How is solve If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues [closed] The Next CEO of Stack OverflowSimilar matrices have the same eigenvalues with the same geometric multiplicityElegant proofs that similar matrices have the same characteristic polynomial?similar matrices have the same eigenvaluesIf two matrices have the same characteristic polynomials, determinant and trace, are they similar?Proving that Matrices with same minimal and characteristic polynomial are similarIs this circular reasoning, when trying to prove that similar matrices have the same eigenvalues and characteristic polynomial?Can two matrices with the same characteristic polynomial have different eigenvalues?Show that similar matrices have same traceSimilar matrices have same Eigenvalues, Is converse true?Do similar matrices have same characteristic equations?
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How is solve If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues [closed]
The Next CEO of Stack OverflowSimilar matrices have the same eigenvalues with the same geometric multiplicityElegant proofs that similar matrices have the same characteristic polynomial?similar matrices have the same eigenvaluesIf two matrices have the same characteristic polynomials, determinant and trace, are they similar?Proving that Matrices with same minimal and characteristic polynomial are similarIs this circular reasoning, when trying to prove that similar matrices have the same eigenvalues and characteristic polynomial?Can two matrices with the same characteristic polynomial have different eigenvalues?Show that similar matrices have same traceSimilar matrices have same Eigenvalues, Is converse true?Do similar matrices have same characteristic equations?
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If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues.
matrices eigenvalues-eigenvectors characteristics
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
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closed as off-topic by Gerry Myerson, mrtaurho, Paul Frost, José Carlos Santos, rtybase Mar 24 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Paul Frost, rtybase
add a comment |
$begingroup$
If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues.
matrices eigenvalues-eigenvectors characteristics
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
$endgroup$
closed as off-topic by Gerry Myerson, mrtaurho, Paul Frost, José Carlos Santos, rtybase Mar 24 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Paul Frost, rtybase
$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
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– Gerry Myerson
Mar 20 at 12:05
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Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
1
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49
add a comment |
$begingroup$
If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues.
matrices eigenvalues-eigenvectors characteristics
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
$endgroup$
If A and B are similar n a matrices, then show that A and B have the same characteristic equation and therefore have the same eigenvalues.
matrices eigenvalues-eigenvectors characteristics
matrices eigenvalues-eigenvectors characteristics
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
asked Mar 20 at 11:05
Tharindu LakshanTharindu Lakshan
1
1
closed as off-topic by Gerry Myerson, mrtaurho, Paul Frost, José Carlos Santos, rtybase Mar 24 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Paul Frost, rtybase
closed as off-topic by Gerry Myerson, mrtaurho, Paul Frost, José Carlos Santos, rtybase Mar 24 at 19:09
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – mrtaurho, Paul Frost, rtybase
$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
$endgroup$
– Gerry Myerson
Mar 20 at 12:05
$begingroup$
Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
1
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49
add a comment |
$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
$endgroup$
– Gerry Myerson
Mar 20 at 12:05
$begingroup$
Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
1
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49
$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
$endgroup$
– Gerry Myerson
Mar 20 at 12:05
$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
$endgroup$
– Gerry Myerson
Mar 20 at 12:05
$begingroup$
Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
$begingroup$
Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
1
1
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49
add a comment |
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$begingroup$
If $f$ is a polynomial, then $P^-1f(A)P=f(P^-1AP)$.
$endgroup$
– Gerry Myerson
Mar 20 at 12:05
$begingroup$
Did that help, Tharindu?
$endgroup$
– Gerry Myerson
Mar 21 at 22:35
1
$begingroup$
I'm voting to close this question as off-topic because OP has abandoned it.
$endgroup$
– Gerry Myerson
Mar 24 at 8:49