Spectrum of $T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$ in $ell^2$ for $varepsilon>0$ The 2019 Stack Overflow Developer Survey Results Are InSpectrum of shift-operatorWhat's the spectrum of this operator in $ell^2$?Find the spectrum of the operator $T: ell^2(mathbbC) to ell^2(mathbbC)$ defined by $(Tx)_n = fracx_nn$Point spectrum of operator on $ell^2$?What is the spectrum of the sequence operator $B: (x_1,x_2,ldots) rightarrow (0,x_1,frac12x_2,ldots,frac1nx_n,ldots)$?range of weighted shift operatorA Spectrum of a compact operator in $ell^p$Spectrum of Diagonal Operator in $ell^2$Spectrum of $ T(x_1, x_2, x_3, x_4, ldots )=(-x_2, x_1, -x_4, x_3, ldots). $Spectrum of the right-shift operator on $ell ^2 (mathbbC)$, and a general spectrum question
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Spectrum of $T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$ in $ell^2$ for $varepsilon>0$
The 2019 Stack Overflow Developer Survey Results Are InSpectrum of shift-operatorWhat's the spectrum of this operator in $ell^2$?Find the spectrum of the operator $T: ell^2(mathbbC) to ell^2(mathbbC)$ defined by $(Tx)_n = fracx_nn$Point spectrum of operator on $ell^2$?What is the spectrum of the sequence operator $B: (x_1,x_2,ldots) rightarrow (0,x_1,frac12x_2,ldots,frac1nx_n,ldots)$?range of weighted shift operatorA Spectrum of a compact operator in $ell^p$Spectrum of Diagonal Operator in $ell^2$Spectrum of $ T(x_1, x_2, x_3, x_4, ldots )=(-x_2, x_1, -x_4, x_3, ldots). $Spectrum of the right-shift operator on $ell ^2 (mathbbC)$, and a general spectrum question
$begingroup$
Let $varepsilon>0$ and $T:ell^2to ell^2 $ the operator defined by
$$T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$$
Can you help me to calculate the spectrum of $T$, please?.
I think that $sigma_p(T)=varepsilon$ if $varepsilongeq1$ and that $sigma_p(T)=emptyset$ otherwise, because $(T-lambda I)(x_1,ldots) =0$ for $(x_1,ldots) neq 0$ iff $lambda neq 0, x_1 neq 0, lambda=varepsilon$ and
$$x_n+1=fracx_1varepsilon^n, n>1.$$
Am I right?
functional-analysis operator-theory spectral-theory
asked Apr 6 at 20:40
krenickkrenick
374
$endgroup$
add a comment |
$begingroup$
Let $varepsilon>0$ and $T:ell^2to ell^2 $ the operator defined by
$$T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$$
Can you help me to calculate the spectrum of $T$, please?.
I think that $sigma_p(T)=varepsilon$ if $varepsilongeq1$ and that $sigma_p(T)=emptyset$ otherwise, because $(T-lambda I)(x_1,ldots) =0$ for $(x_1,ldots) neq 0$ iff $lambda neq 0, x_1 neq 0, lambda=varepsilon$ and
$$x_n+1=fracx_1varepsilon^n, n>1.$$
Am I right?
functional-analysis operator-theory spectral-theory
asked Apr 6 at 20:40
krenickkrenick
374
$endgroup$
3
$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
1
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54
add a comment |
$begingroup$
Let $varepsilon>0$ and $T:ell^2to ell^2 $ the operator defined by
$$T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$$
Can you help me to calculate the spectrum of $T$, please?.
I think that $sigma_p(T)=varepsilon$ if $varepsilongeq1$ and that $sigma_p(T)=emptyset$ otherwise, because $(T-lambda I)(x_1,ldots) =0$ for $(x_1,ldots) neq 0$ iff $lambda neq 0, x_1 neq 0, lambda=varepsilon$ and
$$x_n+1=fracx_1varepsilon^n, n>1.$$
Am I right?
functional-analysis operator-theory spectral-theory
asked Apr 6 at 20:40
krenickkrenick
374
$endgroup$
Let $varepsilon>0$ and $T:ell^2to ell^2 $ the operator defined by
$$T(x_1, x_2, ldots)=(varepsilon x_1, x_1,x_2,ldots)$$
Can you help me to calculate the spectrum of $T$, please?.
I think that $sigma_p(T)=varepsilon$ if $varepsilongeq1$ and that $sigma_p(T)=emptyset$ otherwise, because $(T-lambda I)(x_1,ldots) =0$ for $(x_1,ldots) neq 0$ iff $lambda neq 0, x_1 neq 0, lambda=varepsilon$ and
$$x_n+1=fracx_1varepsilon^n, n>1.$$
Am I right?
functional-analysis operator-theory spectral-theory
functional-analysis operator-theory spectral-theory
asked Apr 6 at 20:40
krenickkrenick
374
asked Apr 6 at 20:40
krenickkrenick
374
edited Apr 6 at 21:01
krenick
asked Apr 6 at 20:40
krenickkrenick
374
asked Apr 6 at 20:40
krenickkrenick
374
asked Apr 6 at 20:40
krenickkrenick
374
374
3
$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
1
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54
add a comment |
3
$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
1
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54
3
3
$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
1
1
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your conclusion regarding the point spectrum is true, with the slight alteration that $sigma_p (T)=epsilon$ when $epsilon > 1$, rather than $geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-lambda I$ for $lambda neq epsilon$.
Clearly $T$ is not surjective, so suppose $lambda neq 0$. We have $(T-lambda I)(x) = ((epsilon-lambda)x_1,x_1-lambda x_2,...) =y implies x_1 = fracy_1epsilon - lambda$, $x_n+1 = fracx_n-y_n+1lambda$, so
$$x_n =fracx_1-sum_k=2^n lambda^k-2 y_k lambda^n-1$$
In particular, if $y_k$ is nonzero for only finitely many $k in mathbbN$, so $exists$ $N in mathbbN$ such that $y_n =0$ for $n>N$, then $forall n>N$ $x_n=fracx_Nlambda^n-N$. Choosing $y in ell^2$ such that $x_N neq 0$ ( $y_k = delta_N,k$ works), we have $x in ell^2 iff |lambda| > 1$. That is to say, $T-lambda I$ is not surjective for $|lambda| leq 1$.
Further, this shows that for $|lambda|>1$, im$(T-lambda I)$ includes all sequences $y in ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-lambda I$ for $|lambda|>1$, then, it suffices to show $x in ell^2$ when $y in ell^2$ and $y_1=0$. In this case, $x_n = -sum_k=2^n lambda^k-n-1y_k$, so
$|x_n|^2 leq sum_k=1^n |lambda|^2(k-n-1)|y_k|^2$, and thus
$$sum_n=1^N |x_n|^2 leq sum_n=1^N sum_k=1^n |lambda|^2(k-n-1)|y_k|^2 = sum_k=1^N |y_k|^2 sum_n=k^N |lambda|^2(k-n-1) \ = sum_k=1^N |y_k|^2 sum_n=1^N+1-k |lambda|^-2n leq frac1lambdasum_k=1^N |y_k|^2 leq frac^2lambda$$
Showing $x in ell^2$, and hence that $sigma(T)=lambda cup epsilon$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
answered Apr 7 at 6:07
jawheelejawheele
48439
$endgroup$
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
add a comment |
Your Answer
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$begingroup$
Your conclusion regarding the point spectrum is true, with the slight alteration that $sigma_p (T)=epsilon$ when $epsilon > 1$, rather than $geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-lambda I$ for $lambda neq epsilon$.
Clearly $T$ is not surjective, so suppose $lambda neq 0$. We have $(T-lambda I)(x) = ((epsilon-lambda)x_1,x_1-lambda x_2,...) =y implies x_1 = fracy_1epsilon - lambda$, $x_n+1 = fracx_n-y_n+1lambda$, so
$$x_n =fracx_1-sum_k=2^n lambda^k-2 y_k lambda^n-1$$
In particular, if $y_k$ is nonzero for only finitely many $k in mathbbN$, so $exists$ $N in mathbbN$ such that $y_n =0$ for $n>N$, then $forall n>N$ $x_n=fracx_Nlambda^n-N$. Choosing $y in ell^2$ such that $x_N neq 0$ ( $y_k = delta_N,k$ works), we have $x in ell^2 iff |lambda| > 1$. That is to say, $T-lambda I$ is not surjective for $|lambda| leq 1$.
Further, this shows that for $|lambda|>1$, im$(T-lambda I)$ includes all sequences $y in ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-lambda I$ for $|lambda|>1$, then, it suffices to show $x in ell^2$ when $y in ell^2$ and $y_1=0$. In this case, $x_n = -sum_k=2^n lambda^k-n-1y_k$, so
$|x_n|^2 leq sum_k=1^n |lambda|^2(k-n-1)|y_k|^2$, and thus
$$sum_n=1^N |x_n|^2 leq sum_n=1^N sum_k=1^n |lambda|^2(k-n-1)|y_k|^2 = sum_k=1^N |y_k|^2 sum_n=k^N |lambda|^2(k-n-1) \ = sum_k=1^N |y_k|^2 sum_n=1^N+1-k |lambda|^-2n leq frac1lambdasum_k=1^N |y_k|^2 leq frac^2lambda$$
Showing $x in ell^2$, and hence that $sigma(T)=lambda cup epsilon$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
answered Apr 7 at 6:07
jawheelejawheele
48439
$endgroup$
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
add a comment |
$begingroup$
Your conclusion regarding the point spectrum is true, with the slight alteration that $sigma_p (T)=epsilon$ when $epsilon > 1$, rather than $geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-lambda I$ for $lambda neq epsilon$.
Clearly $T$ is not surjective, so suppose $lambda neq 0$. We have $(T-lambda I)(x) = ((epsilon-lambda)x_1,x_1-lambda x_2,...) =y implies x_1 = fracy_1epsilon - lambda$, $x_n+1 = fracx_n-y_n+1lambda$, so
$$x_n =fracx_1-sum_k=2^n lambda^k-2 y_k lambda^n-1$$
In particular, if $y_k$ is nonzero for only finitely many $k in mathbbN$, so $exists$ $N in mathbbN$ such that $y_n =0$ for $n>N$, then $forall n>N$ $x_n=fracx_Nlambda^n-N$. Choosing $y in ell^2$ such that $x_N neq 0$ ( $y_k = delta_N,k$ works), we have $x in ell^2 iff |lambda| > 1$. That is to say, $T-lambda I$ is not surjective for $|lambda| leq 1$.
Further, this shows that for $|lambda|>1$, im$(T-lambda I)$ includes all sequences $y in ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-lambda I$ for $|lambda|>1$, then, it suffices to show $x in ell^2$ when $y in ell^2$ and $y_1=0$. In this case, $x_n = -sum_k=2^n lambda^k-n-1y_k$, so
$|x_n|^2 leq sum_k=1^n |lambda|^2(k-n-1)|y_k|^2$, and thus
$$sum_n=1^N |x_n|^2 leq sum_n=1^N sum_k=1^n |lambda|^2(k-n-1)|y_k|^2 = sum_k=1^N |y_k|^2 sum_n=k^N |lambda|^2(k-n-1) \ = sum_k=1^N |y_k|^2 sum_n=1^N+1-k |lambda|^-2n leq frac1lambdasum_k=1^N |y_k|^2 leq frac^2lambda$$
Showing $x in ell^2$, and hence that $sigma(T)=lambda cup epsilon$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
answered Apr 7 at 6:07
jawheelejawheele
48439
$endgroup$
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
add a comment |
$begingroup$
Your conclusion regarding the point spectrum is true, with the slight alteration that $sigma_p (T)=epsilon$ when $epsilon > 1$, rather than $geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-lambda I$ for $lambda neq epsilon$.
Clearly $T$ is not surjective, so suppose $lambda neq 0$. We have $(T-lambda I)(x) = ((epsilon-lambda)x_1,x_1-lambda x_2,...) =y implies x_1 = fracy_1epsilon - lambda$, $x_n+1 = fracx_n-y_n+1lambda$, so
$$x_n =fracx_1-sum_k=2^n lambda^k-2 y_k lambda^n-1$$
In particular, if $y_k$ is nonzero for only finitely many $k in mathbbN$, so $exists$ $N in mathbbN$ such that $y_n =0$ for $n>N$, then $forall n>N$ $x_n=fracx_Nlambda^n-N$. Choosing $y in ell^2$ such that $x_N neq 0$ ( $y_k = delta_N,k$ works), we have $x in ell^2 iff |lambda| > 1$. That is to say, $T-lambda I$ is not surjective for $|lambda| leq 1$.
Further, this shows that for $|lambda|>1$, im$(T-lambda I)$ includes all sequences $y in ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-lambda I$ for $|lambda|>1$, then, it suffices to show $x in ell^2$ when $y in ell^2$ and $y_1=0$. In this case, $x_n = -sum_k=2^n lambda^k-n-1y_k$, so
$|x_n|^2 leq sum_k=1^n |lambda|^2(k-n-1)|y_k|^2$, and thus
$$sum_n=1^N |x_n|^2 leq sum_n=1^N sum_k=1^n |lambda|^2(k-n-1)|y_k|^2 = sum_k=1^N |y_k|^2 sum_n=k^N |lambda|^2(k-n-1) \ = sum_k=1^N |y_k|^2 sum_n=1^N+1-k |lambda|^-2n leq frac1lambdasum_k=1^N |y_k|^2 leq frac^2lambda$$
Showing $x in ell^2$, and hence that $sigma(T)=lambda cup epsilon$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
answered Apr 7 at 6:07
jawheelejawheele
48439
$endgroup$
Your conclusion regarding the point spectrum is true, with the slight alteration that $sigma_p (T)=epsilon$ when $epsilon > 1$, rather than $geq 1$. To determine the residual and continuous spectra, we investigate the surjectivity of $T-lambda I$ for $lambda neq epsilon$.
Clearly $T$ is not surjective, so suppose $lambda neq 0$. We have $(T-lambda I)(x) = ((epsilon-lambda)x_1,x_1-lambda x_2,...) =y implies x_1 = fracy_1epsilon - lambda$, $x_n+1 = fracx_n-y_n+1lambda$, so
$$x_n =fracx_1-sum_k=2^n lambda^k-2 y_k lambda^n-1$$
In particular, if $y_k$ is nonzero for only finitely many $k in mathbbN$, so $exists$ $N in mathbbN$ such that $y_n =0$ for $n>N$, then $forall n>N$ $x_n=fracx_Nlambda^n-N$. Choosing $y in ell^2$ such that $x_N neq 0$ ( $y_k = delta_N,k$ works), we have $x in ell^2 iff |lambda| > 1$. That is to say, $T-lambda I$ is not surjective for $|lambda| leq 1$.
Further, this shows that for $|lambda|>1$, im$(T-lambda I)$ includes all sequences $y in ell^2$ with only finitely many nonzero terms. To show surjectivity of $T-lambda I$ for $|lambda|>1$, then, it suffices to show $x in ell^2$ when $y in ell^2$ and $y_1=0$. In this case, $x_n = -sum_k=2^n lambda^k-n-1y_k$, so
$|x_n|^2 leq sum_k=1^n |lambda|^2(k-n-1)|y_k|^2$, and thus
$$sum_n=1^N |x_n|^2 leq sum_n=1^N sum_k=1^n |lambda|^2(k-n-1)|y_k|^2 = sum_k=1^N |y_k|^2 sum_n=k^N |lambda|^2(k-n-1) \ = sum_k=1^N |y_k|^2 sum_n=1^N+1-k |lambda|^-2n leq frac1lambdasum_k=1^N |y_k|^2 leq frac^2lambda$$
Showing $x in ell^2$, and hence that $sigma(T)=lambda cup epsilon$. I'll leave it to you to determine the breakdown into the residual/continuous spectra, if that's of concern.
answered Apr 7 at 6:07
jawheelejawheele
48439
answered Apr 7 at 6:07
jawheelejawheele
48439
answered Apr 7 at 6:07
jawheelejawheele
48439
answered Apr 7 at 6:07
jawheelejawheele
48439
48439
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
add a comment |
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
Thank you so much!. I think that if $varepsilon=1$, then $(T-I)(1/2,1/2,1/2,ldots)=(0,0,0, ldots)$ and because of that $sigma_p(T)=1$.
$endgroup$
– krenick
Apr 7 at 14:16
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
@krenick $(1/2,1/2,...) notin ell^2$.
$endgroup$
– jawheele
Apr 7 at 14:35
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
$begingroup$
You are right. Sorry, I don't know why I was thinking about the series $sum_n=0^infty frac12^n$ instead of $sum_n=0^infty frac12^2$.
$endgroup$
– krenick
Apr 7 at 15:03
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$begingroup$
What have you done so far ?
$endgroup$
– Victoria M
Apr 6 at 20:43
1
$begingroup$
What is special about $epsilon > 1$ for your expectation regarding the point spectrum?
$endgroup$
– jawheele
Apr 6 at 20:54