Mixing
Quadratic forms, change of variables
Clash Royale CLAN TAG#URR8PPP
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If one has a symmetric matrix $A$, one can diagonalize it with an orthonormal change of basis vectors, e.g. $S^TAS$ is diagonal.
Now lets consider the following matrix $$A=beginbmatrix
1&1\
1&2
endbmatrix.
$$
This matrix corresponds to the symmetric form
$$x_1^2+2x_1x_2+2x_2^2=(x_1+x_2)^2+x_2^2.$$
For me this looks like there has to be a way of determining some $S$ as above (without having to diagonalize etc., just by completing of the square) by taking some change of variables $x_1leadsto x_1+x_2, x_2leadsto x_2$. This would be done by the matrix $$S=beginbmatrix
1&1\
0&1
endbmatrix$$
but this doesn't work out for me...
Any help will be gratefully appreciated.
Edit:
Let me reformulate my question. By Sylvester's law of inertia there exists for every symmetric matrix $A$ some basis such that $S^TAS$ is diagonal with only 1,-1 and 0 on the diagonal, where $S$ is a (not necessarily orthogonal) invertible matrix. I want to determine $S$ without having to calculate all the eigenvalues and diagonalize $A$, because the eigenvalues dont occur in the wanted form.
quadratics diagonalization symmetric-matrices quadrics
asked 21 hours ago
Luke Mathwalker
188110
 |Â
show 3 more comments
up vote
0
down vote
favorite
If one has a symmetric matrix $A$, one can diagonalize it with an orthonormal change of basis vectors, e.g. $S^TAS$ is diagonal.
Now lets consider the following matrix $$A=beginbmatrix
1&1\
1&2
endbmatrix.
$$
This matrix corresponds to the symmetric form
$$x_1^2+2x_1x_2+2x_2^2=(x_1+x_2)^2+x_2^2.$$
For me this looks like there has to be a way of determining some $S$ as above (without having to diagonalize etc., just by completing of the square) by taking some change of variables $x_1leadsto x_1+x_2, x_2leadsto x_2$. This would be done by the matrix $$S=beginbmatrix
1&1\
0&1
endbmatrix$$
but this doesn't work out for me...
Any help will be gratefully appreciated.
Edit:
Let me reformulate my question. By Sylvester's law of inertia there exists for every symmetric matrix $A$ some basis such that $S^TAS$ is diagonal with only 1,-1 and 0 on the diagonal, where $S$ is a (not necessarily orthogonal) invertible matrix. I want to determine $S$ without having to calculate all the eigenvalues and diagonalize $A$, because the eigenvalues dont occur in the wanted form.
quadratics diagonalization symmetric-matrices quadrics
asked 21 hours ago
Luke Mathwalker
188110
Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago
 |Â
show 3 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
If one has a symmetric matrix $A$, one can diagonalize it with an orthonormal change of basis vectors, e.g. $S^TAS$ is diagonal.
Now lets consider the following matrix $$A=beginbmatrix
1&1\
1&2
endbmatrix.
$$
This matrix corresponds to the symmetric form
$$x_1^2+2x_1x_2+2x_2^2=(x_1+x_2)^2+x_2^2.$$
For me this looks like there has to be a way of determining some $S$ as above (without having to diagonalize etc., just by completing of the square) by taking some change of variables $x_1leadsto x_1+x_2, x_2leadsto x_2$. This would be done by the matrix $$S=beginbmatrix
1&1\
0&1
endbmatrix$$
but this doesn't work out for me...
Any help will be gratefully appreciated.
Edit:
Let me reformulate my question. By Sylvester's law of inertia there exists for every symmetric matrix $A$ some basis such that $S^TAS$ is diagonal with only 1,-1 and 0 on the diagonal, where $S$ is a (not necessarily orthogonal) invertible matrix. I want to determine $S$ without having to calculate all the eigenvalues and diagonalize $A$, because the eigenvalues dont occur in the wanted form.
quadratics diagonalization symmetric-matrices quadrics
asked 21 hours ago
Luke Mathwalker
188110
If one has a symmetric matrix $A$, one can diagonalize it with an orthonormal change of basis vectors, e.g. $S^TAS$ is diagonal.
Now lets consider the following matrix $$A=beginbmatrix
1&1\
1&2
endbmatrix.
$$
This matrix corresponds to the symmetric form
$$x_1^2+2x_1x_2+2x_2^2=(x_1+x_2)^2+x_2^2.$$
For me this looks like there has to be a way of determining some $S$ as above (without having to diagonalize etc., just by completing of the square) by taking some change of variables $x_1leadsto x_1+x_2, x_2leadsto x_2$. This would be done by the matrix $$S=beginbmatrix
1&1\
0&1
endbmatrix$$
but this doesn't work out for me...
Any help will be gratefully appreciated.
Edit:
Let me reformulate my question. By Sylvester's law of inertia there exists for every symmetric matrix $A$ some basis such that $S^TAS$ is diagonal with only 1,-1 and 0 on the diagonal, where $S$ is a (not necessarily orthogonal) invertible matrix. I want to determine $S$ without having to calculate all the eigenvalues and diagonalize $A$, because the eigenvalues dont occur in the wanted form.
quadratics diagonalization symmetric-matrices quadrics
asked 21 hours ago
Luke Mathwalker
188110
edited 19 hours ago
asked 21 hours ago
Luke Mathwalker
188110
asked 21 hours ago
Luke Mathwalker
188110
asked 21 hours ago
Luke Mathwalker
188110
188110
Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago
 |Â
show 3 more comments
Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago
Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
up vote
0
down vote
We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $det P = pm 1.$ If it is imperative to have diagonal elements restricted to $pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrr
1 & 0 \
- 1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrr
1 & 0 \
1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
answered 18 hours ago
Will Jagy
96.7k594195
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $det P = pm 1.$ If it is imperative to have diagonal elements restricted to $pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrr
1 & 0 \
- 1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrr
1 & 0 \
1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
answered 18 hours ago
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $det P = pm 1.$ If it is imperative to have diagonal elements restricted to $pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrr
1 & 0 \
- 1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrr
1 & 0 \
1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
answered 18 hours ago
Will Jagy
96.7k594195
add a comment |Â
up vote
0
down vote
up vote
0
down vote
We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $det P = pm 1.$ If it is imperative to have diagonal elements restricted to $pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrr
1 & 0 \
- 1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrr
1 & 0 \
1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
answered 18 hours ago
Will Jagy
96.7k594195
We can, given a symmetric matrix of integers or rational numbers, construct $P^THP = D$ with rational $P$ and $det P = pm 1.$ If it is imperative to have diagonal elements restricted to $pm 1, 0,$ we may take the final $D$ and construct a further diagonal $F$ with elements the reciprocals of some square roots to get $FDF$ the way you want. In this particular problem, no need.
Algorithm discussed at http://math.stackexchange.com/questions/1388421/reference-for-linear-algebra-books-that-teach-reverse-hermite-method-for-symmetr
https://en.wikipedia.org/wiki/Sylvester%27s_law_of_inertia
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
$$ D_0 = H $$
$$ E_j^T D_j-1 E_j = D_j $$
$$ P_j-1 E_j = P_j $$
$$ E_j^-1 Q_j-1 = Q_j $$
$$ P_j Q_j = Q_j P_j = I $$
$$ P_j^T H P_j = D_j $$
$$ Q_j^T D_j Q_j = H $$
$$ H = left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
==============================================
$$ E_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
$$
$$ P_1 = left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
, ; ; ; Q_1 = left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
, ; ; ; D_1 = left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
==============================================
$$ P^T H P = D $$
$$left(
beginarrayrr
1 & 0 \
- 1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
left(
beginarrayrr
1 & - 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
$$
$$ Q^T D Q = H $$
$$left(
beginarrayrr
1 & 0 \
1 & 1 \
endarray
right)
left(
beginarrayrr
1 & 0 \
0 & 1 \
endarray
right)
left(
beginarrayrr
1 & 1 \
0 & 1 \
endarray
right)
= left(
beginarrayrr
1 & 1 \
1 & 2 \
endarray
right)
$$
answered 18 hours ago
Will Jagy
96.7k594195
answered 18 hours ago
Will Jagy
96.7k594195
answered 18 hours ago
Will Jagy
96.7k594195
answered 18 hours ago
Will Jagy
96.7k594195
96.7k594195
add a comment |Â
add a comment |Â
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Generally the method of completing squares may not deduce the same result as orthogonal diagonalization. Maybe you could try another way to complete squares.
– xbh
21 hours ago
Why not? Isnt there any way to find $S$ by this procedure?
– Luke Mathwalker
21 hours ago
Completing squares you remain in the same field containing the coefficients, since you only need to divide by $2$ sometimes, take squares, add and multiply. On the other hand, in some cases, and your example is one of those, the eigenvalues don't belong to the same field. In your example, the coefficients are all rational numbers, while the eigenvalues are not. The algorithm will need to have some step that takes you out of the same field.
– spiralstotheleft
20 hours ago
Ok I see. I dont want to get the eigenvalues, just some diagonal matrix, I'm interested in the signature of the form and a change of basis vectors, so that $S^TAS$ is in the "signature form" with 1,-1 and 0 on the diagonal, so we can reduce modulo square numbers which we write in $S$ to leave the field in the last step. I still believe that this should be possible without having to calculate all the eigenvalues.
– Luke Mathwalker
20 hours ago
Your $S$ isn't orthogonal. The special thing about symmetric matrices is that they are orthogonally (or unitarily) diagonalizable (that's why you see $S^TAS$, not $S^-1AS$).
– Arthur
20 hours ago