Mixing
Inner product involving operators
Clash Royale CLAN TAG#URR8PPP
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I was wondering about inner product spaces and adjoints. The more I think about this the more I seem to be confusing myself.
I am not sure if given T and S as operators and an inner product space if the following are true:
A) $langle Tx,Txrangle = langle x,T^*Txrangle$ or $langle x,TT^*xrangle$
B) $langle Tx,Txrangle = langle T^*Tx,xrangle$ or $langle TT^*x,xrangle$
Does anything still hold if we have a $y neq x$?
C) $langle Tx,Tyrangle = langle x,T^*Tyrangle$ or $langle x,TT^*yrangle$
D) $langle Tx,Tyrangle = langle T^*Tx,yrangle$ or $langle TT^*x,yrangle$
Does anything still hold if we have a $S neq T$?
E) $langle Tx,Sxrangle = langle x,T^*Sxrangle$ or $langle x,ST^*xrangle$
F) $langle Tx,Sxrangle = langle TS^*x,xrangle$ or $langle S^*Tx,xrangle$
G) $langle Tx,Syrangle = langle x,T^*Syrangle$ or $langle x,ST^*yrangle$
H) $langle Tx,Syrangle = langle TS^*x,yrangle$ or $langle S^*Tx,yrangle$
I am sorry if this is a basic question, I just can't seem to find it in my book.
functional-analysis inner-product-space adjoint-operators
edited Jul 16 at 19:40
Javi
2,1631725
asked Jul 16 at 19:19
MathIsHard
1,122415
add a comment |Â
up vote
0
down vote
favorite
I was wondering about inner product spaces and adjoints. The more I think about this the more I seem to be confusing myself.
I am not sure if given T and S as operators and an inner product space if the following are true:
A) $langle Tx,Txrangle = langle x,T^*Txrangle$ or $langle x,TT^*xrangle$
B) $langle Tx,Txrangle = langle T^*Tx,xrangle$ or $langle TT^*x,xrangle$
Does anything still hold if we have a $y neq x$?
C) $langle Tx,Tyrangle = langle x,T^*Tyrangle$ or $langle x,TT^*yrangle$
D) $langle Tx,Tyrangle = langle T^*Tx,yrangle$ or $langle TT^*x,yrangle$
Does anything still hold if we have a $S neq T$?
E) $langle Tx,Sxrangle = langle x,T^*Sxrangle$ or $langle x,ST^*xrangle$
F) $langle Tx,Sxrangle = langle TS^*x,xrangle$ or $langle S^*Tx,xrangle$
G) $langle Tx,Syrangle = langle x,T^*Syrangle$ or $langle x,ST^*yrangle$
H) $langle Tx,Syrangle = langle TS^*x,yrangle$ or $langle S^*Tx,yrangle$
I am sorry if this is a basic question, I just can't seem to find it in my book.
functional-analysis inner-product-space adjoint-operators
edited Jul 16 at 19:40
Javi
2,1631725
asked Jul 16 at 19:19
MathIsHard
1,122415
4
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I was wondering about inner product spaces and adjoints. The more I think about this the more I seem to be confusing myself.
I am not sure if given T and S as operators and an inner product space if the following are true:
A) $langle Tx,Txrangle = langle x,T^*Txrangle$ or $langle x,TT^*xrangle$
B) $langle Tx,Txrangle = langle T^*Tx,xrangle$ or $langle TT^*x,xrangle$
Does anything still hold if we have a $y neq x$?
C) $langle Tx,Tyrangle = langle x,T^*Tyrangle$ or $langle x,TT^*yrangle$
D) $langle Tx,Tyrangle = langle T^*Tx,yrangle$ or $langle TT^*x,yrangle$
Does anything still hold if we have a $S neq T$?
E) $langle Tx,Sxrangle = langle x,T^*Sxrangle$ or $langle x,ST^*xrangle$
F) $langle Tx,Sxrangle = langle TS^*x,xrangle$ or $langle S^*Tx,xrangle$
G) $langle Tx,Syrangle = langle x,T^*Syrangle$ or $langle x,ST^*yrangle$
H) $langle Tx,Syrangle = langle TS^*x,yrangle$ or $langle S^*Tx,yrangle$
I am sorry if this is a basic question, I just can't seem to find it in my book.
functional-analysis inner-product-space adjoint-operators
edited Jul 16 at 19:40
Javi
2,1631725
asked Jul 16 at 19:19
MathIsHard
1,122415
I was wondering about inner product spaces and adjoints. The more I think about this the more I seem to be confusing myself.
I am not sure if given T and S as operators and an inner product space if the following are true:
A) $langle Tx,Txrangle = langle x,T^*Txrangle$ or $langle x,TT^*xrangle$
B) $langle Tx,Txrangle = langle T^*Tx,xrangle$ or $langle TT^*x,xrangle$
Does anything still hold if we have a $y neq x$?
C) $langle Tx,Tyrangle = langle x,T^*Tyrangle$ or $langle x,TT^*yrangle$
D) $langle Tx,Tyrangle = langle T^*Tx,yrangle$ or $langle TT^*x,yrangle$
Does anything still hold if we have a $S neq T$?
E) $langle Tx,Sxrangle = langle x,T^*Sxrangle$ or $langle x,ST^*xrangle$
F) $langle Tx,Sxrangle = langle TS^*x,xrangle$ or $langle S^*Tx,xrangle$
G) $langle Tx,Syrangle = langle x,T^*Syrangle$ or $langle x,ST^*yrangle$
H) $langle Tx,Syrangle = langle TS^*x,yrangle$ or $langle S^*Tx,yrangle$
I am sorry if this is a basic question, I just can't seem to find it in my book.
functional-analysis inner-product-space adjoint-operators
edited Jul 16 at 19:40
Javi
2,1631725
asked Jul 16 at 19:19
MathIsHard
1,122415
edited Jul 16 at 19:40
Javi
2,1631725
edited Jul 16 at 19:40
Javi
2,1631725
edited Jul 16 at 19:40
Javi
2,1631725
2,1631725
asked Jul 16 at 19:19
MathIsHard
1,122415
asked Jul 16 at 19:19
MathIsHard
1,122415
asked Jul 16 at 19:19
MathIsHard
1,122415
1,122415
4
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42
add a comment |Â
4
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42
4
4
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
The adjoint $A^*$ of a linear map $A:Uto V$ satisfies (is defined by) the following equation, for all $uin U, , vin V$
$$langle v, Aurangle=langle A^*v, urangle$$
Using symmetry of the inner product (and taking complex conjugates in the complex case), this also means
$$langle Au, vrangle=langle u, A^*vrangle$$
Applying these properly answers each of your questions. E.g. take $A=T, u=x, v=Tx$ for A), B)
answered Jul 16 at 20:31
Berci
56.4k23570
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
The adjoint $A^*$ of a linear map $A:Uto V$ satisfies (is defined by) the following equation, for all $uin U, , vin V$
$$langle v, Aurangle=langle A^*v, urangle$$
Using symmetry of the inner product (and taking complex conjugates in the complex case), this also means
$$langle Au, vrangle=langle u, A^*vrangle$$
Applying these properly answers each of your questions. E.g. take $A=T, u=x, v=Tx$ for A), B)
answered Jul 16 at 20:31
Berci
56.4k23570
add a comment |Â
up vote
2
down vote
accepted
The adjoint $A^*$ of a linear map $A:Uto V$ satisfies (is defined by) the following equation, for all $uin U, , vin V$
$$langle v, Aurangle=langle A^*v, urangle$$
Using symmetry of the inner product (and taking complex conjugates in the complex case), this also means
$$langle Au, vrangle=langle u, A^*vrangle$$
Applying these properly answers each of your questions. E.g. take $A=T, u=x, v=Tx$ for A), B)
answered Jul 16 at 20:31
Berci
56.4k23570
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
The adjoint $A^*$ of a linear map $A:Uto V$ satisfies (is defined by) the following equation, for all $uin U, , vin V$
$$langle v, Aurangle=langle A^*v, urangle$$
Using symmetry of the inner product (and taking complex conjugates in the complex case), this also means
$$langle Au, vrangle=langle u, A^*vrangle$$
Applying these properly answers each of your questions. E.g. take $A=T, u=x, v=Tx$ for A), B)
answered Jul 16 at 20:31
Berci
56.4k23570
The adjoint $A^*$ of a linear map $A:Uto V$ satisfies (is defined by) the following equation, for all $uin U, , vin V$
$$langle v, Aurangle=langle A^*v, urangle$$
Using symmetry of the inner product (and taking complex conjugates in the complex case), this also means
$$langle Au, vrangle=langle u, A^*vrangle$$
Applying these properly answers each of your questions. E.g. take $A=T, u=x, v=Tx$ for A), B)
answered Jul 16 at 20:31
Berci
56.4k23570
answered Jul 16 at 20:31
Berci
56.4k23570
answered Jul 16 at 20:31
Berci
56.4k23570
answered Jul 16 at 20:31
Berci
56.4k23570
56.4k23570
add a comment |Â
add a comment |Â
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4
Use langle and rangle for the inner product typesetting.
– Adrian Keister
Jul 16 at 19:29
Oops sorry about that! I just saw this. But I see it was fixed already... Thank you for fixing it. :/
– MathIsHard
Jul 16 at 19:42