Skipping indices in a product
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_{lambda_ine lambda_j} frac{A - lambda_i I}{lambda_j-lambda_i}$, 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 linear-algebra operators
edited 2 days ago
Michael E2
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
$endgroup$
|
show 6 more comments
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_{lambda_ine lambda_j} frac{A - lambda_i I}{lambda_j-lambda_i}$, 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 linear-algebra operators
edited 2 days ago
Michael E2
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
$endgroup$
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:12
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34
|
show 6 more comments
$begingroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_{lambda_ine lambda_j} frac{A - lambda_i I}{lambda_j-lambda_i}$, 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 linear-algebra operators
edited 2 days ago
Michael E2
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
$endgroup$
I have a matrix $A$ for which I want to compute the quantity $Tlambda_j = Pi_{lambda_ine lambda_j} frac{A - lambda_i I}{lambda_j-lambda_i}$, 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 linear-algebra operators
matrix linear-algebra operators
edited 2 days ago
Michael E2
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
edited 2 days ago
Michael E2
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
edited 2 days ago
Michael E2
150k12203482
edited 2 days ago
Michael E2
150k12203482
edited 2 days ago
Michael E2
150k12203482
150k12203482
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
asked Mar 29 at 17:54
Tobias FritznTobias Fritzn
1895
1895
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:12
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34
|
show 6 more comments
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:12
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:12
$begingroup$
@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
Mar 29 at 18:12
2
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34
|
show 6 more comments
4 Answers
4
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 Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
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 Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
$endgroup$
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_}]]
answered Mar 29 at 19:04
BillBill
5,89569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
add a comment |
$begingroup$
Another way:
ClearAll[t];
t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
Fold[
#1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
IdentityMatrix[Length@A],
Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
];
Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.
answered 2 days ago
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
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 Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
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 Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
$endgroup$
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
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 Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
$endgroup$
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 Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
answered Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
answered Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
answered Mar 29 at 18:38
MarcoBMarcoB
38.4k556115
38.4k556115
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
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
Mar 29 at 18:46
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
$begingroup$
Thanks, @MarcoB. It leads precisely to the expected result. However, it looks too complicated. Nevertheless, it is fine as it works.
$endgroup$
– Tobias Fritzn
Mar 29 at 18:46
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 Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
$endgroup$
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 Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
$endgroup$
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 Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
$endgroup$
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 Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
answered Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
answered Mar 29 at 18:47
That Gravity GuyThat Gravity Guy
2,1411615
answered Mar 29 at 18:47
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_}]]
answered Mar 29 at 19:04
BillBill
5,89569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
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_}]]
answered Mar 29 at 19:04
BillBill
5,89569
$endgroup$
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
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_}]]
answered Mar 29 at 19:04
BillBill
5,89569
$endgroup$
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_}]]
answered Mar 29 at 19:04
BillBill
5,89569
edited Mar 29 at 19:28
answered Mar 29 at 19:04
BillBill
5,89569
answered Mar 29 at 19:04
BillBill
5,89569
answered Mar 29 at 19:04
BillBill
5,89569
5,89569
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
add a comment |
$begingroup$
Shoulde[[#[[2]]]]-e[[[[1]]]]bee[[#[[2]]]]-e[[#[[1]]]]?
$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
Should
e[[#[[2]]]]-e[[[[1]]]] be e[[#[[2]]]]-e[[#[[1]]]]?$endgroup$
– That Gravity Guy
Mar 29 at 19:12
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
$begingroup$
@ThatGravityGuy Yes! Good catch. Thank you! Corrected.
$endgroup$
– Bill
Mar 29 at 19:14
add a comment |
$begingroup$
Another way:
ClearAll[t];
t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
Fold[
#1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
IdentityMatrix[Length@A],
Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
];
Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.
answered 2 days ago
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
$begingroup$
Another way:
ClearAll[t];
t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
Fold[
#1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
IdentityMatrix[Length@A],
Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
];
Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.
answered 2 days ago
Michael E2Michael E2
150k12203482
$endgroup$
add a comment |
$begingroup$
Another way:
ClearAll[t];
t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
Fold[
#1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
IdentityMatrix[Length@A],
Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
];
Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.
answered 2 days ago
Michael E2Michael E2
150k12203482
$endgroup$
Another way:
ClearAll[t];
t[j_Integer, A_?SquareMatrixQ] := t[j, A, Eigenvalues@A]; (* add the eigenvalues *)
t[j_Integer, A_?SquareMatrixQ, evs_?VectorQ] /; Length@A == Length@evs := (* arg checks *)=
Fold[
#1.(A - #2 IdentityMatrix[Length@A])/(evs[[j]] - #2) &,
IdentityMatrix[Length@A],
Pick[evs, Unitize[evs - evs[[j]]], 1] (* Pick nonzero differences *)
];
Performance tuning: One can use DeleteCases[evs, e_ /; e == evs[[j]]] to pick the eigenvalues that give a nonzero difference. It makes no consistent difference to the timing on a 101 x 101 machine real matrix. One can save a little time by computing the identity matrix once and using With to inject it in the two places it occurs. One can also save time using dot = (dot = Dot; #2) & instead of Dot to skip the multiplication by the identity matrix (the first step of Fold). The differences evs - evs[[j]] appear twice, so they can be replaced by a single computation like the identity matrix. It can make up to a 10% improvement.
answered 2 days ago
Michael E2Michael E2
150k12203482
answered 2 days ago
Michael E2Michael E2
150k12203482
answered 2 days ago
Michael E2Michael E2
150k12203482
answered 2 days ago
Michael E2Michael E2
150k12203482
150k12203482
add a comment |
add a comment |
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$begingroup$
in which part exactly you want to exclude it in Tj !??
$endgroup$
– Alrubaie
Mar 29 at 18:02
$begingroup$
do you want it to be skipped put not Zero right !?
$endgroup$
– Alrubaie
Mar 29 at 18:07
$begingroup$
@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
Mar 29 at 18:08
$begingroup$
@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
Mar 29 at 18:12
2
$begingroup$
That product is presumably a matrix multiplication?
$endgroup$
– J. M. is slightly pensive♦
Mar 29 at 18:34