Mixing
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 yesterday
Aweygan
12.9k21441
asked yesterday
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 yesterday
Aweygan
12.9k21441
asked yesterday
Zombiegit123
304113
1
Have you tried using induction?
– John Douma
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
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
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
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
yesterday
|
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 yesterday
Aweygan
12.9k21441
asked yesterday
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 yesterday
Aweygan
12.9k21441
asked yesterday
Zombiegit123
304113
edited yesterday
Aweygan
12.9k21441
asked yesterday
Zombiegit123
304113
edited yesterday
Aweygan
12.9k21441
edited yesterday
Aweygan
12.9k21441
edited yesterday
Aweygan
12.9k21441
12.9k21441
asked yesterday
Zombiegit123
304113
asked yesterday
Zombiegit123
304113
asked yesterday
Zombiegit123
304113
304113
1
Have you tried using induction?
– John Douma
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
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
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
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
yesterday
|
show 5 more comments
1
Have you tried using induction?
– John Douma
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
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
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
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
yesterday
1
1
Have you tried using induction?
– John Douma
yesterday
Have you tried using induction?
– John Douma
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
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
yesterday
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
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
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
yesterday
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
yesterday
|
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 yesterday
Aweygan
12.9k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
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 yesterday
Aweygan
12.9k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
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 yesterday
Aweygan
12.9k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
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 yesterday
Aweygan
12.9k21441
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 yesterday
Aweygan
12.9k21441
answered yesterday
Aweygan
12.9k21441
answered yesterday
Aweygan
12.9k21441
answered yesterday
Aweygan
12.9k21441
12.9k21441
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
add a comment |
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
I follow this quite well, but how do you get the first $T^2$ from that norm?
– Zombiegit123
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
Read it backwards. Since $|Tx|>0$, $|Tx|^2>0$, and thus $langle T^2x,xrangle>0$.
– Aweygan
yesterday
add a comment |
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1
Have you tried using induction?
– John Douma
yesterday
@JohnDouma Yes but I wasn't sure how to approach it. I don't use induction much.
– Zombiegit123
yesterday
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
yesterday
@JohnDouma Just did it for the base case. Not sure how to conclude it for k+1 though.
– Zombiegit123
yesterday
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
yesterday