Diagonalization without computing the inverse
up vote
0
down vote
favorite
Is there a way to find A = PDP^-1 but without computing the inverse of P?
If PP^-1equal to identity matrix, can I say A = ID ?
matrices inverse diagonalization
asked Nov 19 at 0:40
myadeniboy
11
add a comment |
up vote
0
down vote
favorite
Is there a way to find A = PDP^-1 but without computing the inverse of P?
If PP^-1equal to identity matrix, can I say A = ID ?
matrices inverse diagonalization
asked Nov 19 at 0:40
myadeniboy
11
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Is there a way to find A = PDP^-1 but without computing the inverse of P?
If PP^-1equal to identity matrix, can I say A = ID ?
matrices inverse diagonalization
asked Nov 19 at 0:40
myadeniboy
11
Is there a way to find A = PDP^-1 but without computing the inverse of P?
If PP^-1equal to identity matrix, can I say A = ID ?
matrices inverse diagonalization
matrices inverse diagonalization
asked Nov 19 at 0:40
myadeniboy
11
asked Nov 19 at 0:40
myadeniboy
11
asked Nov 19 at 0:40
myadeniboy
11
asked Nov 19 at 0:40
myadeniboy
11
asked Nov 19 at 0:40
myadeniboy
11
11
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20
add a comment |
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20
add a comment |
1 Answer
1
active
oldest
votes
up vote
0
down vote
Calculate the normalised eigenvectors of your matrix and make these eigenvectors the columns of a new matrix U.
Now calculate U^T * P * U using matrix multiplication and then you will obtain a diagonalised matrix with the eigenvalues of P as the entries on the diagonal.
NB: U^T is U-transpose.
answered Nov 19 at 1:07
Anteater23
62
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Calculate the normalised eigenvectors of your matrix and make these eigenvectors the columns of a new matrix U.
Now calculate U^T * P * U using matrix multiplication and then you will obtain a diagonalised matrix with the eigenvalues of P as the entries on the diagonal.
NB: U^T is U-transpose.
answered Nov 19 at 1:07
Anteater23
62
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
add a comment |
up vote
0
down vote
Calculate the normalised eigenvectors of your matrix and make these eigenvectors the columns of a new matrix U.
Now calculate U^T * P * U using matrix multiplication and then you will obtain a diagonalised matrix with the eigenvalues of P as the entries on the diagonal.
NB: U^T is U-transpose.
answered Nov 19 at 1:07
Anteater23
62
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
add a comment |
up vote
0
down vote
up vote
0
down vote
Calculate the normalised eigenvectors of your matrix and make these eigenvectors the columns of a new matrix U.
Now calculate U^T * P * U using matrix multiplication and then you will obtain a diagonalised matrix with the eigenvalues of P as the entries on the diagonal.
NB: U^T is U-transpose.
answered Nov 19 at 1:07
Anteater23
62
Calculate the normalised eigenvectors of your matrix and make these eigenvectors the columns of a new matrix U.
Now calculate U^T * P * U using matrix multiplication and then you will obtain a diagonalised matrix with the eigenvalues of P as the entries on the diagonal.
NB: U^T is U-transpose.
answered Nov 19 at 1:07
Anteater23
62
answered Nov 19 at 1:07
Anteater23
62
answered Nov 19 at 1:07
Anteater23
62
answered Nov 19 at 1:07
Anteater23
62
62
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
add a comment |
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
1
1
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
No need to normalize eigenvectors, but much more important: in general, multiplying by the traspose won't diagonalize a matrix even if its diagonalizable, so the above is incorrect.
– DonAntonio
Nov 19 at 1:16
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Some of your past answers have not been well-received, and you're in danger of being blocked from answering.
Please pay close attention to the following guidance:
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3004343%2fdiagonalization-without-computing-the-inverse%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
It will always be true that $;PP^{-1}=I;$ , so that should be a hint for you that no: you can't definitely say $;A=I;$ as matrix multiplication isn't, in general, commutative. You can diagonalize $;A;$ though if it is diagonalizable and if you know the matrix's eigenvalues, without any need of $;P;$ or of its inverse.
– DonAntonio
Nov 19 at 0:45
Given a diagonalizable matrix $A$, it is possible to compute both $P$ and $D$ without computing $P^{-1}$. Is that what you're trying to do?
– Omnomnomnom
Nov 19 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 at 1:20