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Do similar matrices have same characteristic equations?

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1

$begingroup$

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

483

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1

$begingroup$

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

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

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

483

$endgroup$

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

483

share|cite|improve this question

asked 2 hours ago

Samurai BaleSamurai Bale

483

asked 2 hours ago

Samurai BaleSamurai Bale

483

asked 2 hours ago

Samurai BaleSamurai Bale

483

2 Answers
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3

$begingroup$

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 | 

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 2 hours ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

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3

$begingroup$

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 | 

3

$begingroup$

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

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 2 hours 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 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
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 2 hours ago

J. W. Tanner

2,9541318

answered 2 hours ago

johnny133253johnny133253

384110

$endgroup$

share|cite|improve this answer

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

answered 2 hours ago

johnny133253johnny133253

384110

answered 2 hours ago

johnny133253johnny133253

384110