Do QR and RQ have the same eigenvalues [duplicate]
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Eigen Values Proof
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Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?
$det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$
so they have the same eigenvalue, is this ok?
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51
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up vote
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This question already has an answer here:
Eigen Values Proof
1 answer
Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?
$det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$
so they have the same eigenvalue, is this ok?
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
This question already has an answer here:
Eigen Values Proof
1 answer
Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?
$det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$
so they have the same eigenvalue, is this ok?
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
This question already has an answer here:
Eigen Values Proof
1 answer
Let $A=QR$ where $Qin$ $M_n(mathbb R)$ orthogonal and $Rin M_n(mathbb R)$ is upper triangular matrix. Do $A$ and $RQ$ have the same eigenvalues?
$det(A-lambda I)=det(QR-lambda I)=det(QR-lambda QQ^T)=det Qdet(R-lambda Q^T)=det Qdet(RI-lambda Q^T)=det Qdet(RQQ^T-lambda Q^T)=det Qdet(RQ-lambda I)det Q^T=det(RQ-lambda I),$
so they have the same eigenvalue, is this ok?
This question already has an answer here:
Eigen Values Proof
1 answer
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
linear-algebra eigenvalues-eigenvectors orthogonal-matrices
edited Sep 4 at 12:16
Bernard
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112k635104
asked Sep 4 at 11:15
Marko à  koriÃÂ
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marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
marked as duplicate by Arnaud D., Strants, Ethan Bolker, Ali Caglayan, A. Pongrácz Sep 4 at 19:51
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
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Both matrices are similar:
$$
A = QR = Q(RQ)Q^-1,
$$
hence they have the same Jordan form.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Both matrices are similar:
$$
A = QR = Q(RQ)Q^-1,
$$
hence they have the same Jordan form.
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up vote
1
down vote
Both matrices are similar:
$$
A = QR = Q(RQ)Q^-1,
$$
hence they have the same Jordan form.
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Both matrices are similar:
$$
A = QR = Q(RQ)Q^-1,
$$
hence they have the same Jordan form.
Both matrices are similar:
$$
A = QR = Q(RQ)Q^-1,
$$
hence they have the same Jordan form.
answered Sep 4 at 11:28
daw
22.2k1542
22.2k1542
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