Normal Operator || T^2|| = ||T||^2 Announcing the arrival of Valued Associate #679: Cesar...
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Normal Operator || T^2|| = ||T||^2
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)If $S$ and $T$ are commuting, normal operators, then $ST$ is normal$T^2=I$ implies that $T$ is a normal operatorProving an operator is Self-adjoint using the Spectral TheoremEigenvalues of adjoint operator [General Case]diagonalizability implies existence of an inner product wrt an operator is normalNormal operator over real inner product spaceNormal matrix over real inner product space with real eigenvalues is Hermitianpolar form of unitary operatorWe have a linear operator T. Show $T^2=Id$ implies $T=T^*$Some property of Normal Operator
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
add a comment |
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
add a comment |
$begingroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
asked 1 hour ago
EricEric
798
$endgroup$
Given a complex inner product space X, and an operator T: X $rightarrow$ X is normal i.e. T$^*$T = TT$^*$.
How can we show ||T$^2$|| = ||T||$^2$?
By the definition of operator norm, it follows that ||T|| = sup $frac{||Tx||}{||x||}$ and ||T$^2$|| = sup $frac{||T^2x||}{||x||}$. Then I can express the numerator as a form of inner product. But I still am not able to make these two equal. Any good ideas?
linear-algebra
linear-algebra
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
asked 1 hour ago
EricEric
798
798
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
add a comment |
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
add a comment |
$begingroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
$endgroup$
If $T$ is normal, then $|Tx|^2=left<Tx,Txright>=left<x,T^*Txright>
=left<x,TT^*xright>=|T^*x|^2$, so $|Tx|=|T^*x|$ (and therefore
$|T|=|T^*|$).
Then (replacing $x$ by $Tx$)
$|T^2x|=|T^*Tx|$ so that $|T^2|=|T^*T|$. But also
$|Tx|^2=left<x,T^*Txright>le|T^*T||x|^2$ so that $|T|^2le|T^*T|
=|T^2|$. But $|T^2|le|T|^2$. We conclude that $|T^2|=|T|^2$
whenever $T$ is normal.
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
answered 19 mins ago
Lord Shark the UnknownLord Shark the Unknown
109k1163136
109k1163136
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
add a comment |
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
Thanks! By the way, how can we get (i) ||Tx|| = ||T*x|| implies that ||T|| = ||T*||?
$endgroup$
– Eric
3 mins ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
(ii) why can we have <x, T$^*$Tx> $leq$ ||T$^*$T|| ||x||$^2$?
$endgroup$
– Eric
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
$begingroup$
By the definition: $|T|=sup_{|x|=1}|Tx|$. @Eric
$endgroup$
– Lord Shark the Unknown
1 min ago
add a comment |
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