Diagonalization without computing the inverse
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 '18 at 0:40
myadeniboy
11
add a comment |
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 '18 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 '18 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 '18 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 at 1:20
add a comment |
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 '18 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 '18 at 0:40
myadeniboy
11
asked Nov 19 '18 at 0:40
myadeniboy
11
asked Nov 19 '18 at 0:40
myadeniboy
11
asked Nov 19 '18 at 0:40
myadeniboy
11
asked Nov 19 '18 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 '18 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 '18 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 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 '18 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 '18 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 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 '18 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 '18 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 '18 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 '18 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 at 1:20
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 at 1:20
add a comment |
1 Answer
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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 '18 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 '18 at 1:16
add a comment |
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1 Answer
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1 Answer
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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 '18 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 '18 at 1:16
add a comment |
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 '18 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 '18 at 1:16
add a comment |
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 '18 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 '18 at 1:07
Anteater23
62
answered Nov 19 '18 at 1:07
Anteater23
62
answered Nov 19 '18 at 1:07
Anteater23
62
answered Nov 19 '18 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 '18 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 '18 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 '18 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 '18 at 1:16
add a comment |
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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 '18 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 '18 at 0:51
@Omnomnomnom Yes. Now I understand after DonAntonio's comment.
– myadeniboy
Nov 19 '18 at 1:20