Characteristic Polynomial of a matrix and its transposeShow that arbitrary $A$ and $A^T$ have same eigenvalue, algebraic and geometric multiplicityMinimal polynomials and characteristic polynomialseigenproblem and characteristic equationThe eigenvectors of a matrix and its transpose that correspond to the same eigenvalue are not orthogonalShow that minimal polynomial for $ntimes n$ matrix and its transpose is the same$A$ is singular and normal matrix, what must be its characteristic polynomial?Characteristic polynomial of a unitary matrix.How can we tell if a matrix is diagonalizabel based on its characteristic polynomial?A question on characteristic polynomial and minimal polynomialMatrices similar to their inverse or transposeEigenvalues and Eigenvectors of circulant matrix and its transpose
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Characteristic Polynomial of a matrix and its transpose
Show that arbitrary $A$ and $A^T$ have same eigenvalue, algebraic and geometric multiplicityMinimal polynomials and characteristic polynomialseigenproblem and characteristic equationThe eigenvectors of a matrix and its transpose that correspond to the same eigenvalue are not orthogonalShow that minimal polynomial for $ntimes n$ matrix and its transpose is the same$A$ is singular and normal matrix, what must be its characteristic polynomial?Characteristic polynomial of a unitary matrix.How can we tell if a matrix is diagonalizabel based on its characteristic polynomial?A question on characteristic polynomial and minimal polynomialMatrices similar to their inverse or transposeEigenvalues and Eigenvectors of circulant matrix and its transpose
$begingroup$
Does a matrix and and its transpose have the same characteristic polynomial? I know that they have the same eigenvalues but different eigenvectors. Does having the same eigenvalues mean they share the same characteristic polynomial?
linear-algebra
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
$endgroup$
add a comment |
$begingroup$
Does a matrix and and its transpose have the same characteristic polynomial? I know that they have the same eigenvalues but different eigenvectors. Does having the same eigenvalues mean they share the same characteristic polynomial?
linear-algebra
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
$endgroup$
$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
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A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50
add a comment |
$begingroup$
Does a matrix and and its transpose have the same characteristic polynomial? I know that they have the same eigenvalues but different eigenvectors. Does having the same eigenvalues mean they share the same characteristic polynomial?
linear-algebra
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
$endgroup$
Does a matrix and and its transpose have the same characteristic polynomial? I know that they have the same eigenvalues but different eigenvectors. Does having the same eigenvalues mean they share the same characteristic polynomial?
linear-algebra
linear-algebra
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
asked Mar 10 at 19:45
Samurai BaleSamurai Bale
645
645
$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
$begingroup$
A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50
add a comment |
$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
$begingroup$
A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50
$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
$begingroup$
A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50
add a comment |
1 Answer
1
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$begingroup$
For all $A in mathcalM_n(mathbbK)$,
$$det(A-XI_n)=det(^t(A-XI_n))=det(^tA - XI_n)$$
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For all $A in mathcalM_n(mathbbK)$,
$$det(A-XI_n)=det(^t(A-XI_n))=det(^tA - XI_n)$$
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
$begingroup$
For all $A in mathcalM_n(mathbbK)$,
$$det(A-XI_n)=det(^t(A-XI_n))=det(^tA - XI_n)$$
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
add a comment |
$begingroup$
For all $A in mathcalM_n(mathbbK)$,
$$det(A-XI_n)=det(^t(A-XI_n))=det(^tA - XI_n)$$
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
$endgroup$
For all $A in mathcalM_n(mathbbK)$,
$$det(A-XI_n)=det(^t(A-XI_n))=det(^tA - XI_n)$$
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
answered Mar 10 at 19:50
TheSilverDoeTheSilverDoe
3,837112
3,837112
add a comment |
add a comment |
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$begingroup$
See the proof here.
$endgroup$
– Dietrich Burde
Mar 10 at 19:47
$begingroup$
A matrix over a field is always similar to its transpose.
$endgroup$
– Lord Shark the Unknown
Mar 10 at 19:47
$begingroup$
Yes. (Presumably you're talking about square matrices)
$endgroup$
– J. W. Tanner
Mar 10 at 19:48
$begingroup$
See also here
$endgroup$
– J. W. Tanner
Mar 10 at 19:50