Mixing
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Image and pseduo-inverse of an operator
Clash Royale CLAN TAG#URR8PPP
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Let $mathcalH$ be a Hilbert space and $(e_n)_nin Bbb N$ be an orthonormal basis for $mathcalH$.
Define the surjective operator $Tin B(mathcalH)$ such that $Te_2n-1=frac12^ne_1$ and $Te_2n= e_n+1$ for each $ninBbb N$.
There are several questions:
$bullet$ What is the pseudo-inverse $T^dagger$ of $T$?
How is $T^dagger$ define?
$bullet$Decomposing $e_2n-1$ to $ f_n,1oplus f_n,2$ where $f_n,1 in R(T^*)$ and $f_n,2in ker T$ for each $n$. Could we conclude $f_n,1, f_n,2$? ($R(T^*)$ is the image of $T^*$ and $ker T$ is the kernel of $T$)
functional-analysis operator-theory hilbert-spaces
asked Jul 25 at 20:56
saeed
13810
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up vote
2
down vote
favorite
Let $mathcalH$ be a Hilbert space and $(e_n)_nin Bbb N$ be an orthonormal basis for $mathcalH$.
Define the surjective operator $Tin B(mathcalH)$ such that $Te_2n-1=frac12^ne_1$ and $Te_2n= e_n+1$ for each $ninBbb N$.
There are several questions:
$bullet$ What is the pseudo-inverse $T^dagger$ of $T$?
How is $T^dagger$ define?
$bullet$Decomposing $e_2n-1$ to $ f_n,1oplus f_n,2$ where $f_n,1 in R(T^*)$ and $f_n,2in ker T$ for each $n$. Could we conclude $f_n,1, f_n,2$? ($R(T^*)$ is the image of $T^*$ and $ker T$ is the kernel of $T$)
functional-analysis operator-theory hilbert-spaces
asked Jul 25 at 20:56
saeed
13810
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $mathcalH$ be a Hilbert space and $(e_n)_nin Bbb N$ be an orthonormal basis for $mathcalH$.
Define the surjective operator $Tin B(mathcalH)$ such that $Te_2n-1=frac12^ne_1$ and $Te_2n= e_n+1$ for each $ninBbb N$.
There are several questions:
$bullet$ What is the pseudo-inverse $T^dagger$ of $T$?
How is $T^dagger$ define?
$bullet$Decomposing $e_2n-1$ to $ f_n,1oplus f_n,2$ where $f_n,1 in R(T^*)$ and $f_n,2in ker T$ for each $n$. Could we conclude $f_n,1, f_n,2$? ($R(T^*)$ is the image of $T^*$ and $ker T$ is the kernel of $T$)
functional-analysis operator-theory hilbert-spaces
asked Jul 25 at 20:56
saeed
13810
Let $mathcalH$ be a Hilbert space and $(e_n)_nin Bbb N$ be an orthonormal basis for $mathcalH$.
Define the surjective operator $Tin B(mathcalH)$ such that $Te_2n-1=frac12^ne_1$ and $Te_2n= e_n+1$ for each $ninBbb N$.
There are several questions:
$bullet$ What is the pseudo-inverse $T^dagger$ of $T$?
How is $T^dagger$ define?
$bullet$Decomposing $e_2n-1$ to $ f_n,1oplus f_n,2$ where $f_n,1 in R(T^*)$ and $f_n,2in ker T$ for each $n$. Could we conclude $f_n,1, f_n,2$? ($R(T^*)$ is the image of $T^*$ and $ker T$ is the kernel of $T$)
functional-analysis operator-theory hilbert-spaces
asked Jul 25 at 20:56
saeed
13810
edited Jul 25 at 21:49
asked Jul 25 at 20:56
saeed
13810
asked Jul 25 at 20:56
saeed
13810
asked Jul 25 at 20:56
saeed
13810
13810
add a comment |Â
add a comment |Â
1 Answer
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The answer to the previous version of the question, which was is $e_2k-1 in R(T^*)$, is no.
We have
beginalign
T^*x &= sum_n=1^infty langle T^*x, e_nrangle e_n \
&= sum_n=1^infty langle x, Te_nrangle e_n \
&= sum_n=1^infty langle x, Te_2n-1rangle e_2n-1 + sum_n=1^infty langle x, Te_2n-1rangle e_2n\
&= langle x, e_1ranglesum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n
endalign
so if $T^*x = e_2k-1$, we must have $$1 = langle T^*x, e_2k-1rangle = langle x, Te_1rangle = frac12^klangle x, e_1rangle$$ so $langle x, e_1rangle = 2^k$ but then
$$T^*x = 2^ksum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n ne e_2k-1$$
answered Jul 25 at 21:58
mechanodroid
22.2k52041
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The answer to the previous version of the question, which was is $e_2k-1 in R(T^*)$, is no.
We have
beginalign
T^*x &= sum_n=1^infty langle T^*x, e_nrangle e_n \
&= sum_n=1^infty langle x, Te_nrangle e_n \
&= sum_n=1^infty langle x, Te_2n-1rangle e_2n-1 + sum_n=1^infty langle x, Te_2n-1rangle e_2n\
&= langle x, e_1ranglesum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n
endalign
so if $T^*x = e_2k-1$, we must have $$1 = langle T^*x, e_2k-1rangle = langle x, Te_1rangle = frac12^klangle x, e_1rangle$$ so $langle x, e_1rangle = 2^k$ but then
$$T^*x = 2^ksum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n ne e_2k-1$$
answered Jul 25 at 21:58
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
The answer to the previous version of the question, which was is $e_2k-1 in R(T^*)$, is no.
We have
beginalign
T^*x &= sum_n=1^infty langle T^*x, e_nrangle e_n \
&= sum_n=1^infty langle x, Te_nrangle e_n \
&= sum_n=1^infty langle x, Te_2n-1rangle e_2n-1 + sum_n=1^infty langle x, Te_2n-1rangle e_2n\
&= langle x, e_1ranglesum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n
endalign
so if $T^*x = e_2k-1$, we must have $$1 = langle T^*x, e_2k-1rangle = langle x, Te_1rangle = frac12^klangle x, e_1rangle$$ so $langle x, e_1rangle = 2^k$ but then
$$T^*x = 2^ksum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n ne e_2k-1$$
answered Jul 25 at 21:58
mechanodroid
22.2k52041
add a comment |Â
up vote
0
down vote
up vote
0
down vote
The answer to the previous version of the question, which was is $e_2k-1 in R(T^*)$, is no.
We have
beginalign
T^*x &= sum_n=1^infty langle T^*x, e_nrangle e_n \
&= sum_n=1^infty langle x, Te_nrangle e_n \
&= sum_n=1^infty langle x, Te_2n-1rangle e_2n-1 + sum_n=1^infty langle x, Te_2n-1rangle e_2n\
&= langle x, e_1ranglesum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n
endalign
so if $T^*x = e_2k-1$, we must have $$1 = langle T^*x, e_2k-1rangle = langle x, Te_1rangle = frac12^klangle x, e_1rangle$$ so $langle x, e_1rangle = 2^k$ but then
$$T^*x = 2^ksum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n ne e_2k-1$$
answered Jul 25 at 21:58
mechanodroid
22.2k52041
The answer to the previous version of the question, which was is $e_2k-1 in R(T^*)$, is no.
We have
beginalign
T^*x &= sum_n=1^infty langle T^*x, e_nrangle e_n \
&= sum_n=1^infty langle x, Te_nrangle e_n \
&= sum_n=1^infty langle x, Te_2n-1rangle e_2n-1 + sum_n=1^infty langle x, Te_2n-1rangle e_2n\
&= langle x, e_1ranglesum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n
endalign
so if $T^*x = e_2k-1$, we must have $$1 = langle T^*x, e_2k-1rangle = langle x, Te_1rangle = frac12^klangle x, e_1rangle$$ so $langle x, e_1rangle = 2^k$ but then
$$T^*x = 2^ksum_n=1^infty frac12^n e_2n-1 + sum_n=1^infty langle x, e_n+1rangle e_2n ne e_2k-1$$
answered Jul 25 at 21:58
mechanodroid
22.2k52041
answered Jul 25 at 21:58
mechanodroid
22.2k52041
answered Jul 25 at 21:58
mechanodroid
22.2k52041
answered Jul 25 at 21:58
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
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