Do similar matrices have same characteristic equations?Jordan Canonical Form - Similar matrices and same minimal polynomialsSimilar Matrices and Characteristic PolynomialsSame characteristic polynomial $iff$ same eigenvalues?If two matrices have the same characteristic polynomials, determinant and trace, are they similar?$3 times 3$ matrices are similar if and only if they have the same characteristic and minimal polynomialHow to prove two diagonalizable matrices are similar iff they have same eigenvalue with same multiplicity.Can two matrices with the same characteristic polynomial have different eigenvalues?Are these $4$ by $4$ matrices similar?are all these matrices similar?Matrices with given characteristic polynomial are similar iff the characteristic polynomial have no repeated roots
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Do similar matrices have same characteristic equations?
Jordan Canonical Form - Similar matrices and same minimal polynomialsSimilar Matrices and Characteristic PolynomialsSame characteristic polynomial $iff$ same eigenvalues?If two matrices have the same characteristic polynomials, determinant and trace, are they similar?$3 times 3$ matrices are similar if and only if they have the same characteristic and minimal polynomialHow to prove two diagonalizable matrices are similar iff they have same eigenvalue with same multiplicity.Can two matrices with the same characteristic polynomial have different eigenvalues?Are these $4$ by $4$ matrices similar?are all these matrices similar?Matrices with given characteristic polynomial are similar iff the characteristic polynomial have no repeated roots
$begingroup$
Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?
linear-algebra
asked 2 hours ago
Samurai BaleSamurai Bale
483
$endgroup$
add a comment |
$begingroup$
Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?
linear-algebra
asked 2 hours ago
Samurai BaleSamurai Bale
483
$endgroup$
add a comment |
$begingroup$
Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?
linear-algebra
asked 2 hours ago
Samurai BaleSamurai Bale
483
$endgroup$
Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?
linear-algebra
linear-algebra
asked 2 hours ago
Samurai BaleSamurai Bale
483
asked 2 hours ago
Samurai BaleSamurai Bale
483
asked 2 hours ago
Samurai BaleSamurai Bale
483
asked 2 hours ago
Samurai BaleSamurai Bale
483
asked 2 hours ago
Samurai BaleSamurai Bale
483
483
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Note that $det (lambda I -A) = det S det (lambda I -A) det S^-1 = det (lambda I - S A S^-1)$.
answered 2 hours ago
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ are square matrices such that $A = P B P^-1$ for some invertible matrix $P$. Then
beginalign*textcharpoly(A,t) & = det(A - tI)\
& = det(PBP^-1 - tI)\
& = det(PBP^-1-tPP^-1)\
& = det(P(B-tI)P^-1)\
& = det(P)det(B - tI) det(P^-1)\
& = det(P)det(B - tI) frac1det(P)\
& = det(B-tI)\
& = textcharpoly(B,t).
endalign*
This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.
edited 2 hours ago
J. W. Tanner
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that $det (lambda I -A) = det S det (lambda I -A) det S^-1 = det (lambda I - S A S^-1)$.
answered 2 hours ago
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Note that $det (lambda I -A) = det S det (lambda I -A) det S^-1 = det (lambda I - S A S^-1)$.
answered 2 hours ago
copper.hatcopper.hat
127k559160
$endgroup$
add a comment |
$begingroup$
Note that $det (lambda I -A) = det S det (lambda I -A) det S^-1 = det (lambda I - S A S^-1)$.
answered 2 hours ago
copper.hatcopper.hat
127k559160
$endgroup$
Note that $det (lambda I -A) = det S det (lambda I -A) det S^-1 = det (lambda I - S A S^-1)$.
answered 2 hours ago
copper.hatcopper.hat
127k559160
answered 2 hours ago
copper.hatcopper.hat
127k559160
answered 2 hours ago
copper.hatcopper.hat
127k559160
answered 2 hours ago
copper.hatcopper.hat
127k559160
127k559160
add a comment |
add a comment |
$begingroup$
Suppose $A$ and $B$ are square matrices such that $A = P B P^-1$ for some invertible matrix $P$. Then
beginalign*textcharpoly(A,t) & = det(A - tI)\
& = det(PBP^-1 - tI)\
& = det(PBP^-1-tPP^-1)\
& = det(P(B-tI)P^-1)\
& = det(P)det(B - tI) det(P^-1)\
& = det(P)det(B - tI) frac1det(P)\
& = det(B-tI)\
& = textcharpoly(B,t).
endalign*
This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.
edited 2 hours ago
J. W. Tanner
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ are square matrices such that $A = P B P^-1$ for some invertible matrix $P$. Then
beginalign*textcharpoly(A,t) & = det(A - tI)\
& = det(PBP^-1 - tI)\
& = det(PBP^-1-tPP^-1)\
& = det(P(B-tI)P^-1)\
& = det(P)det(B - tI) det(P^-1)\
& = det(P)det(B - tI) frac1det(P)\
& = det(B-tI)\
& = textcharpoly(B,t).
endalign*
This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.
edited 2 hours ago
J. W. Tanner
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
$endgroup$
add a comment |
$begingroup$
Suppose $A$ and $B$ are square matrices such that $A = P B P^-1$ for some invertible matrix $P$. Then
beginalign*textcharpoly(A,t) & = det(A - tI)\
& = det(PBP^-1 - tI)\
& = det(PBP^-1-tPP^-1)\
& = det(P(B-tI)P^-1)\
& = det(P)det(B - tI) det(P^-1)\
& = det(P)det(B - tI) frac1det(P)\
& = det(B-tI)\
& = textcharpoly(B,t).
endalign*
This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.
edited 2 hours ago
J. W. Tanner
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
$endgroup$
Suppose $A$ and $B$ are square matrices such that $A = P B P^-1$ for some invertible matrix $P$. Then
beginalign*textcharpoly(A,t) & = det(A - tI)\
& = det(PBP^-1 - tI)\
& = det(PBP^-1-tPP^-1)\
& = det(P(B-tI)P^-1)\
& = det(P)det(B - tI) det(P^-1)\
& = det(P)det(B - tI) frac1det(P)\
& = det(B-tI)\
& = textcharpoly(B,t).
endalign*
This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.
edited 2 hours ago
J. W. Tanner
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
edited 2 hours ago
J. W. Tanner
2,9541318
edited 2 hours ago
J. W. Tanner
2,9541318
edited 2 hours ago
J. W. Tanner
2,9541318
2,9541318
answered 2 hours ago
johnny133253johnny133253
384110
answered 2 hours ago
johnny133253johnny133253
384110
answered 2 hours ago
johnny133253johnny133253
384110
384110
add a comment |
add a comment |
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