Problem with EigenvectorsOptimization problem with matrix positivity constraintsEigenvector AnomalyEasier way to calculate Taylor remainder in 2nd order seriesWhat does it mean when Mathematica returns a zero “eigenvector”?DEigenvalues with Robin B.C. sign problemHow to use DEigensystem with periodic boundary conditions on the derivative?Where do two bicubic-interpolated surfaces intersect a plane?Solving 4th order polynomialWhy SVD cannot recover the original matrix and suffer from numerical instability?Eigenvectors of Hermitian matrices
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Problem with Eigenvectors
Optimization problem with matrix positivity constraintsEigenvector AnomalyEasier way to calculate Taylor remainder in 2nd order seriesWhat does it mean when Mathematica returns a zero “eigenvector”?DEigenvalues with Robin B.C. sign problemHow to use DEigensystem with periodic boundary conditions on the derivative?Where do two bicubic-interpolated surfaces intersect a plane?Solving 4th order polynomialWhy SVD cannot recover the original matrix and suffer from numerical instability?Eigenvectors of Hermitian matrices
.everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty margin-bottom:0;
$begingroup$
When I want to calculate eigenvectors of the following matrix in Mathematica the only answer it gives me is zero vector, anybody knows how to fix this?
here's my matrix :
beginequation
X=left(beginarraycccc
0 & 1 & 0 & 0&0&0\
1 & 0 & sqrt2 & 0&0&0\
0 & sqrt2 & 0 & sqrt3&0&0\
0 & 0 & sqrt3 & 0& sqrt4 &0\0&0&0&sqrt4 &0&sqrt5 \ 0&0&0&0&sqrt5 &0
endarrayright)
endequation
heres the code:
X = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[X]
The point is I don't want to find them numerically I want analytical expression for eigenvectors.
matrix eigenvalues
edited Jul 30 at 11:56
user64494
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
$endgroup$
add a comment |
$begingroup$
When I want to calculate eigenvectors of the following matrix in Mathematica the only answer it gives me is zero vector, anybody knows how to fix this?
here's my matrix :
beginequation
X=left(beginarraycccc
0 & 1 & 0 & 0&0&0\
1 & 0 & sqrt2 & 0&0&0\
0 & sqrt2 & 0 & sqrt3&0&0\
0 & 0 & sqrt3 & 0& sqrt4 &0\0&0&0&sqrt4 &0&sqrt5 \ 0&0&0&0&sqrt5 &0
endarrayright)
endequation
heres the code:
X = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[X]
The point is I don't want to find them numerically I want analytical expression for eigenvectors.
matrix eigenvalues
edited Jul 30 at 11:56
user64494
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
$endgroup$
add a comment |
$begingroup$
When I want to calculate eigenvectors of the following matrix in Mathematica the only answer it gives me is zero vector, anybody knows how to fix this?
here's my matrix :
beginequation
X=left(beginarraycccc
0 & 1 & 0 & 0&0&0\
1 & 0 & sqrt2 & 0&0&0\
0 & sqrt2 & 0 & sqrt3&0&0\
0 & 0 & sqrt3 & 0& sqrt4 &0\0&0&0&sqrt4 &0&sqrt5 \ 0&0&0&0&sqrt5 &0
endarrayright)
endequation
heres the code:
X = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[X]
The point is I don't want to find them numerically I want analytical expression for eigenvectors.
matrix eigenvalues
edited Jul 30 at 11:56
user64494
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
$endgroup$
When I want to calculate eigenvectors of the following matrix in Mathematica the only answer it gives me is zero vector, anybody knows how to fix this?
here's my matrix :
beginequation
X=left(beginarraycccc
0 & 1 & 0 & 0&0&0\
1 & 0 & sqrt2 & 0&0&0\
0 & sqrt2 & 0 & sqrt3&0&0\
0 & 0 & sqrt3 & 0& sqrt4 &0\0&0&0&sqrt4 &0&sqrt5 \ 0&0&0&0&sqrt5 &0
endarrayright)
endequation
heres the code:
X = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[X]
The point is I don't want to find them numerically I want analytical expression for eigenvectors.
matrix eigenvalues
matrix eigenvalues
edited Jul 30 at 11:56
user64494
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
edited Jul 30 at 11:56
user64494
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
edited Jul 30 at 11:56
user64494
1
edited Jul 30 at 11:56
user64494
1
edited Jul 30 at 11:56
user64494
1
1
asked Jul 30 at 8:12
JasonJason
375 bronze badges
asked Jul 30 at 8:12
JasonJason
375 bronze badges
asked Jul 30 at 8:12
JasonJason
375 bronze badges
375 bronze badges
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
the only answer it gives me is zero vector
It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.
On V12:
mat = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[mat];
MatrixForm[% // N]
Update
OP wants solution to be analytical solution and not numerical.
Mathematica gives answer using Roots objects.
Using
SetSystemOptions[
"TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]
Gives
TO obtain numerical values, the command N can be applied to the above.
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
$endgroup$
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expressionFirst, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove//N. Answer will be in terms ofRootobjects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..
$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
add a comment |
$begingroup$
Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!
When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.
matA = 0, 1, 0, 0, 0, 0, 1, 0, Sqrt[2], 0, 0, 0, 0, Sqrt[2], 0, Sqrt[3], 0, 0, 0, 0, Sqrt[3], 0, Sqrt[4], 0, 0, 0, 0, Sqrt[4],0, Sqrt[5], 0, 0, 0, 0, Sqrt[5], 0;
poly = Det[matA - IdentityMatrix[6] [Lambda]]
$$lambda ^6-15 lambda ^4+45 lambda ^2-15$$
sol = Solve[poly == 0 , [Lambda]];
eigen = matA - IdentityMatrix[6] [Lambda] /. sol[[1]]
Using RowReduce we can find one of our vectors.
sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm
The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
the only answer it gives me is zero vector
It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.
On V12:
mat = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[mat];
MatrixForm[% // N]
Update
OP wants solution to be analytical solution and not numerical.
Mathematica gives answer using Roots objects.
Using
SetSystemOptions[
"TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]
Gives
TO obtain numerical values, the command N can be applied to the above.
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
$endgroup$
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expressionFirst, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove//N. Answer will be in terms ofRootobjects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..
$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
add a comment |
$begingroup$
the only answer it gives me is zero vector
It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.
On V12:
mat = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[mat];
MatrixForm[% // N]
Update
OP wants solution to be analytical solution and not numerical.
Mathematica gives answer using Roots objects.
Using
SetSystemOptions[
"TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]
Gives
TO obtain numerical values, the command N can be applied to the above.
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
$endgroup$
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expressionFirst, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove//N. Answer will be in terms ofRootobjects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..
$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
add a comment |
$begingroup$
the only answer it gives me is zero vector
It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.
On V12:
mat = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[mat];
MatrixForm[% // N]
Update
OP wants solution to be analytical solution and not numerical.
Mathematica gives answer using Roots objects.
Using
SetSystemOptions[
"TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]
Gives
TO obtain numerical values, the command N can be applied to the above.
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
$endgroup$
the only answer it gives me is zero vector
It works for me. Please post exact code you used and show the output. Not just Latex. And give which version you used. You might have made a mistake in the input.
On V12:
mat = 0, 1, 0, 0, 0, 0,
1, 0, Sqrt[2], 0, 0, 0,
0, Sqrt[2], 0, Sqrt[3], 0, 0,
0, 0, Sqrt[3], 0, Sqrt[4], 0,
0, 0, 0, Sqrt[4], 0, Sqrt[5],
0, 0, 0, 0, Sqrt[5], 0
;
Eigenvectors[mat];
MatrixForm[% // N]
Update
OP wants solution to be analytical solution and not numerical.
Mathematica gives answer using Roots objects.
Using
SetSystemOptions[
"TypesetOptions" -> "NumericalApproximationForms" -> False];
Eigenvectors[mat]
Gives
TO obtain numerical values, the command N can be applied to the above.
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
edited Jul 30 at 11:08
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
answered Jul 30 at 8:37
NasserNasser
61.1k4 gold badges93 silver badges214 bronze badges
61.1k4 gold badges93 silver badges214 bronze badges
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expressionFirst, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove//N. Answer will be in terms ofRootobjects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..
$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
add a comment |
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expressionFirst, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove//N. Answer will be in terms ofRootobjects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..
$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression.
$endgroup$
– Jason
Jul 30 at 10:57
1
1
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression First, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove //N. Answer will be in terms of Root objects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
The point is I dont want to find the mnumerically i want their analytical expression First, this was not what your question was about. You said you got zero as answer. So I do not know how this "numerical vs. analytical" now became the "point" of the question. But if you do not want numerical, you can remove //N. Answer will be in terms of Root objects though. I do not think there is a way to obtain analytical solution to roots of polynomials of order 6. So mathematica gives answer using Root objects. I'll add the code now..$endgroup$
– Nasser
Jul 30 at 11:04
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
$begingroup$
@ Nasser yes you are right I wasn't considering the fact that it could be solved numerically, my mistake, anyway thanks a lot that worked
$endgroup$
– Jason
Jul 30 at 12:48
add a comment |
$begingroup$
Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!
When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.
matA = 0, 1, 0, 0, 0, 0, 1, 0, Sqrt[2], 0, 0, 0, 0, Sqrt[2], 0, Sqrt[3], 0, 0, 0, 0, Sqrt[3], 0, Sqrt[4], 0, 0, 0, 0, Sqrt[4],0, Sqrt[5], 0, 0, 0, 0, Sqrt[5], 0;
poly = Det[matA - IdentityMatrix[6] [Lambda]]
$$lambda ^6-15 lambda ^4+45 lambda ^2-15$$
sol = Solve[poly == 0 , [Lambda]];
eigen = matA - IdentityMatrix[6] [Lambda] /. sol[[1]]
Using RowReduce we can find one of our vectors.
sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm
The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!
When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.
matA = 0, 1, 0, 0, 0, 0, 1, 0, Sqrt[2], 0, 0, 0, 0, Sqrt[2], 0, Sqrt[3], 0, 0, 0, 0, Sqrt[3], 0, Sqrt[4], 0, 0, 0, 0, Sqrt[4],0, Sqrt[5], 0, 0, 0, 0, Sqrt[5], 0;
poly = Det[matA - IdentityMatrix[6] [Lambda]]
$$lambda ^6-15 lambda ^4+45 lambda ^2-15$$
sol = Solve[poly == 0 , [Lambda]];
eigen = matA - IdentityMatrix[6] [Lambda] /. sol[[1]]
Using RowReduce we can find one of our vectors.
sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm
The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
$endgroup$
add a comment |
$begingroup$
Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!
When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.
matA = 0, 1, 0, 0, 0, 0, 1, 0, Sqrt[2], 0, 0, 0, 0, Sqrt[2], 0, Sqrt[3], 0, 0, 0, 0, Sqrt[3], 0, Sqrt[4], 0, 0, 0, 0, Sqrt[4],0, Sqrt[5], 0, 0, 0, 0, Sqrt[5], 0;
poly = Det[matA - IdentityMatrix[6] [Lambda]]
$$lambda ^6-15 lambda ^4+45 lambda ^2-15$$
sol = Solve[poly == 0 , [Lambda]];
eigen = matA - IdentityMatrix[6] [Lambda] /. sol[[1]]
Using RowReduce we can find one of our vectors.
sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm
The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
$endgroup$
Since you didn't post any code, it's hard to know what you did wrong. Please always post your code!
When in doubt if mathematica is correct about what it's doing, do it by hand..this of course really only works when the question in hand is small like yours.
matA = 0, 1, 0, 0, 0, 0, 1, 0, Sqrt[2], 0, 0, 0, 0, Sqrt[2], 0, Sqrt[3], 0, 0, 0, 0, Sqrt[3], 0, Sqrt[4], 0, 0, 0, 0, Sqrt[4],0, Sqrt[5], 0, 0, 0, 0, Sqrt[5], 0;
poly = Det[matA - IdentityMatrix[6] [Lambda]]
$$lambda ^6-15 lambda ^4+45 lambda ^2-15$$
sol = Solve[poly == 0 , [Lambda]];
eigen = matA - IdentityMatrix[6] [Lambda] /. sol[[1]]
Using RowReduce we can find one of our vectors.
sol2 = RowReduce[eigen]
sol2[[All, 6]] // MatrixForm
The Last column being our first vector. The last element is a zero...this should be a one...an artifact of RowReduce I'm not sure why it does that. Regardless, a lot of work when one probably just used Eigenvectors[] the function incorrectly.
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
answered Jul 30 at 9:09
morbomorbo
7323 silver badges9 bronze badges
7323 silver badges9 bronze badges
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