Determine all real numbers that can be eigenvalues of operator A.
$begingroup$
Problem:
Let $A$ be some linear operator such that:
$$ (A^{2006}-I)^{2006}-I=0. $$
Determine all real numbers that can be eigenvalues of operator $A$.
Question: How to solve this problem?
My attempt:
$(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$
$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$
Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?
Thank you for any help.
linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A$ be some linear operator such that:
$$ (A^{2006}-I)^{2006}-I=0. $$
Determine all real numbers that can be eigenvalues of operator $A$.
Question: How to solve this problem?
My attempt:
$(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$
$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$
Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?
Thank you for any help.
linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization
$endgroup$
add a comment |
$begingroup$
Problem:
Let $A$ be some linear operator such that:
$$ (A^{2006}-I)^{2006}-I=0. $$
Determine all real numbers that can be eigenvalues of operator $A$.
Question: How to solve this problem?
My attempt:
$(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$
$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$
Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?
Thank you for any help.
linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization
$endgroup$
Problem:
Let $A$ be some linear operator such that:
$$ (A^{2006}-I)^{2006}-I=0. $$
Determine all real numbers that can be eigenvalues of operator $A$.
Question: How to solve this problem?
My attempt:
$(A^{2006}-I)^{2006}-I=0 implies (A^{2006}-I)^{2006}=I implies det((A^{2006}-I)^{2006})=1 implies [det(A^{2006}-I)]^{2006} = 1$
$A^{2006} - I = (A-x_1I)...(A-x_{2006}I) implies [det(A-x_1I)...det(A-x_{2006}I)]^{2006}=1$
Let's say that $A$ can have $n$ real eigenvalues and let's call them $lambda_j$ ($j=1,...n$). That means $det(A-lambda_jI)=0$. From this follow that no eigenvalue can be $x_j$ ($j=1,...2006$). That means that real numbers which satisfy equation $x^{2006}=1$ can not be eigenvalues. But it would seem to me that my "solution" is incomplete, for how do I know that there are no more restrictions on eigenvalues? And in my attempt I prejudiced that there is finate number $n$ of eigenvalues, how do I know there is not infinitley many eigenvalues?
Thank you for any help.
linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization
linear-algebra vector-spaces eigenvalues-eigenvectors linear-transformations diagonalization
asked Dec 16 '18 at 20:55
ThomThom
322110
322110
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2 Answers
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$begingroup$
$textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)
$endgroup$
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
add a comment |
$begingroup$
If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$
$endgroup$
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
$begingroup$
$textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)
$endgroup$
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
add a comment |
$begingroup$
$textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)
$endgroup$
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
add a comment |
$begingroup$
$textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)
$endgroup$
$textbf{Hint:}$ Suppose $lambda$ is an eigenvalue of $A$. Is $p(lambda)$ an eigenvalue of $p(A)$? ($p(cdot)$ denotes a polynomial.)
answered Dec 16 '18 at 21:05
SongSong
14.4k1635
14.4k1635
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
add a comment |
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
$begingroup$
Thanks for help. Does that theorem hold if functions $p$ is not polynomial? If not in general, what conditions must we put on $p$ so that it holds?
$endgroup$
– Thom
Dec 16 '18 at 21:23
1
1
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
$begingroup$
Well, $p$ can be any entire function or more generally, function which is analytic on a neighborhood of the spectrum of $A$. (A bit of knowledge on complex analysis may be needed to fully understand the concepts.) Under these conditions, $p(A)$ can be rigorously defined and $p(lambda)$ are exactly all eigenvalues of $p(A)$. It is called the spectral mapping theorem.
$endgroup$
– Song
Dec 16 '18 at 21:45
add a comment |
$begingroup$
If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$
$endgroup$
add a comment |
$begingroup$
If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$
$endgroup$
add a comment |
$begingroup$
If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$
$endgroup$
If $x$ is an eigenvalue it must satisfy $(x^{2006}-1)^{2006}-1=0$, so $x^{2006}-1$ is a $2006^{th}$ root of $1$. The only reals that satisfy that are $pm 1$. We can have $x=0$ or $x^{2006}=2, x=pm2^{1/2006}$
answered Dec 16 '18 at 21:08
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
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