Proof regarding self-adjoint linear operators.
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Let $H$ be a Hilbert space, and let $T : H rightarrow H$ be a bounded self-adjoint linear operator, with $T neq 0.$
I need to show that $T^{2^k} neq 0$ $forall k in mathbb{N}$.
Here's what I've done so far:
$T^2x = T(Tx)$ and so $T^{2^k}x = T(T^{2k-1}x)$.
Hence, as T is self-adjoint, $<Tx,y> = <x,T^*y>$ and so $<T^{2k}x,y> = <x,(T^{2k})^*y>$.
However, I'm struggling to go from here, any help is appreciated.
functional-analysis hilbert-spaces adjoint-operators
edited Nov 19 at 16:46
Aweygan
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
|
show 5 more comments
up vote
0
down vote
favorite
Let $H$ be a Hilbert space, and let $T : H rightarrow H$ be a bounded self-adjoint linear operator, with $T neq 0.$
I need to show that $T^{2^k} neq 0$ $forall k in mathbb{N}$.
Here's what I've done so far:
$T^2x = T(Tx)$ and so $T^{2^k}x = T(T^{2k-1}x)$.
Hence, as T is self-adjoint, $<Tx,y> = <x,T^*y>$ and so $<T^{2k}x,y> = <x,(T^{2k})^*y>$.
However, I'm struggling to go from here, any help is appreciated.
functional-analysis hilbert-spaces adjoint-operators
edited Nov 19 at 16:46
Aweygan
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
1
Have you tried using induction?
– John Douma
Nov 19 at 15:43
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
1
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
2
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49
|
show 5 more comments
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $H$ be a Hilbert space, and let $T : H rightarrow H$ be a bounded self-adjoint linear operator, with $T neq 0.$
I need to show that $T^{2^k} neq 0$ $forall k in mathbb{N}$.
Here's what I've done so far:
$T^2x = T(Tx)$ and so $T^{2^k}x = T(T^{2k-1}x)$.
Hence, as T is self-adjoint, $<Tx,y> = <x,T^*y>$ and so $<T^{2k}x,y> = <x,(T^{2k})^*y>$.
However, I'm struggling to go from here, any help is appreciated.
functional-analysis hilbert-spaces adjoint-operators
edited Nov 19 at 16:46
Aweygan
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
Let $H$ be a Hilbert space, and let $T : H rightarrow H$ be a bounded self-adjoint linear operator, with $T neq 0.$
I need to show that $T^{2^k} neq 0$ $forall k in mathbb{N}$.
Here's what I've done so far:
$T^2x = T(Tx)$ and so $T^{2^k}x = T(T^{2k-1}x)$.
Hence, as T is self-adjoint, $<Tx,y> = <x,T^*y>$ and so $<T^{2k}x,y> = <x,(T^{2k})^*y>$.
However, I'm struggling to go from here, any help is appreciated.
functional-analysis hilbert-spaces adjoint-operators
functional-analysis hilbert-spaces adjoint-operators
edited Nov 19 at 16:46
Aweygan
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
edited Nov 19 at 16:46
Aweygan
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
edited Nov 19 at 16:46
Aweygan
13.1k21441
edited Nov 19 at 16:46
Aweygan
13.1k21441
edited Nov 19 at 16:46
Aweygan
13.1k21441
13.1k21441
asked Nov 19 at 15:41
Zombiegit123
304113
asked Nov 19 at 15:41
Zombiegit123
304113
asked Nov 19 at 15:41
Zombiegit123
304113
304113
1
Have you tried using induction?
– John Douma
Nov 19 at 15:43
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
1
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
2
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49
|
show 5 more comments
1
Have you tried using induction?
– John Douma
Nov 19 at 15:43
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
1
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
2
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49
1
1
Have you tried using induction?
– John Douma
Nov 19 at 15:43
Have you tried using induction?
– John Douma
Nov 19 at 15:43
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
1
1
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
2
2
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49
|
show 5 more comments
1 Answer
1
active
oldest
votes
up vote
4
down vote
accepted
Induction certainly does make this proof go by much smoother. Here's how one should proceed:
Since $Tneq0$, there is some nonzero $xin H$ such that $Txneq0$. Hence $|Tx|>0$,
$$langle T^2x,xrangle=langle Tx,Txrangle=|Tx|^2>0,$$
and thus $T^2neq0$. (Can you see how self-adjointness is used? )
For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.
answered Nov 19 at 16:51
Aweygan
13.1k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Induction certainly does make this proof go by much smoother. Here's how one should proceed:
Since $Tneq0$, there is some nonzero $xin H$ such that $Txneq0$. Hence $|Tx|>0$,
$$langle T^2x,xrangle=langle Tx,Txrangle=|Tx|^2>0,$$
and thus $T^2neq0$. (Can you see how self-adjointness is used? )
For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.
answered Nov 19 at 16:51
Aweygan
13.1k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
add a comment |
up vote
4
down vote
accepted
Induction certainly does make this proof go by much smoother. Here's how one should proceed:
Since $Tneq0$, there is some nonzero $xin H$ such that $Txneq0$. Hence $|Tx|>0$,
$$langle T^2x,xrangle=langle Tx,Txrangle=|Tx|^2>0,$$
and thus $T^2neq0$. (Can you see how self-adjointness is used? )
For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.
answered Nov 19 at 16:51
Aweygan
13.1k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
add a comment |
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Induction certainly does make this proof go by much smoother. Here's how one should proceed:
Since $Tneq0$, there is some nonzero $xin H$ such that $Txneq0$. Hence $|Tx|>0$,
$$langle T^2x,xrangle=langle Tx,Txrangle=|Tx|^2>0,$$
and thus $T^2neq0$. (Can you see how self-adjointness is used? )
For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.
answered Nov 19 at 16:51
Aweygan
13.1k21441
Induction certainly does make this proof go by much smoother. Here's how one should proceed:
Since $Tneq0$, there is some nonzero $xin H$ such that $Txneq0$. Hence $|Tx|>0$,
$$langle T^2x,xrangle=langle Tx,Txrangle=|Tx|^2>0,$$
and thus $T^2neq0$. (Can you see how self-adjointness is used? )
For the induction step, just repeat the same proof with $T^{2^k}$ taking the place of $T$.
answered Nov 19 at 16:51
Aweygan
13.1k21441
answered Nov 19 at 16:51
Aweygan
13.1k21441
answered Nov 19 at 16:51
Aweygan
13.1k21441
answered Nov 19 at 16:51
Aweygan
13.1k21441
13.1k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
add a comment |
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
Nov 19 at 17:44
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
Nov 19 at 17:46
add a comment |
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Have you tried using induction?
– John Douma
Nov 19 at 15:43
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
Nov 19 at 15:45
1
Can you see why the result is true for $k=0$? Assume it is true for arbitrary $k$ and show it must be true for $k+1$.
– John Douma
Nov 19 at 15:46
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
Nov 19 at 15:49
2
Additionally, you should really use self-adjointness. For the equality $langle Tx,yrangle=langle x,T^ast yrangle$ you do not need self-adjointness, it's just the definition of the adjoint. But the statement you want to prove does not hold for arbitary bounded operators.
– MaoWao
Nov 19 at 15:49