What is the procedure finding the Orthogonal Basis of a Quadratic Form?Graphing quadratic form, which eigenvalue should be chosen first?Zero as an Eigenvalue of stress tensorfind real numbers so that a specific linear isometry existscondition for a binary quadratic form to be positive at infinityWhat is wrong with my calculation of Jordan canonical form?Find a positive-definite scalar product for which $F:mathbbR^3to mathbbR^3$ is symmetricalConfusion about orthogonal basis of a quadratic form.If $ax^2_1+by^2_1+cz^2_1=ax^2_2+by^2_2+cz^2_2=ax^2_3+by^2_3+cz^2_3=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$,Find basis of fundamental subspaces with given eigenvalues and eigenvectorsShow that $P^TAP$ reduces to the diagonal matrix $lambda_idelta_ij$
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What is the procedure finding the Orthogonal Basis of a Quadratic Form?
Graphing quadratic form, which eigenvalue should be chosen first?Zero as an Eigenvalue of stress tensorfind real numbers so that a specific linear isometry existscondition for a binary quadratic form to be positive at infinityWhat is wrong with my calculation of Jordan canonical form?Find a positive-definite scalar product for which $F:mathbbR^3to mathbbR^3$ is symmetricalConfusion about orthogonal basis of a quadratic form.If $ax^2_1+by^2_1+cz^2_1=ax^2_2+by^2_2+cz^2_2=ax^2_3+by^2_3+cz^2_3=d$ and $ax_2x_3+by_2y_3+cz_2z_3=ax_3x_1+by_3y_1+cz_3z_1=ax_1x_2+by_1y_2+cz_1z_2=f$,Find basis of fundamental subspaces with given eigenvalues and eigenvectorsShow that $P^TAP$ reduces to the diagonal matrix $lambda_idelta_ij$
$begingroup$
We have the following quadratic form:
$$q := 6 x^2_1 + 3 x^2_2 + 3 x^2_3 - 4 x_1 x_2 + 4 x_1 x_3 - 2 x_2 x_3$$
whose eigenvalues are $lambda_1=lambda_2=2$ and $lambda_3=8$
I am not really sure how could we get the procedure to find the orthogonal basis for this form...
For now, I know two things:
1. The Orthogonal Basis of the quadratic form was assumed randomly
2. Somehow we used some sort of "Witchcraft" from the eigenvectors.
Which one is true
linear-algebra quadratic-forms
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
We have the following quadratic form:
$$q := 6 x^2_1 + 3 x^2_2 + 3 x^2_3 - 4 x_1 x_2 + 4 x_1 x_3 - 2 x_2 x_3$$
whose eigenvalues are $lambda_1=lambda_2=2$ and $lambda_3=8$
I am not really sure how could we get the procedure to find the orthogonal basis for this form...
For now, I know two things:
1. The Orthogonal Basis of the quadratic form was assumed randomly
2. Somehow we used some sort of "Witchcraft" from the eigenvectors.
Which one is true
linear-algebra quadratic-forms
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday
add a comment |
$begingroup$
We have the following quadratic form:
$$q := 6 x^2_1 + 3 x^2_2 + 3 x^2_3 - 4 x_1 x_2 + 4 x_1 x_3 - 2 x_2 x_3$$
whose eigenvalues are $lambda_1=lambda_2=2$ and $lambda_3=8$
I am not really sure how could we get the procedure to find the orthogonal basis for this form...
For now, I know two things:
1. The Orthogonal Basis of the quadratic form was assumed randomly
2. Somehow we used some sort of "Witchcraft" from the eigenvectors.
Which one is true
linear-algebra quadratic-forms
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
We have the following quadratic form:
$$q := 6 x^2_1 + 3 x^2_2 + 3 x^2_3 - 4 x_1 x_2 + 4 x_1 x_3 - 2 x_2 x_3$$
whose eigenvalues are $lambda_1=lambda_2=2$ and $lambda_3=8$
I am not really sure how could we get the procedure to find the orthogonal basis for this form...
For now, I know two things:
1. The Orthogonal Basis of the quadratic form was assumed randomly
2. Somehow we used some sort of "Witchcraft" from the eigenvectors.
Which one is true
linear-algebra quadratic-forms
linear-algebra quadratic-forms
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
edited Mar 10 at 22:27
Rodrigo de Azevedo
13k41960
13k41960
asked Mar 10 at 21:02
EternalKoreEternalKore
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 21:02
EternalKoreEternalKore
61
asked Mar 10 at 21:02
EternalKoreEternalKore
61
61
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
EternalKore is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday
add a comment |
$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday
$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Not quite sure exactly what it is you’re asking here. There are a couple of ways that you can find an orthogonal eigenbasis.
One way is to go through the normal process of finding a basis for each eigenspace. The eigenspace of $2$ will be two-dimensional, so orthogonalize the basis that you found for it using the Gram-Schmidt process.
Alternatively, recall that the eigenspaces are pairwise orthogonal, so since the ambient space is three-dimensional, the eigenspace of $2$ must be the orthogonal complement of the eigenspace of $8$. Find an eigenvector for the latter, generate a vector that’s orthogonal to it (I hope you can do at least that), and then take the cross product of these two vectors to complete the basis.
answered Mar 10 at 21:58
amdamd
30.9k21051
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Not quite sure exactly what it is you’re asking here. There are a couple of ways that you can find an orthogonal eigenbasis.
One way is to go through the normal process of finding a basis for each eigenspace. The eigenspace of $2$ will be two-dimensional, so orthogonalize the basis that you found for it using the Gram-Schmidt process.
Alternatively, recall that the eigenspaces are pairwise orthogonal, so since the ambient space is three-dimensional, the eigenspace of $2$ must be the orthogonal complement of the eigenspace of $8$. Find an eigenvector for the latter, generate a vector that’s orthogonal to it (I hope you can do at least that), and then take the cross product of these two vectors to complete the basis.
answered Mar 10 at 21:58
amdamd
30.9k21051
$endgroup$
add a comment |
$begingroup$
Not quite sure exactly what it is you’re asking here. There are a couple of ways that you can find an orthogonal eigenbasis.
One way is to go through the normal process of finding a basis for each eigenspace. The eigenspace of $2$ will be two-dimensional, so orthogonalize the basis that you found for it using the Gram-Schmidt process.
Alternatively, recall that the eigenspaces are pairwise orthogonal, so since the ambient space is three-dimensional, the eigenspace of $2$ must be the orthogonal complement of the eigenspace of $8$. Find an eigenvector for the latter, generate a vector that’s orthogonal to it (I hope you can do at least that), and then take the cross product of these two vectors to complete the basis.
answered Mar 10 at 21:58
amdamd
30.9k21051
$endgroup$
add a comment |
$begingroup$
Not quite sure exactly what it is you’re asking here. There are a couple of ways that you can find an orthogonal eigenbasis.
One way is to go through the normal process of finding a basis for each eigenspace. The eigenspace of $2$ will be two-dimensional, so orthogonalize the basis that you found for it using the Gram-Schmidt process.
Alternatively, recall that the eigenspaces are pairwise orthogonal, so since the ambient space is three-dimensional, the eigenspace of $2$ must be the orthogonal complement of the eigenspace of $8$. Find an eigenvector for the latter, generate a vector that’s orthogonal to it (I hope you can do at least that), and then take the cross product of these two vectors to complete the basis.
answered Mar 10 at 21:58
amdamd
30.9k21051
$endgroup$
Not quite sure exactly what it is you’re asking here. There are a couple of ways that you can find an orthogonal eigenbasis.
One way is to go through the normal process of finding a basis for each eigenspace. The eigenspace of $2$ will be two-dimensional, so orthogonalize the basis that you found for it using the Gram-Schmidt process.
Alternatively, recall that the eigenspaces are pairwise orthogonal, so since the ambient space is three-dimensional, the eigenspace of $2$ must be the orthogonal complement of the eigenspace of $8$. Find an eigenvector for the latter, generate a vector that’s orthogonal to it (I hope you can do at least that), and then take the cross product of these two vectors to complete the basis.
answered Mar 10 at 21:58
amdamd
30.9k21051
answered Mar 10 at 21:58
amdamd
30.9k21051
answered Mar 10 at 21:58
amdamd
30.9k21051
answered Mar 10 at 21:58
amdamd
30.9k21051
30.9k21051
add a comment |
add a comment |
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$begingroup$
This might be helpful: batty.mullikin.org/quadratic.pdf
$endgroup$
– enedil
Mar 10 at 21:29
$begingroup$
Have you ever heard of the Gram-Schmidt process?
$endgroup$
– amd
Mar 10 at 21:53
$begingroup$
Isn't a quadratic form a polynomial? How can it have eigenvalues? If you're referring to the symmetric matrix one can use to write the quadratic form, please mention it.
$endgroup$
– Rodrigo de Azevedo
Mar 10 at 22:28
$begingroup$
After I read my notes, I realised that I have to use the Gram-Schmidt process Much obliged @amd
$endgroup$
– EternalKore
yesterday