What are the approximate eigenvalues of the right shift operator $R$ on $ell_infty$ The Next CEO of Stack OverflowApproximate eigenvalue and continuous spectrumWhat are the Eigenvectors of the curl operator?Spectral theory - continuous spectrumEigenvalues of Left Shift + Right Shift in $l_2([0,infty))$Eigenvalues of the circle over the Laplacian operatorIf an operator has a discrete spectrum eigenvalues, but infinitely many eigenvalues, are these necessarily countable?Spectrum of the right-shift operator on $ell ^2 (mathbbC)$, and a general spectrum questionAre the poles of the resolvent of a Hermitian operator real?The spectrum of a matrix operator is the set of its eigenvalues. But how to prove?Intuitive meaning of the members of the spectrum of a linear operator which are not eigenvalues
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What are the approximate eigenvalues of the right shift operator $R$ on $ell_infty$
The Next CEO of Stack OverflowApproximate eigenvalue and continuous spectrumWhat are the Eigenvectors of the curl operator?Spectral theory - continuous spectrumEigenvalues of Left Shift + Right Shift in $l_2([0,infty))$Eigenvalues of the circle over the Laplacian operatorIf an operator has a discrete spectrum eigenvalues, but infinitely many eigenvalues, are these necessarily countable?Spectrum of the right-shift operator on $ell ^2 (mathbbC)$, and a general spectrum questionAre the poles of the resolvent of a Hermitian operator real?The spectrum of a matrix operator is the set of its eigenvalues. But how to prove?Intuitive meaning of the members of the spectrum of a linear operator which are not eigenvalues
$begingroup$
I have shown that the spectrum of $R=zin C$.
Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?
If they are not how do I find the rest?
Thanks.
functional-analysis operator-theory spectral-theory
$endgroup$
add a comment |
$begingroup$
I have shown that the spectrum of $R=zin C$.
Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?
If they are not how do I find the rest?
Thanks.
functional-analysis operator-theory spectral-theory
$endgroup$
add a comment |
$begingroup$
I have shown that the spectrum of $R=zin C$.
Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?
If they are not how do I find the rest?
Thanks.
functional-analysis operator-theory spectral-theory
$endgroup$
I have shown that the spectrum of $R=zin C$.
Also, elements on the boundary of the spectrum are approximate eigenvalues, i.e. $forall |z|=1$, $z$ is an approx. eigenvalue. However, are these the only ones?
If they are not how do I find the rest?
Thanks.
functional-analysis operator-theory spectral-theory
functional-analysis operator-theory spectral-theory
edited Mar 18 at 11:59
Andrews
1,2812422
1,2812422
asked Mar 18 at 11:09
Jhon DoeJhon Doe
688414
688414
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add a comment |
1 Answer
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$begingroup$
For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have
$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$
Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.
From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$
$endgroup$
add a comment |
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1 Answer
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1 Answer
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$begingroup$
For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have
$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$
Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.
From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$
$endgroup$
add a comment |
$begingroup$
For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have
$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$
Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.
From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$
$endgroup$
add a comment |
$begingroup$
For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have
$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$
Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.
From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$
$endgroup$
For all $x in ell_infty$ we have $||Rx||=||x||.$ Thus, for $lambda$ with $|lambda | le 1$ we have
$$(*) ||Rx-lambda x|| ge | quad ||Rx||-|lambda| ||x|| quad|=(1-|lambda|)||x||.$$
Now suppose that $ lambda $ is an approximate eigenvalue . Then there is a sequence $(x_n)$ in $ell_infty$ such that $||x_n||=1$ for all $n$ and $(R-lambda I )x_n to 0$.
From $(*)$ we get: $(1-|lambda|)||x_n|| to 0.$ Since $||x_n||=1$ for all $n$, we derive $|lambda|=1.$
answered Mar 18 at 12:18
FredFred
48.8k11849
48.8k11849
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