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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

2

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Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?

linear-algebra

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asked 2 hours ago

Samurai BaleSamurai Bale

503

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2

$begingroup$

Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?

linear-algebra

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

503

$endgroup$

add a comment | 

$begingroup$

Since similar matrices have same eigenvalues and characteristic polynomials, then they must have the same characteristic equation, right?

linear-algebra

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

503

$endgroup$

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

503

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

503

asked 2 hours ago

Samurai BaleSamurai Bale

503

asked 2 hours ago

Samurai BaleSamurai Bale

503

2 Answers
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3

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Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
begin{align*}text{charpoly}(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) frac{1}{det(P)}\
& = det(B-tI)\
& = text{charpoly}(B,t).
end{align*}

This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.

share|cite|improve this answer

edited 1 hour ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

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add a comment | 

2

$begingroup$

Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.

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answered 2 hours ago

copper.hatcopper.hat

127k559160

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3

$begingroup$

Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
begin{align*}text{charpoly}(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) frac{1}{det(P)}\
& = det(B-tI)\
& = text{charpoly}(B,t).
end{align*}

This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.

share|cite|improve this answer

edited 1 hour ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

$endgroup$

add a comment | 

3

$begingroup$

Suppose $A$ and $B$ are square matrices such that $A = P B P^{-1}$ for some invertible matrix $P$. Then
begin{align*}text{charpoly}(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) frac{1}{det(P)}\
& = det(B-tI)\
& = text{charpoly}(B,t).
end{align*}

This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.

share|cite|improve this answer

edited 1 hour 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
begin{align*}text{charpoly}(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) frac{1}{det(P)}\
& = det(B-tI)\
& = text{charpoly}(B,t).
end{align*}

This shows that similar matrices have the same characteristic polynomial. Note that this proof relies on several facts. In particular, the determinant is multiplicative.

share|cite|improve this answer

edited 1 hour ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

$endgroup$

share|cite|improve this answer

edited 1 hour ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

edited 1 hour ago

J. W. Tanner

2,9541318

edited 1 hour ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

answered 2 hours ago

johnny133253johnny133253

384110

2

$begingroup$

Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.

share|cite|improve this answer

answered 2 hours ago

copper.hatcopper.hat

127k559160

$endgroup$

add a comment | 

2

$begingroup$

Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.

share|cite|improve this answer

answered 2 hours ago

copper.hatcopper.hat

127k559160

$endgroup$

add a comment | 

$begingroup$

Note that $det (lambda I -A) = det S det (lambda I -A) det S^{-1} = det (lambda I - S A S^{-1})$.

share|cite|improve this answer

answered 2 hours ago

copper.hatcopper.hat

127k559160

$endgroup$

share|cite|improve this answer

answered 2 hours ago

copper.hatcopper.hat

127k559160

answered 2 hours ago

copper.hatcopper.hat

127k559160

answered 2 hours ago

copper.hatcopper.hat

127k559160