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Skipping indices in a product
Skipping indices in a product
The Next CEO of Stack OverflowWhat's the best way to generate all the upper triangular matrix whose singular values are given?What is the fastest way to obtain the eigenvalues of a Wishart matrix?Evaluating the product of a matrix sequenceMapping over two indices with a conditionOuter product using the quantum mathematica packageWhy does Eigenvalues work for a matrix $M$ but not $M$?Conditions on a productHow to put conditions on indices of productProduct of matrices with symbolic boundsInverting a matrix when its elements are given by difficult expressions?
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]
2, -1, 1, 1
Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]
matrix operators products
edited 25 mins ago
That Gravity Guy
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
$endgroup$
|
show 6 more comments
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]
2, -1, 1, 1
Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]
matrix operators products
edited 25 mins ago
That Gravity Guy
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
$endgroup$
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
1 hour ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
49 mins ago
|
show 6 more comments
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]
2, -1, 1, 1
Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]
matrix operators products
edited 25 mins ago
That Gravity Guy
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
$endgroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_lambda_ine lambda_j fracA - lambda_i Ilambda_j-lambda_i$, where $lambda_i$ ($lambda_j$) denote the eigenvalues of $A$. How can this be implemented in Mathematica? Just gave a try here:
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
Eigenvalues[A]
2, -1, 1, 1
Tj = Product[(A - Eigenvalues[A][[i]] IdentityMatrix[4])/(
Eigenvalues[A][[j]] - Eigenvalues[A][[i]]), i, 1, 4]
matrix operators products
matrix operators products
edited 25 mins ago
That Gravity Guy
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
edited 25 mins ago
That Gravity Guy
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
edited 25 mins ago
That Gravity Guy
2,1311615
edited 25 mins ago
That Gravity Guy
2,1311615
edited 25 mins ago
That Gravity Guy
2,1311615
2,1311615
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
asked 1 hour ago
Tobias FritznTobias Fritzn
1745
1745
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
1 hour ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
49 mins ago
|
show 6 more comments
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
1 hour ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
49 mins ago
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
1 hour ago
$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
1 hour ago
$begingroup$
@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
$endgroup$
– Tobias Fritzn
1 hour ago
2
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
49 mins ago
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
49 mins ago
|
show 6 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Here is my pedestrian implementation of your formula:
a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]
t[a, 4]
You do not provide an example of desired output, so I will let you check whether this is what you expect.
answered 45 mins ago
MarcoBMarcoB
38.1k556114
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
add a comment |
$begingroup$
Something like this?
Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]
MatrixForm /@ Array[T[A, #] &, 4]
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
$endgroup$
add a comment |
$begingroup$
This
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
e=Eigenvalues[A];
Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
generates your twelve matricies with i not equal to j.
Put Dot@@ in front of that Map to form the dot product of the 12 matricies.
That works by forming every possible distinct i,j pair and then using those in the Map
answered 19 mins ago
BillBill
5,87569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Here is my pedestrian implementation of your formula:
a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]
t[a, 4]
You do not provide an example of desired output, so I will let you check whether this is what you expect.
answered 45 mins ago
MarcoBMarcoB
38.1k556114
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
add a comment |
$begingroup$
Here is my pedestrian implementation of your formula:
a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]
t[a, 4]
You do not provide an example of desired output, so I will let you check whether this is what you expect.
answered 45 mins ago
MarcoBMarcoB
38.1k556114
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
add a comment |
$begingroup$
Here is my pedestrian implementation of your formula:
a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]
t[a, 4]
You do not provide an example of desired output, so I will let you check whether this is what you expect.
answered 45 mins ago
MarcoBMarcoB
38.1k556114
$endgroup$
Here is my pedestrian implementation of your formula:
a = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
ClearAll[t]
t[amat_, j_] := Module[
evals, usable,
evals = Eigenvalues[amat];
usable = DeleteDuplicates@Cases[evals, Except@evals[[j]] ];
Dot @@
Table[
(amat - i IdentityMatrix[Length[amat]])/(evals[[j]] - i),
i, usable
]
]
t[a, 4]
You do not provide an example of desired output, so I will let you check whether this is what you expect.
answered 45 mins ago
MarcoBMarcoB
38.1k556114
answered 45 mins ago
MarcoBMarcoB
38.1k556114
answered 45 mins ago
MarcoBMarcoB
38.1k556114
answered 45 mins ago
MarcoBMarcoB
38.1k556114
38.1k556114
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
add a comment |
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
37 mins ago
add a comment |
$begingroup$
Something like this?
Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]
MatrixForm /@ Array[T[A, #] &, 4]
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
$endgroup$
add a comment |
$begingroup$
Something like this?
Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]
MatrixForm /@ Array[T[A, #] &, 4]
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
$endgroup$
add a comment |
$begingroup$
Something like this?
Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]
MatrixForm /@ Array[T[A, #] &, 4]
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
$endgroup$
Something like this?
Clear[A, evals, T]
A = 1, 0, 0, 1, 0, 1, 2, 0, 1, 1, 0, 2, 0, 0, 0, 1;
T[A_?MatrixQ, j_Integer] := With[
evals = Eigenvalues[A], id = IdentityMatrix@Length@A,
Dot @@ Table[
If[evals[[j]] - evals[[i]] == 0, id, (A - evals[[i]] id)/(evals[[j]] - evals[[i]])],
i, Length@A
]
]
MatrixForm /@ Array[T[A, #] &, 4]
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
answered 36 mins ago
That Gravity GuyThat Gravity Guy
2,1311615
2,1311615
add a comment |
add a comment |
$begingroup$
This
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
e=Eigenvalues[A];
Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
generates your twelve matricies with i not equal to j.
Put Dot@@ in front of that Map to form the dot product of the 12 matricies.
That works by forming every possible distinct i,j pair and then using those in the Map
answered 19 mins ago
BillBill
5,87569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
add a comment |
$begingroup$
This
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
e=Eigenvalues[A];
Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
generates your twelve matricies with i not equal to j.
Put Dot@@ in front of that Map to form the dot product of the 12 matricies.
That works by forming every possible distinct i,j pair and then using those in the Map
answered 19 mins ago
BillBill
5,87569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
add a comment |
$begingroup$
This
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
e=Eigenvalues[A];
Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
generates your twelve matricies with i not equal to j.
Put Dot@@ in front of that Map to form the dot product of the 12 matricies.
That works by forming every possible distinct i,j pair and then using those in the Map
answered 19 mins ago
BillBill
5,87569
$endgroup$
This
A = 1, 0, 0, 1,0, 1, 2, 0,1, 1, 0, 2,0, 0, 0, 1;
e=Eigenvalues[A];
Map[(A-e[[#[[1]]]]*IdentityMatrix[4])/(e[[#[[2]]]]-e[[#[[1]]]])&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
generates your twelve matricies with i not equal to j.
Put Dot@@ in front of that Map to form the dot product of the 12 matricies.
That works by forming every possible distinct i,j pair and then using those in the Map
answered 19 mins ago
BillBill
5,87569
edited 11 mins ago
answered 19 mins ago
BillBill
5,87569
answered 19 mins ago
BillBill
5,87569
answered 19 mins ago
BillBill
5,87569
5,87569
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
add a comment |
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
12 mins ago
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
12 mins ago
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@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
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– Bill
10 mins ago
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@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
10 mins ago
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$begingroup$
in which part exactly you want to exclude it in Tj !??
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– Alrubaie
1 hour ago
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do you want it to be skipped put not Zero right !?
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– Alrubaie
1 hour ago
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@Alrubaie, there was a typo in my post. Just edited it. I want the denominator to be non-zero and hence avoid the case for which $i=j$.
$endgroup$
– Tobias Fritzn
1 hour ago
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@Alrubaie, my $i$ and $j$ are not the indices in my question. They are the eigenvalues. I should have used something like $lambda_i$ and $lambda_j$.
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– Tobias Fritzn
1 hour ago
2
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That product is presumably a matrix multiplication?
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– J. M. is slightly pensive♦
49 mins ago