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Numerical value of Determinant far from what it is supposed to be
Find Determinant/or Row Reduce parameter dependent matrixEigenvalues and Determinant of a large matrixSolving determinant of a Kronecker product of matrices gives a numerical error - why?Analytic determinant of a sparse 25x25 matrix?Eigenvectors of numerical matrixmore numerically accurate inverse matrixSolving simultaneous and determinant given constant value and variable T (temperature)Determinant of Large Sparse MatrixObtaining the determinant from a LinearSolveFunction objectBad result from evaluating a determinant
$begingroup$
I have a large matrix with numerical components and want to set the determinant to zero using the parameter h (see below). Naively, I would have expected that h sets the determinant to (approximately) zero, which isn't the case. On top of that, the order of applying the rule sol seems to affects the final outcome for a reason to don't see.
My output of the code below is:
h -> -0.744736 + 4.42008 I
0.0445865 - 0.0285418 I
0.0545654 - 0.114258 I
I am not familiar with how Mathematica handles floating point numbers so that's probably where my error lies. I have also tried to increase the precision with SetPrecision, but without success.
mat=0.16 - (0.36 + 0.001 I) h - (1.35808 -
0.00120116 I) h^2 - (0.49603 - 0.00137214 I) h^3 - (0.11307 -
0.00105331 I) h^4 + (0.249794 - 0.000384238 I) h^5 -
0.39204 h^6, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h - (1.15528 +
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000353051 - 1.67323*10^-6 I) h^4,
0, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.6394 -
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (11.3534 -
0.00119507 I) h^2 - (0.484268 - 0.00140481 I) h^3 - (5.0714 -
0.00114074 I) h^4 + (0.27061 - 0.000416258 I) h^5 -
0.42471 h^6, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (4.95742 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000484431 - 2.29589*10^-6 I) h^4, (0.0000353051 -
1.67323*10^-6 I) h^4, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (41.4016 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (29.348 -
0.00118803 I) h^2 - (0.470698 - 0.00144251 I) h^3 - (13.9095 -
0.00124106 I) h^4 + (0.294629 - 0.000453204 I) h^5 -
0.462406 h^6, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.0123 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), 0, (0.0000484431 -
2.29589*10^-6 I) h^4, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (71.32 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (55.3462 -
0.00118568 I) h^2 - (0.466163 - 0.0014551 I) h^3 - (26.6556 -
0.00127449 I) h^4 + (0.302655 - 0.000465551 I) h^5 -
0.475003 h^6;
sol = Part[NSolve[Det[%] == 0, h], 1]
Det[mat /. sol]
Det[mat] /. sol
linear-algebra
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
$begingroup$
I have a large matrix with numerical components and want to set the determinant to zero using the parameter h (see below). Naively, I would have expected that h sets the determinant to (approximately) zero, which isn't the case. On top of that, the order of applying the rule sol seems to affects the final outcome for a reason to don't see.
My output of the code below is:
h -> -0.744736 + 4.42008 I
0.0445865 - 0.0285418 I
0.0545654 - 0.114258 I
I am not familiar with how Mathematica handles floating point numbers so that's probably where my error lies. I have also tried to increase the precision with SetPrecision, but without success.
mat=0.16 - (0.36 + 0.001 I) h - (1.35808 -
0.00120116 I) h^2 - (0.49603 - 0.00137214 I) h^3 - (0.11307 -
0.00105331 I) h^4 + (0.249794 - 0.000384238 I) h^5 -
0.39204 h^6, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h - (1.15528 +
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000353051 - 1.67323*10^-6 I) h^4,
0, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.6394 -
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (11.3534 -
0.00119507 I) h^2 - (0.484268 - 0.00140481 I) h^3 - (5.0714 -
0.00114074 I) h^4 + (0.27061 - 0.000416258 I) h^5 -
0.42471 h^6, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (4.95742 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000484431 - 2.29589*10^-6 I) h^4, (0.0000353051 -
1.67323*10^-6 I) h^4, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (41.4016 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (29.348 -
0.00118803 I) h^2 - (0.470698 - 0.00144251 I) h^3 - (13.9095 -
0.00124106 I) h^4 + (0.294629 - 0.000453204 I) h^5 -
0.462406 h^6, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.0123 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), 0, (0.0000484431 -
2.29589*10^-6 I) h^4, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (71.32 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (55.3462 -
0.00118568 I) h^2 - (0.466163 - 0.0014551 I) h^3 - (26.6556 -
0.00127449 I) h^4 + (0.302655 - 0.000465551 I) h^5 -
0.475003 h^6;
sol = Part[NSolve[Det[%] == 0, h], 1]
Det[mat /. sol]
Det[mat] /. sol
linear-algebra
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
Correction: I get the output0.118714 - 0.0526506 I(as the second output) and0.106201 - 0.0979004 I(as the third output); sorry, used a different matrix. But the problem still stands.
$endgroup$
– Nils
4 hours ago
add a comment |
$begingroup$
I have a large matrix with numerical components and want to set the determinant to zero using the parameter h (see below). Naively, I would have expected that h sets the determinant to (approximately) zero, which isn't the case. On top of that, the order of applying the rule sol seems to affects the final outcome for a reason to don't see.
My output of the code below is:
h -> -0.744736 + 4.42008 I
0.0445865 - 0.0285418 I
0.0545654 - 0.114258 I
I am not familiar with how Mathematica handles floating point numbers so that's probably where my error lies. I have also tried to increase the precision with SetPrecision, but without success.
mat=0.16 - (0.36 + 0.001 I) h - (1.35808 -
0.00120116 I) h^2 - (0.49603 - 0.00137214 I) h^3 - (0.11307 -
0.00105331 I) h^4 + (0.249794 - 0.000384238 I) h^5 -
0.39204 h^6, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h - (1.15528 +
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000353051 - 1.67323*10^-6 I) h^4,
0, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.6394 -
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (11.3534 -
0.00119507 I) h^2 - (0.484268 - 0.00140481 I) h^3 - (5.0714 -
0.00114074 I) h^4 + (0.27061 - 0.000416258 I) h^5 -
0.42471 h^6, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (4.95742 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000484431 - 2.29589*10^-6 I) h^4, (0.0000353051 -
1.67323*10^-6 I) h^4, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (41.4016 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (29.348 -
0.00118803 I) h^2 - (0.470698 - 0.00144251 I) h^3 - (13.9095 -
0.00124106 I) h^4 + (0.294629 - 0.000453204 I) h^5 -
0.462406 h^6, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.0123 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), 0, (0.0000484431 -
2.29589*10^-6 I) h^4, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (71.32 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (55.3462 -
0.00118568 I) h^2 - (0.466163 - 0.0014551 I) h^3 - (26.6556 -
0.00127449 I) h^4 + (0.302655 - 0.000465551 I) h^5 -
0.475003 h^6;
sol = Part[NSolve[Det[%] == 0, h], 1]
Det[mat /. sol]
Det[mat] /. sol
linear-algebra
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I have a large matrix with numerical components and want to set the determinant to zero using the parameter h (see below). Naively, I would have expected that h sets the determinant to (approximately) zero, which isn't the case. On top of that, the order of applying the rule sol seems to affects the final outcome for a reason to don't see.
My output of the code below is:
h -> -0.744736 + 4.42008 I
0.0445865 - 0.0285418 I
0.0545654 - 0.114258 I
I am not familiar with how Mathematica handles floating point numbers so that's probably where my error lies. I have also tried to increase the precision with SetPrecision, but without success.
mat=0.16 - (0.36 + 0.001 I) h - (1.35808 -
0.00120116 I) h^2 - (0.49603 - 0.00137214 I) h^3 - (0.11307 -
0.00105331 I) h^4 + (0.249794 - 0.000384238 I) h^5 -
0.39204 h^6, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h - (1.15528 +
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000353051 - 1.67323*10^-6 I) h^4,
0, -0.1711 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.6394 -
0.00267142 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (11.3534 -
0.00119507 I) h^2 - (0.484268 - 0.00140481 I) h^3 - (5.0714 -
0.00114074 I) h^4 + (0.27061 - 0.000416258 I) h^5 -
0.42471 h^6, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (4.95742 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), (0.0000484431 - 2.29589*10^-6 I) h^4, (0.0000353051 -
1.67323*10^-6 I) h^4, -0.223386 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (41.4016 -
0.00267502 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (29.348 -
0.00118803 I) h^2 - (0.470698 - 0.00144251 I) h^3 - (13.9095 -
0.00124106 I) h^4 + (0.294629 - 0.000453204 I) h^5 -
0.462406 h^6, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (19.0123 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 +
1. h^4), 0, (0.0000484431 -
2.29589*10^-6 I) h^4, -0.234771 h^2 ((-0.143205 +
0.000186623 I) - (0.36 + 0.001 I) h + (71.32 -
0.00267319 I) h^2 - (0.637164 - 0.0009801 I) h^3 + 1. h^4),
0.16 - (0.36 + 0.001 I) h - (55.3462 -
0.00118568 I) h^2 - (0.466163 - 0.0014551 I) h^3 - (26.6556 -
0.00127449 I) h^4 + (0.302655 - 0.000465551 I) h^5 -
0.475003 h^6;
sol = Part[NSolve[Det[%] == 0, h], 1]
Det[mat /. sol]
Det[mat] /. sol
linear-algebra
linear-algebra
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
edited 1 hour ago
J. M. is computer-less♦
97.3k10303463
97.3k10303463
asked 4 hours ago
NilsNils
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
NilsNils
111
asked 4 hours ago
NilsNils
111
111
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Nils is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
Correction: I get the output0.118714 - 0.0526506 I(as the second output) and0.106201 - 0.0979004 I(as the third output); sorry, used a different matrix. But the problem still stands.
$endgroup$
– Nils
4 hours ago
add a comment |
$begingroup$
Correction: I get the output0.118714 - 0.0526506 I(as the second output) and0.106201 - 0.0979004 I(as the third output); sorry, used a different matrix. But the problem still stands.
$endgroup$
– Nils
4 hours ago
$begingroup$
Correction: I get the output
0.118714 - 0.0526506 I (as the second output) and 0.106201 - 0.0979004 I (as the third output); sorry, used a different matrix. But the problem still stands.$endgroup$
– Nils
4 hours ago
$begingroup$
Correction: I get the output
0.118714 - 0.0526506 I (as the second output) and 0.106201 - 0.0979004 I (as the third output); sorry, used a different matrix. But the problem still stands.$endgroup$
– Nils
4 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
As you suspected when you mentioned SetPrecision, you are encountering numerical errors, probably catastrophic loss of precision when calculating the determinant; your calculations do in fact need to be carried out at higher precision.
If possible, you would want to use exact numbers in your matrix, or take advantage of the arbitrary-precision capabilities of Mathematica. For instance, we can convert all machine-precision numbers to arbitrary-precision ones with a number of digits of precision equal to that of common machine-precision numbers on your machine using SetPrecision (see also $MachinePrecision in the documentation):
det = Det[SetPrecision[mat, $MachinePrecision]];
sol = NSolve[det == 0, h];
det /. sol // PossibleZeroQ
(* Out:
True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True, True,
True, True
*)
As you can see, all those values of $h$ do bring your determinant reasonably close to zero, within machine-precision approximations.
answered 4 hours ago
MarcoBMarcoB
37.3k556113
$endgroup$
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
As you suspected when you mentioned SetPrecision, you are encountering numerical errors, probably catastrophic loss of precision when calculating the determinant; your calculations do in fact need to be carried out at higher precision.
If possible, you would want to use exact numbers in your matrix, or take advantage of the arbitrary-precision capabilities of Mathematica. For instance, we can convert all machine-precision numbers to arbitrary-precision ones with a number of digits of precision equal to that of common machine-precision numbers on your machine using SetPrecision (see also $MachinePrecision in the documentation):
det = Det[SetPrecision[mat, $MachinePrecision]];
sol = NSolve[det == 0, h];
det /. sol // PossibleZeroQ
(* Out:
True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True, True,
True, True
*)
As you can see, all those values of $h$ do bring your determinant reasonably close to zero, within machine-precision approximations.
answered 4 hours ago
MarcoBMarcoB
37.3k556113
$endgroup$
add a comment |
$begingroup$
As you suspected when you mentioned SetPrecision, you are encountering numerical errors, probably catastrophic loss of precision when calculating the determinant; your calculations do in fact need to be carried out at higher precision.
If possible, you would want to use exact numbers in your matrix, or take advantage of the arbitrary-precision capabilities of Mathematica. For instance, we can convert all machine-precision numbers to arbitrary-precision ones with a number of digits of precision equal to that of common machine-precision numbers on your machine using SetPrecision (see also $MachinePrecision in the documentation):
det = Det[SetPrecision[mat, $MachinePrecision]];
sol = NSolve[det == 0, h];
det /. sol // PossibleZeroQ
(* Out:
True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True, True,
True, True
*)
As you can see, all those values of $h$ do bring your determinant reasonably close to zero, within machine-precision approximations.
answered 4 hours ago
MarcoBMarcoB
37.3k556113
$endgroup$
add a comment |
$begingroup$
As you suspected when you mentioned SetPrecision, you are encountering numerical errors, probably catastrophic loss of precision when calculating the determinant; your calculations do in fact need to be carried out at higher precision.
If possible, you would want to use exact numbers in your matrix, or take advantage of the arbitrary-precision capabilities of Mathematica. For instance, we can convert all machine-precision numbers to arbitrary-precision ones with a number of digits of precision equal to that of common machine-precision numbers on your machine using SetPrecision (see also $MachinePrecision in the documentation):
det = Det[SetPrecision[mat, $MachinePrecision]];
sol = NSolve[det == 0, h];
det /. sol // PossibleZeroQ
(* Out:
True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True, True,
True, True
*)
As you can see, all those values of $h$ do bring your determinant reasonably close to zero, within machine-precision approximations.
answered 4 hours ago
MarcoBMarcoB
37.3k556113
$endgroup$
As you suspected when you mentioned SetPrecision, you are encountering numerical errors, probably catastrophic loss of precision when calculating the determinant; your calculations do in fact need to be carried out at higher precision.
If possible, you would want to use exact numbers in your matrix, or take advantage of the arbitrary-precision capabilities of Mathematica. For instance, we can convert all machine-precision numbers to arbitrary-precision ones with a number of digits of precision equal to that of common machine-precision numbers on your machine using SetPrecision (see also $MachinePrecision in the documentation):
det = Det[SetPrecision[mat, $MachinePrecision]];
sol = NSolve[det == 0, h];
det /. sol // PossibleZeroQ
(* Out:
True, True, True, True, True, True, True, True, True, True, True,
True, True, True, True, True, True, True, True, True, True, True,
True, True
*)
As you can see, all those values of $h$ do bring your determinant reasonably close to zero, within machine-precision approximations.
answered 4 hours ago
MarcoBMarcoB
37.3k556113
edited 4 hours ago
answered 4 hours ago
MarcoBMarcoB
37.3k556113
answered 4 hours ago
MarcoBMarcoB
37.3k556113
answered 4 hours ago
MarcoBMarcoB
37.3k556113
37.3k556113
add a comment |
add a comment |
Nils is a new contributor. Be nice, and check out our Code of Conduct.
Nils is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
Correction: I get the output
0.118714 - 0.0526506 I(as the second output) and0.106201 - 0.0979004 I(as the third output); sorry, used a different matrix. But the problem still stands.$endgroup$
– Nils
4 hours ago