Is there an easy way to find the (analytical form of) remaining eigenvalues of a 4x4 matrix?
$begingroup$
Consider the following 4x4 real symmetric matrix:
$$
M = 2 ,
begin{pmatrix}
1 & c_{xy} & c_{xz} & c_{yz} \
c_{xy} & 1 & c_{yoverline{z}} & c_{xoverline{z}} \
c_{xz} & c_{yoverline{z}} & 1 & c_{xoverline{y}}\
c_{yz} & c_{xoverline{z}} & c_{xoverline{y}} & 1
end{pmatrix}
$$
where shortcuts $c_{xy} = cos(frac{x+ y}{4})$ and $c_{xoverline{y}} = cos(frac{x - y}{4})$ etc. are introduced.
I know the eigenvalues $lambda$ of $M$:
$$
lambda = 0, 0, 4pm 2sqrt{1+Q}
$$
where $Q = (c_{xy}^2 + c_{xoverline{y}}^2 + c_{yz}^2 + c_{yoverline{z}}^2 + c_{xz}^2 + c_{xoverline{z}}^2 - 3)$, i.e. two zeros and two non-zeros.
Question
Now my question is, if I know for a fact that 2 of the eigenvalues are 0, how can I arrive simply at the other two non-zero eigenvalues that I shown above?
Note I am interested in the analytical form (I know I can solve these easily with numerical methods).
I'm thinking maybe there is a way to reduce $M$ to a 2x2 matrix block (?)
My attempt
Not much of an attempt, but the usual trick of trace of matrix is the sum of eigenvalues and determinant is the product of eigenvalues doesn't really help here.
I solved for the eigenvalues symbolically using sympy in Python (I believe one can do the same in Mathematica), but this method doesn't scale well (see below context).
Context
The point of asking this is that I have another, more complicated, 4x4 real symmetric matrix filled with trigonometric expressions, that I know has 2 zero eigenvalues (like $M$).
I want to find the analytical form of the other two eigenvalues, but solving for all 4 of the eigenvalues symbolically is too computationally expensive. I was hoping I can make use of the fact that two of the eigenvalues are (known and) zero in some way.
Hence, a useful answer here would describe a method in which I can obtain the two non-zero $lambda$'s from $M$ -- then hopefully I will be able to generalize it for my own, complicated matrix.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 5 '18 at 1:02
TroyTroy
4181519
$endgroup$
add a comment |
$begingroup$
Consider the following 4x4 real symmetric matrix:
$$
M = 2 ,
begin{pmatrix}
1 & c_{xy} & c_{xz} & c_{yz} \
c_{xy} & 1 & c_{yoverline{z}} & c_{xoverline{z}} \
c_{xz} & c_{yoverline{z}} & 1 & c_{xoverline{y}}\
c_{yz} & c_{xoverline{z}} & c_{xoverline{y}} & 1
end{pmatrix}
$$
where shortcuts $c_{xy} = cos(frac{x+ y}{4})$ and $c_{xoverline{y}} = cos(frac{x - y}{4})$ etc. are introduced.
I know the eigenvalues $lambda$ of $M$:
$$
lambda = 0, 0, 4pm 2sqrt{1+Q}
$$
where $Q = (c_{xy}^2 + c_{xoverline{y}}^2 + c_{yz}^2 + c_{yoverline{z}}^2 + c_{xz}^2 + c_{xoverline{z}}^2 - 3)$, i.e. two zeros and two non-zeros.
Question
Now my question is, if I know for a fact that 2 of the eigenvalues are 0, how can I arrive simply at the other two non-zero eigenvalues that I shown above?
Note I am interested in the analytical form (I know I can solve these easily with numerical methods).
I'm thinking maybe there is a way to reduce $M$ to a 2x2 matrix block (?)
My attempt
Not much of an attempt, but the usual trick of trace of matrix is the sum of eigenvalues and determinant is the product of eigenvalues doesn't really help here.
I solved for the eigenvalues symbolically using sympy in Python (I believe one can do the same in Mathematica), but this method doesn't scale well (see below context).
Context
The point of asking this is that I have another, more complicated, 4x4 real symmetric matrix filled with trigonometric expressions, that I know has 2 zero eigenvalues (like $M$).
I want to find the analytical form of the other two eigenvalues, but solving for all 4 of the eigenvalues symbolically is too computationally expensive. I was hoping I can make use of the fact that two of the eigenvalues are (known and) zero in some way.
Hence, a useful answer here would describe a method in which I can obtain the two non-zero $lambda$'s from $M$ -- then hopefully I will be able to generalize it for my own, complicated matrix.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 5 '18 at 1:02
TroyTroy
4181519
$endgroup$
add a comment |
$begingroup$
Consider the following 4x4 real symmetric matrix:
$$
M = 2 ,
begin{pmatrix}
1 & c_{xy} & c_{xz} & c_{yz} \
c_{xy} & 1 & c_{yoverline{z}} & c_{xoverline{z}} \
c_{xz} & c_{yoverline{z}} & 1 & c_{xoverline{y}}\
c_{yz} & c_{xoverline{z}} & c_{xoverline{y}} & 1
end{pmatrix}
$$
where shortcuts $c_{xy} = cos(frac{x+ y}{4})$ and $c_{xoverline{y}} = cos(frac{x - y}{4})$ etc. are introduced.
I know the eigenvalues $lambda$ of $M$:
$$
lambda = 0, 0, 4pm 2sqrt{1+Q}
$$
where $Q = (c_{xy}^2 + c_{xoverline{y}}^2 + c_{yz}^2 + c_{yoverline{z}}^2 + c_{xz}^2 + c_{xoverline{z}}^2 - 3)$, i.e. two zeros and two non-zeros.
Question
Now my question is, if I know for a fact that 2 of the eigenvalues are 0, how can I arrive simply at the other two non-zero eigenvalues that I shown above?
Note I am interested in the analytical form (I know I can solve these easily with numerical methods).
I'm thinking maybe there is a way to reduce $M$ to a 2x2 matrix block (?)
My attempt
Not much of an attempt, but the usual trick of trace of matrix is the sum of eigenvalues and determinant is the product of eigenvalues doesn't really help here.
I solved for the eigenvalues symbolically using sympy in Python (I believe one can do the same in Mathematica), but this method doesn't scale well (see below context).
Context
The point of asking this is that I have another, more complicated, 4x4 real symmetric matrix filled with trigonometric expressions, that I know has 2 zero eigenvalues (like $M$).
I want to find the analytical form of the other two eigenvalues, but solving for all 4 of the eigenvalues symbolically is too computationally expensive. I was hoping I can make use of the fact that two of the eigenvalues are (known and) zero in some way.
Hence, a useful answer here would describe a method in which I can obtain the two non-zero $lambda$'s from $M$ -- then hopefully I will be able to generalize it for my own, complicated matrix.
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 5 '18 at 1:02
TroyTroy
4181519
$endgroup$
Consider the following 4x4 real symmetric matrix:
$$
M = 2 ,
begin{pmatrix}
1 & c_{xy} & c_{xz} & c_{yz} \
c_{xy} & 1 & c_{yoverline{z}} & c_{xoverline{z}} \
c_{xz} & c_{yoverline{z}} & 1 & c_{xoverline{y}}\
c_{yz} & c_{xoverline{z}} & c_{xoverline{y}} & 1
end{pmatrix}
$$
where shortcuts $c_{xy} = cos(frac{x+ y}{4})$ and $c_{xoverline{y}} = cos(frac{x - y}{4})$ etc. are introduced.
I know the eigenvalues $lambda$ of $M$:
$$
lambda = 0, 0, 4pm 2sqrt{1+Q}
$$
where $Q = (c_{xy}^2 + c_{xoverline{y}}^2 + c_{yz}^2 + c_{yoverline{z}}^2 + c_{xz}^2 + c_{xoverline{z}}^2 - 3)$, i.e. two zeros and two non-zeros.
Question
Now my question is, if I know for a fact that 2 of the eigenvalues are 0, how can I arrive simply at the other two non-zero eigenvalues that I shown above?
Note I am interested in the analytical form (I know I can solve these easily with numerical methods).
I'm thinking maybe there is a way to reduce $M$ to a 2x2 matrix block (?)
My attempt
Not much of an attempt, but the usual trick of trace of matrix is the sum of eigenvalues and determinant is the product of eigenvalues doesn't really help here.
I solved for the eigenvalues symbolically using sympy in Python (I believe one can do the same in Mathematica), but this method doesn't scale well (see below context).
Context
The point of asking this is that I have another, more complicated, 4x4 real symmetric matrix filled with trigonometric expressions, that I know has 2 zero eigenvalues (like $M$).
I want to find the analytical form of the other two eigenvalues, but solving for all 4 of the eigenvalues symbolically is too computationally expensive. I was hoping I can make use of the fact that two of the eigenvalues are (known and) zero in some way.
Hence, a useful answer here would describe a method in which I can obtain the two non-zero $lambda$'s from $M$ -- then hopefully I will be able to generalize it for my own, complicated matrix.
linear-algebra matrices eigenvalues-eigenvectors
linear-algebra matrices eigenvalues-eigenvectors
asked Dec 5 '18 at 1:02
TroyTroy
4181519
asked Dec 5 '18 at 1:02
TroyTroy
4181519
edited Dec 5 '18 at 1:08
Troy
asked Dec 5 '18 at 1:02
TroyTroy
4181519
asked Dec 5 '18 at 1:02
TroyTroy
4181519
asked Dec 5 '18 at 1:02
TroyTroy
4181519
4181519
add a comment |
add a comment |
1 Answer
1
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oldest
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$begingroup$
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
$endgroup$
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
$endgroup$
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
add a comment |
$begingroup$
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
$endgroup$
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
add a comment |
$begingroup$
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
$endgroup$
Possible idea. Since you know two eigenvalues are $0$ you know the $0$-eigenspace is $2$ dimensional. If you can (symbolically) see the null space, then its orthogonal complement will be $2$ dimensional and invariant. It will be spanned by the two missing eigenvectors. Finding the eigenvalues should be easier in just two dimensions.
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
edited Dec 5 '18 at 1:34
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
answered Dec 5 '18 at 1:15
Ethan BolkerEthan Bolker
42.5k549113
42.5k549113
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
add a comment |
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
hi, thanks for the input. would you happen to know how to do this numerically? I can calculate nullspace of $M$ easily (I get 2 vectors that should span the space), how do I get the orthogonal complement from that? how do I link this to the missing eigenvalues? it would be nice if you could show that you can reproduce my $lambda$ expressions by the method you are describing.
$endgroup$
– Troy
Dec 5 '18 at 1:23
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
$begingroup$
I'm glad this might be helpful, but don't know how to do the next steps. Gram-Schmidt would find the orthogonal complement, but I think that's not numerically stable. But lots is known about numerical linear algebra. Perhaps get as far as you can and ask a new question (link to this one) or seriously edit to show progress and the new sticking point.
$endgroup$
– Ethan Bolker
Dec 5 '18 at 1:33
add a comment |
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