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Using good method to produce a regular matrix
How to turn a random graph into a matrixHow 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 elements'MatrixForm' problemA problem about matrix/vector extension via $tt ArrayPad$Integral approximation using a matrix operatorUsing ZeroTest method with RowReduceDAE Jacobian matrix StiffnessSwitching alternativeHow to write a matrix of $ntimes 2$ where the first column is made of only 1's?
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 17 hours ago
KarryMaKarryMa
112
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
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 17 hours ago
KarryMaKarryMa
112
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
17 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
16 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
16 hours ago
|
show 4 more comments
$begingroup$
The matrixform is as follow, and how can I use good method to produce it?
matrix
asked 17 hours ago
KarryMaKarryMa
112
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?
matrix
matrix
asked 17 hours ago
KarryMaKarryMa
112
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 17 hours ago
KarryMaKarryMa
112
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 17 hours ago
KarryMaKarryMa
112
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 17 hours ago
KarryMaKarryMa
112
asked 17 hours ago
KarryMaKarryMa
112
112
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
17 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
16 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
16 hours ago
|
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
17 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]
$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
OrNormal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].
$endgroup$
– Henrik Schumacher
16 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]
$endgroup$
– LouisB
16 hours ago
$begingroup$
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
17 hours ago
$begingroup$
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
17 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
16 hours ago
3
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
16 hours ago
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
16 hours ago
|
show 4 more comments
2 Answers
2
active
oldest
votes
$begingroup$
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 16 hours ago
J42161217J42161217
4,248324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
add a comment |
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
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
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 13 hours ago
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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
11 hours ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
10 hours ago
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
$begingroup$
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 16 hours ago
J42161217J42161217
4,248324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
add a comment |
$begingroup$
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 16 hours ago
J42161217J42161217
4,248324
$endgroup$
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
add a comment |
$begingroup$
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 16 hours ago
J42161217J42161217
4,248324
$endgroup$
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 16 hours ago
J42161217J42161217
4,248324
edited 11 hours ago
answered 16 hours ago
J42161217J42161217
4,248324
answered 16 hours ago
J42161217J42161217
4,248324
answered 16 hours ago
J42161217J42161217
4,248324
4,248324
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
add a comment |
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively toPadLeftyou can add3as 3rd argumentIntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].
$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
2
2
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
I like the answer,thanks!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].$endgroup$
– Coolwater
11 hours ago
$begingroup$
@KarryMa Alternatively to
PadLeft you can add 3 as 3rd argument IntegerDigits[{12, 24, 4, 8, 10, 20}, 3, 3].$endgroup$
– Coolwater
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
$begingroup$
yes, you are so right!
$endgroup$
– J42161217
11 hours ago
add a comment |
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
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
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 13 hours ago
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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
11 hours ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
10 hours ago
add a comment |
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
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
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 13 hours ago
MikeYMikeY
3,768916
$endgroup$
$begingroup$
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
11 hours ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
10 hours ago
add a comment |
$begingroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
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
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 13 hours ago
MikeYMikeY
3,768916
$endgroup$
A nice tool for this job is ArrayFlatten[ ]
a = {{1}, {2}}
$left(
begin{array}{c}
1 \
2 \
end{array}
right)$
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
$left(
begin{array}{ccc}
1 & 1 & 0 \
2 & 2 & 0 \
0 & 1 & 1 \
0 & 2 & 2 \
1 & 0 & 1 \
2 & 0 & 2 \
end{array}
right)$
answered 13 hours ago
MikeYMikeY
3,768916
edited 10 hours ago
answered 13 hours ago
MikeYMikeY
3,768916
answered 13 hours ago
MikeYMikeY
3,768916
answered 13 hours ago
MikeYMikeY
3,768916
3,768916
$begingroup$
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
11 hours ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
10 hours ago
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
11 hours ago
$begingroup$
My purpose was to introduceArrayFlatten. Not enough info to algorithmically determine the order of perms.
$endgroup$
– MikeY
10 hours ago
$begingroup$
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
11 hours ago
$begingroup$
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
11 hours ago
$begingroup$
My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
10 hours ago
$begingroup$
My purpose was to introduce
ArrayFlatten. Not enough info to algorithmically determine the order of perms.$endgroup$
– MikeY
10 hours ago
add a comment |
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.
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
17 hours ago
$begingroup$
KroneckerProduct[ MapThread[ ReplacePart[#1, #2 -> 0] &, {ConstantArray[1, {3, 3}], RotateRight[Range[3]]}], {{1}, {2}} ]$endgroup$
– Henrik Schumacher
16 hours ago
$begingroup$
Great,thank you very much!
$endgroup$
– KarryMa
16 hours ago
$begingroup$
Or
Normal@KroneckerProduct[ SparseArray[{Band[{1, 1}] -> 1, Band[{1, 2}] -> 1, Band[{3, 1}] -> 1}, {3, 3}], {{1}, {2}} ].$endgroup$
– Henrik Schumacher
16 hours ago
3
$begingroup$
Transpose@KroneckerProduct[Permutations[{1, 0, 1}], {1, 2}]$endgroup$
– LouisB
16 hours ago