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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
$endgroup$
|
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
$endgroup$
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours 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
5 hours 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
5 hours ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours 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
$endgroup$
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 4 hours ago
That Gravity Guy
2,1411615
2,1411615
asked 5 hours ago
Tobias FritznTobias Fritzn
1795
1795
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours 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
5 hours 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
5 hours ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours ago
|
show 6 more comments
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours 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
5 hours 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
5 hours ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours ago
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours 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
5 hours 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
5 hours 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
5 hours 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
5 hours ago
2
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours ago
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours 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.
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
4 hours 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]

$endgroup$
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
If it might be easier to read you can also write it this way
Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
(A-ei*IdentityMatrix[4])/(ej-ei))&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours 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.
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
4 hours 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.
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
4 hours 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.
$endgroup$
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 4 hours 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
4 hours 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
4 hours 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
4 hours 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
4 hours 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]

$endgroup$
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]

$endgroup$
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]

$endgroup$
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 4 hours ago
That Gravity GuyThat Gravity Guy
2,1411615
2,1411615
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
If it might be easier to read you can also write it this way
Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
(A-ei*IdentityMatrix[4])/(ej-ei))&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours 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
If it might be easier to read you can also write it this way
Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
(A-ei*IdentityMatrix[4])/(ej-ei))&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours 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
If it might be easier to read you can also write it this way
Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
(A-ei*IdentityMatrix[4])/(ej-ei))&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
$endgroup$
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
If it might be easier to read you can also write it this way
Map[(ei=e[[#[[1]]]];ej=e[[#[[2]]]];
(A-ei*IdentityMatrix[4])/(ej-ei))&,
DeleteCases[Tuples[Range[4],2],i_,i_]]
edited 4 hours ago
answered 4 hours ago
BillBill
5,88569
5,88569
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours ago
add a comment |
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours ago
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours ago
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
4 hours ago
add a comment |
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$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
5 hours ago
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
5 hours 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
5 hours 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
5 hours ago
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
4 hours ago