Using good method to produce a regular matrix The 2019 Stack Overflow Developer Survey Results Are InHow can I fit a matrix function with multiple variables to given eigenvalues?What is the best way to produce a symmetric semi-definite matrix using as few variables as possible?Produce a 1-dimensional list of all unique matrix elementsEfficient method for Inserting arrays into arraysCreate a matrix of matrices using Band and ArrayFlatten'MatrixForm' problemElegant method of generating this matrix?A problem about matrix/vector extension via $tt ArrayPad$replacement in a nested matrixBuild matrix from smaller matrices or rearange list of lists
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Using good method to produce a regular matrix
The 2019 Stack Overflow Developer Survey Results Are InHow can I fit a matrix function with multiple variables to given eigenvalues?What is the best way to produce a symmetric semi-definite matrix using as few variables as possible?Produce a 1-dimensional list of all unique matrix elementsEfficient method for Inserting arrays into arraysCreate a matrix of matrices using Band and ArrayFlatten'MatrixForm' problemElegant method of generating this matrix?A problem about matrix/vector extension via $tt ArrayPad$replacement in a nested matrixBuild matrix from smaller matrices or rearange list of lists
2
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
H = 1, 1, 0, 2, 2, 0, 0, 1, 1,
0, 2, 2, 1, 0, 1, 2, 0, 2
list-manipulation matrix array
edited Apr 7 at 13:06
MarcoB
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
|
show 4 more comments
2
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
H = 1, 1, 0, 2, 2, 0, 0, 1, 1,
0, 2, 2, 1, 0, 1, 2, 0, 2
list-manipulation matrix array
edited Apr 7 at 13:06
MarcoB
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]
$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].
$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
3
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]
$endgroup$
– LouisB
Apr 6 at 9:45
|
show 4 more comments
2
2
2
0
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
H = 1, 1, 0, 2, 2, 0, 0, 1, 1,
0, 2, 2, 1, 0, 1, 2, 0, 2
list-manipulation matrix array
edited Apr 7 at 13:06
MarcoB
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The matrixform is as follow, and how can I use good method to produce it?
H = 1, 1, 0, 2, 2, 0, 0, 1, 1,
0, 2, 2, 1, 0, 1, 2, 0, 2
list-manipulation matrix array
list-manipulation matrix array
edited Apr 7 at 13:06
MarcoB
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 7 at 13:06
MarcoB
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Apr 7 at 13:06
MarcoB
38.6k557115
edited Apr 7 at 13:06
MarcoB
38.6k557115
edited Apr 7 at 13:06
MarcoB
38.6k557115
38.6k557115
asked Apr 6 at 9:18
KarryMaKarryMa
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Apr 6 at 9:18
KarryMaKarryMa
163
asked Apr 6 at 9:18
KarryMaKarryMa
163
163
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
KarryMa is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]
$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].
$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
3
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]
$endgroup$
– LouisB
Apr 6 at 9:45
|
show 4 more comments
$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]
$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].
$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
3
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]
$endgroup$
– LouisB
Apr 6 at 9:45
$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
3
3
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]$endgroup$
– LouisB
Apr 6 at 9:45
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]$endgroup$
– LouisB
Apr 6 at 9:45
|
show 4 more comments
2 Answers
2
active
oldest
votes
2
$begingroup$
IntegerDigits[12,24,4,8,10,20,3,3]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
also..
s = Transpose[Permutations /@ 1, 1, 0, 2, 2, 0];
Flatten[s[[1]],Reverse@Rest@s,2]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
answered Apr 6 at 9:29
J42161217J42161217
4,273324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
add a comment |
1
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = 1, 2
$left(
beginarrayc
1 \
2 \
endarray
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
a,a,0,0,a,a,a,0,a// ArrayFlatten
$left(
beginarrayccc
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
endarray
right)$
answered Apr 6 at 13:21
MikeYMikeY
3,828916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27
$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
$begingroup$
IntegerDigits[12,24,4,8,10,20,3,3]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
also..
s = Transpose[Permutations /@ 1, 1, 0, 2, 2, 0];
Flatten[s[[1]],Reverse@Rest@s,2]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
answered Apr 6 at 9:29
J42161217J42161217
4,273324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
add a comment |
2
$begingroup$
IntegerDigits[12,24,4,8,10,20,3,3]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
also..
s = Transpose[Permutations /@ 1, 1, 0, 2, 2, 0];
Flatten[s[[1]],Reverse@Rest@s,2]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
answered Apr 6 at 9:29
J42161217J42161217
4,273324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
add a comment |
2
2
2
$begingroup$
IntegerDigits[12,24,4,8,10,20,3,3]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
also..
s = Transpose[Permutations /@ 1, 1, 0, 2, 2, 0];
Flatten[s[[1]],Reverse@Rest@s,2]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
answered Apr 6 at 9:29
J42161217J42161217
4,273324
$endgroup$
IntegerDigits[12,24,4,8,10,20,3,3]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
also..
s = Transpose[Permutations /@ 1, 1, 0, 2, 2, 0];
Flatten[s[[1]],Reverse@Rest@s,2]
1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
answered Apr 6 at 9:29
J42161217J42161217
4,273324
edited Apr 6 at 14:52
answered Apr 6 at 9:29
J42161217J42161217
4,273324
answered Apr 6 at 9:29
J42161217J42161217
4,273324
answered Apr 6 at 9:29
J42161217J42161217
4,273324
4,273324
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
add a comment |
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].
$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
2
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
Apr 6 at 9:55
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[12, 24, 4, 8, 10, 20, 3, 3].$endgroup$
– Coolwater
Apr 6 at 14:49
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
Apr 6 at 14:51
add a comment |
1
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = 1, 2
$left(
beginarrayc
1 \
2 \
endarray
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
a,a,0,0,a,a,a,0,a// ArrayFlatten
$left(
beginarrayccc
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
endarray
right)$
answered Apr 6 at 13:21
MikeYMikeY
3,828916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27
$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
add a comment |
1
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = 1, 2
$left(
beginarrayc
1 \
2 \
endarray
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
a,a,0,0,a,a,a,0,a// ArrayFlatten
$left(
beginarrayccc
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
endarray
right)$
answered Apr 6 at 13:21
MikeYMikeY
3,828916
$endgroup$
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27
$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
add a comment |
1
1
1
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = 1, 2
$left(
beginarrayc
1 \
2 \
endarray
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
a,a,0,0,a,a,a,0,a// ArrayFlatten
$left(
beginarrayccc
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
endarray
right)$
answered Apr 6 at 13:21
MikeYMikeY
3,828916
$endgroup$
A nice tool for this job is ArrayFlatten[ ]
a = 1, 2
$left(
beginarrayc
1 \
2 \
endarray
right)$
Not sure why your rows are ordered the way they are. Are you trying to have a non-zero diagonal?
a,a,0,0,a,a,a,0,a// ArrayFlatten
$left(
beginarrayccc
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
endarray
right)$
answered Apr 6 at 13:21
MikeYMikeY
3,828916
edited Apr 6 at 15:25
answered Apr 6 at 13:21
MikeYMikeY
3,828916
answered Apr 6 at 13:21
MikeYMikeY
3,828916
answered Apr 6 at 13:21
MikeYMikeY
3,828916
3,828916
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27
$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
add a comment |
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
Apr 6 at 15:27
$begingroup$
Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
$begingroup$
This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
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This is not the exact result the OP is asking. The order of the elements is different. Also the "permutation" solution is already posted in my answer
$endgroup$
– J42161217
Apr 6 at 14:58
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My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
Apr 6 at 15:27
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My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
Apr 6 at 15:27
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Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
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Actually,this is the answer I originally wanted,then I will use ''PauliMatrix'' and ''Map'' to get what I want.Thank you for your help!
$endgroup$
– KarryMa
Apr 7 at 2:35
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KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
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KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
KarryMa is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
If the picture can't show,then here is the matrixform:H=1,1,0,2,2,0,0,1,1,0,2,2,1,0,1,2,0,2
$endgroup$
– KarryMa
Apr 6 at 9:22
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KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, ConstantArray[1, 3, 3], RotateRight[Range[3]]], 1, 2 ]$endgroup$
– Henrik Schumacher
Apr 6 at 9:25
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Great,thank you very much!
$endgroup$
– KarryMa
Apr 6 at 9:26
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[Band[1, 1] -> 1, Band[1, 2] -> 1, Band[3, 1] -> 1, 3, 3], 1, 2 ].$endgroup$
– Henrik Schumacher
Apr 6 at 9:27
3
$begingroup$
Transpose@KroneckerProduct[Permutations[1, 0, 1], 1, 2]$endgroup$
– LouisB
Apr 6 at 9:45