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If $T:H rightarrow H$ is a bounded, self adjoint linear operator and $Tneq 0$ then $T^nneq 0$
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If $T:H rightarrow H$ is a bounded, self adjoint linear operator and $Tneq 0$ then $T^nneq 0$.
Could someone please expand on the general case of the problem here for me?
I see in the $n=2,4,8,dots$ they used what looks like a contrapositive proof. I.e., they seemed to show that if $T^2 = 0$ then $T=0$.
My questions:
1) This doesn't seem very intuitive to me. I would think that if we have a T so that $|T|<1$ then, as $n$ gets large the norm would go to $0$. The proof seems to be saying that it won't.
2) I don't understand what is meant by
"For the general case, if $T^n=0$, then $T^m=0$ for all $m geq n$,
contradicting what you have shown."
functional-analysis adjoint-operators
edited Aug 1 at 18:19
Adrian Keister
3,49321533
asked Aug 1 at 18:16
MathIsHard
1,122415
add a comment |Â
up vote
1
down vote
favorite
If $T:H rightarrow H$ is a bounded, self adjoint linear operator and $Tneq 0$ then $T^nneq 0$.
Could someone please expand on the general case of the problem here for me?
I see in the $n=2,4,8,dots$ they used what looks like a contrapositive proof. I.e., they seemed to show that if $T^2 = 0$ then $T=0$.
My questions:
1) This doesn't seem very intuitive to me. I would think that if we have a T so that $|T|<1$ then, as $n$ gets large the norm would go to $0$. The proof seems to be saying that it won't.
2) I don't understand what is meant by
"For the general case, if $T^n=0$, then $T^m=0$ for all $m geq n$,
contradicting what you have shown."
functional-analysis adjoint-operators
edited Aug 1 at 18:19
Adrian Keister
3,49321533
asked Aug 1 at 18:16
MathIsHard
1,122415
1
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If $T:H rightarrow H$ is a bounded, self adjoint linear operator and $Tneq 0$ then $T^nneq 0$.
Could someone please expand on the general case of the problem here for me?
I see in the $n=2,4,8,dots$ they used what looks like a contrapositive proof. I.e., they seemed to show that if $T^2 = 0$ then $T=0$.
My questions:
1) This doesn't seem very intuitive to me. I would think that if we have a T so that $|T|<1$ then, as $n$ gets large the norm would go to $0$. The proof seems to be saying that it won't.
2) I don't understand what is meant by
"For the general case, if $T^n=0$, then $T^m=0$ for all $m geq n$,
contradicting what you have shown."
functional-analysis adjoint-operators
edited Aug 1 at 18:19
Adrian Keister
3,49321533
asked Aug 1 at 18:16
MathIsHard
1,122415
If $T:H rightarrow H$ is a bounded, self adjoint linear operator and $Tneq 0$ then $T^nneq 0$.
Could someone please expand on the general case of the problem here for me?
I see in the $n=2,4,8,dots$ they used what looks like a contrapositive proof. I.e., they seemed to show that if $T^2 = 0$ then $T=0$.
My questions:
1) This doesn't seem very intuitive to me. I would think that if we have a T so that $|T|<1$ then, as $n$ gets large the norm would go to $0$. The proof seems to be saying that it won't.
2) I don't understand what is meant by
"For the general case, if $T^n=0$, then $T^m=0$ for all $m geq n$,
contradicting what you have shown."
functional-analysis adjoint-operators
edited Aug 1 at 18:19
Adrian Keister
3,49321533
asked Aug 1 at 18:16
MathIsHard
1,122415
edited Aug 1 at 18:19
Adrian Keister
3,49321533
edited Aug 1 at 18:19
Adrian Keister
3,49321533
edited Aug 1 at 18:19
Adrian Keister
3,49321533
3,49321533
asked Aug 1 at 18:16
MathIsHard
1,122415
asked Aug 1 at 18:16
MathIsHard
1,122415
asked Aug 1 at 18:16
MathIsHard
1,122415
1,122415
1
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22
add a comment |Â
1
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22
1
1
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
3
down vote
accepted
For the second point, it really is only necessary to prove that
$Tne0$ implies $T^2ne0$. As $T$ is self-adjoint, then so is $T^2$,
and iterating the argument gives $T^4ne0$, $T^8ne 0$ etc.
For a general $n$, there is $2^k>n$ and $T^2^kne 0$.
But $T^2^k=T^n T^2^k-n$ so $T^n$ must be nonzero.
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
add a comment |Â
up vote
2
down vote
For your question 1: there's a big difference between a finite $m$ and letting $mtoinfty.$ The proof to which you've linked doesn't say anything about letting $mtoinfty.$
answered Aug 1 at 18:23
Adrian Keister
3,49321533
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
For the second point, it really is only necessary to prove that
$Tne0$ implies $T^2ne0$. As $T$ is self-adjoint, then so is $T^2$,
and iterating the argument gives $T^4ne0$, $T^8ne 0$ etc.
For a general $n$, there is $2^k>n$ and $T^2^kne 0$.
But $T^2^k=T^n T^2^k-n$ so $T^n$ must be nonzero.
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
add a comment |Â
up vote
3
down vote
accepted
For the second point, it really is only necessary to prove that
$Tne0$ implies $T^2ne0$. As $T$ is self-adjoint, then so is $T^2$,
and iterating the argument gives $T^4ne0$, $T^8ne 0$ etc.
For a general $n$, there is $2^k>n$ and $T^2^kne 0$.
But $T^2^k=T^n T^2^k-n$ so $T^n$ must be nonzero.
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
For the second point, it really is only necessary to prove that
$Tne0$ implies $T^2ne0$. As $T$ is self-adjoint, then so is $T^2$,
and iterating the argument gives $T^4ne0$, $T^8ne 0$ etc.
For a general $n$, there is $2^k>n$ and $T^2^kne 0$.
But $T^2^k=T^n T^2^k-n$ so $T^n$ must be nonzero.
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
For the second point, it really is only necessary to prove that
$Tne0$ implies $T^2ne0$. As $T$ is self-adjoint, then so is $T^2$,
and iterating the argument gives $T^4ne0$, $T^8ne 0$ etc.
For a general $n$, there is $2^k>n$ and $T^2^kne 0$.
But $T^2^k=T^n T^2^k-n$ so $T^n$ must be nonzero.
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
answered Aug 1 at 18:21
Lord Shark the Unknown
84.2k950111
84.2k950111
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
add a comment |Â
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
Do you think the contrapositive that was used in the other post is not the best approach then? Even for the $n=2,4,8,...$ case?
– MathIsHard
Aug 1 at 18:24
add a comment |Â
up vote
2
down vote
For your question 1: there's a big difference between a finite $m$ and letting $mtoinfty.$ The proof to which you've linked doesn't say anything about letting $mtoinfty.$
answered Aug 1 at 18:23
Adrian Keister
3,49321533
add a comment |Â
up vote
2
down vote
For your question 1: there's a big difference between a finite $m$ and letting $mtoinfty.$ The proof to which you've linked doesn't say anything about letting $mtoinfty.$
answered Aug 1 at 18:23
Adrian Keister
3,49321533
add a comment |Â
up vote
2
down vote
up vote
2
down vote
For your question 1: there's a big difference between a finite $m$ and letting $mtoinfty.$ The proof to which you've linked doesn't say anything about letting $mtoinfty.$
answered Aug 1 at 18:23
Adrian Keister
3,49321533
For your question 1: there's a big difference between a finite $m$ and letting $mtoinfty.$ The proof to which you've linked doesn't say anything about letting $mtoinfty.$
answered Aug 1 at 18:23
Adrian Keister
3,49321533
answered Aug 1 at 18:23
Adrian Keister
3,49321533
answered Aug 1 at 18:23
Adrian Keister
3,49321533
answered Aug 1 at 18:23
Adrian Keister
3,49321533
3,49321533
add a comment |Â
add a comment |Â
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1
What this in effect says is that $T^n$ always has nonzero norm. That doesn't preclude say $|T^n|to0$.
– Lord Shark the Unknown
Aug 1 at 18:22