 1/2 (-1 + Sqrt[1 + 8 n]);
t = {};
For[i = 1, i <= n, i++,
AppendTo[t, l[[i]]];
If[Length[t] == k,
AppendTo[m, Join[Table[0, {i, 1, 1/2 (-1 + Sqrt[1 + 8 n]) - k}], t]];
t = {};
k--;
];
];
Return[m];
];


This is the only method I can figure out. Looking forward to better solutions.

If you don't care the order you can use the following function:

tpart[list_ /; VectorQ[list] && OddQ[Sqrt[1 + 8 Length[list]]]] :=
Module[{m = (-1 + Sqrt[1 + 8 Length[list]])/2},
SparseArray[
Thread[Flatten[Table[{i, j}, {i, 1, m}, {j, i, m}], 1] ->
list], {m, m}]
]


For example

tpart[{a1, a2, a3, a4, a5, a6}] // MatrixForm
tpart[Array[a, 36]] // MatrixForm


gives:

If you do care about your order this one (this function is not robust: be sure to pass an appropriately sized list or add a condition):

tpart2[list_] := Module[{m = (-1 + Sqrt[1 + 8 Length[list]])/2, s},
s = Range[m, 1, -1] //Prepend[1 + Accumulate[#], 1] & //Partition[#, 2, 1] &;
SparseArray[
Thread[Array[Band[{1, #}] &, m] -> (Take[list, # - {0, 1}] & /@ s)]
, {m, m}]
]

tpart2[list]
tpart2[list] // MatrixForm
tpart2[list] // Normal


The result is:

This is if you car about order.

list = {a1, a2, a3, a4, a5, a6, a7, a8, a9, a10};

s = 1/2 (1 + Sqrt[1 + 8 Length@list]);
l = list[[1 + # s - (# (1 + #))/2 ;; (1 + #) s - ((# + 1) (2 + #))/
2]] & /@ Range[0, s - 2];
SparseArray[Band[{1, #}] -> l[[#]] & /@ Range[Length@l]] //
Normal

(* {{a1, a5, a8, a10}, {0, a2, a6, a9}, {0, 0, a3, a7}, {0, 0, 0, a4}}*)


Inspired by Mr. Wizard's solution this is a certain generalization:

rag[v_, w_] :=
MapThread[v[[#1 ;; #2]] &, With[{a = Accumulate[w]}, {a - w + 1, a}]]

tri[n_Integer /; n > 0, order_: True, s_Symbol: a] :=
Module[{x, y},
x = Range[n, 1, -1];
y = rag[ToExpression[ToString@s <> # & /@ ToString /@ Range@Total@x], x];
]

tri[3] // MatrixForm


tri[8, False] // MatrixForm


• Nice extension. Commented Aug 2, 2014 at 18:35
  n = 6
vals = Array[a, {n (n + 1)/2}]
SparseArray[
SortBy[Tuples[Range[n], {2}] ,
#[[2]] - #[[1]] &][[-n (n + 1)/2 ;;]] -> vals ] // MatrixForm


without using SparseArray:

 ReplacePart[ConstantArray[0, {n, n}],
SortBy[Tuples[Range[n], {2}] ,
#[[2]] - #[[1]] &][[-n (n + 1)/2 ;;]], vals}]] // MatrixForm


If the length of your vector can always be guaranteed to be correct (exactly good for an upper triangular matrix), then the following will work.

v = {a1, a2, a3, a4, a5, a6};
d = (-1 + Sqrt[1 + 8*Length[v]])/2;
Array[If[#1 <= #2, v[[(2*d - #1 + 2)*(#1 - 1)/2 + #2 - #1 + 1]], 0] &, {d,d}]
`

Please test it yourself and write it into a function if you need to repeatedly use it.

Regards,

Kuo Kan LIANG

• @Shutao, doesn't it depend on the dialect? Commented Apr 22, 2016 at 4:17
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– 梁國淦
Commented Apr 26, 2016 at 0:59
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– 梁國淦
Commented Apr 26, 2016 at 1:06
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• @xzczd 谢谢你的姓名解释！
– xyz
Commented May 9, 2016 at 0:49