Mixing
Prove that $v in U^perp $.
$begingroup$
Let $U$ be a subspace of $V$. Assume that there exists $vin V$ such that $langle v,urangle +langle u,vrangle le langle u,urangle forall uin U$
Prove that $v in U^perp $.
Let $uin U$ then we need to show that $langle u,vrangle =0$
Now $langle v,urangle +langle u,vrangle le langle u,urangleimplies 2Re langle u,vranglele langle u,urangle$
How to show from here that $langle u,vrangle =0$?
Please help
linear-algebra functional-analysis
$endgroup$
add a comment |
$begingroup$
Let $U$ be a subspace of $V$. Assume that there exists $vin V$ such that $langle v,urangle +langle u,vrangle le langle u,urangle forall uin U$
Prove that $v in U^perp $.
Let $uin U$ then we need to show that $langle u,vrangle =0$
Now $langle v,urangle +langle u,vrangle le langle u,urangleimplies 2Re langle u,vranglele langle u,urangle$
How to show from here that $langle u,vrangle =0$?
Please help
linear-algebra functional-analysis
$endgroup$
$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
1
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
$endgroup$
– Chessanator
Jan 31 at 15:59
$begingroup$
@Chessanator;how is that a hint
$endgroup$
– user596656
Jan 31 at 16:01
add a comment |
$begingroup$
Let $U$ be a subspace of $V$. Assume that there exists $vin V$ such that $langle v,urangle +langle u,vrangle le langle u,urangle forall uin U$
Prove that $v in U^perp $.
Let $uin U$ then we need to show that $langle u,vrangle =0$
Now $langle v,urangle +langle u,vrangle le langle u,urangleimplies 2Re langle u,vranglele langle u,urangle$
How to show from here that $langle u,vrangle =0$?
Please help
linear-algebra functional-analysis
$endgroup$
Let $U$ be a subspace of $V$. Assume that there exists $vin V$ such that $langle v,urangle +langle u,vrangle le langle u,urangle forall uin U$
Prove that $v in U^perp $.
Let $uin U$ then we need to show that $langle u,vrangle =0$
Now $langle v,urangle +langle u,vrangle le langle u,urangleimplies 2Re langle u,vranglele langle u,urangle$
How to show from here that $langle u,vrangle =0$?
Please help
linear-algebra functional-analysis
linear-algebra functional-analysis
edited Jan 31 at 16:25
asked Jan 31 at 15:38
user596656
$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
1
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
$endgroup$
– Chessanator
Jan 31 at 15:59
$begingroup$
@Chessanator;how is that a hint
$endgroup$
– user596656
Jan 31 at 16:01
add a comment |
$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
1
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
$endgroup$
– Chessanator
Jan 31 at 15:59
$begingroup$
@Chessanator;how is that a hint
$endgroup$
– user596656
Jan 31 at 16:01
$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
1
1
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
$endgroup$
– Chessanator
Jan 31 at 15:59
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
$endgroup$
– Chessanator
Jan 31 at 15:59
$begingroup$
@Chessanator;how is that a hint
$endgroup$
– user596656
Jan 31 at 16:01
$begingroup$
@Chessanator;how is that a hint
$endgroup$
– user596656
Jan 31 at 16:01
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
We take the Chessanator hint, and note that $Ucap U^{perp}={0},$ the zero vector. If $v=0,$ then we are done trivially because $0in U^{perp}.$ But now, suppose, by way of establishing a contradiction, that $vin V,$ and $vnot=0$. Then, by assumption, we can write $v=u+u^{perp},$ where $uin U$ and $u^{perp}in U^{perp}$. It must be that
$$langle v,urangle+langle u,vrangle=langle u+u^{perp},urangle+langle u,u+u^{perp}rangle=langle u,urangle+langle u,urangle =2langle u,urangle.$$
If $u=0,$ we are done and $vin U^{perp}.$ If $unot=0,$ then we have a case where $langle v,urangle+langle u,vrangle > langle u,urangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
|
show 3 more comments
$begingroup$
Suppose there is some $uin U$ such that $langle u,vrangleneq 0$. Pick $lambdainmathbb{C}$ and take $w=lambda u$, such that $langle w,vrangle>0$ and $|w|=1$.
Now let $r>0$ and consider
$$langle v,rwrangle+langle rw,vrangleleq |rw|^{2}=r^{2}Leftrightarrow r^{2}-2rlangle v,wranglegeq0$$
which only holds for $rleq0$ and $rgeq 2langle v,wrangle$. So for $r=langle v,wrangle$ we find
$$langle v,rwrangle+langle rw,vrangle=2langle v,wrangle^{2}geq langle v,wrangle^{2}=|rw|^{2}$$
which is a contradiction.
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
$endgroup$
add a comment |
Your Answer
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2 Answers
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active
oldest
votes
2 Answers
2
active
oldest
votes
active
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oldest
votes
$begingroup$
We take the Chessanator hint, and note that $Ucap U^{perp}={0},$ the zero vector. If $v=0,$ then we are done trivially because $0in U^{perp}.$ But now, suppose, by way of establishing a contradiction, that $vin V,$ and $vnot=0$. Then, by assumption, we can write $v=u+u^{perp},$ where $uin U$ and $u^{perp}in U^{perp}$. It must be that
$$langle v,urangle+langle u,vrangle=langle u+u^{perp},urangle+langle u,u+u^{perp}rangle=langle u,urangle+langle u,urangle =2langle u,urangle.$$
If $u=0,$ we are done and $vin U^{perp}.$ If $unot=0,$ then we have a case where $langle v,urangle+langle u,vrangle > langle u,urangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
|
show 3 more comments
$begingroup$
We take the Chessanator hint, and note that $Ucap U^{perp}={0},$ the zero vector. If $v=0,$ then we are done trivially because $0in U^{perp}.$ But now, suppose, by way of establishing a contradiction, that $vin V,$ and $vnot=0$. Then, by assumption, we can write $v=u+u^{perp},$ where $uin U$ and $u^{perp}in U^{perp}$. It must be that
$$langle v,urangle+langle u,vrangle=langle u+u^{perp},urangle+langle u,u+u^{perp}rangle=langle u,urangle+langle u,urangle =2langle u,urangle.$$
If $u=0,$ we are done and $vin U^{perp}.$ If $unot=0,$ then we have a case where $langle v,urangle+langle u,vrangle > langle u,urangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
|
show 3 more comments
$begingroup$
We take the Chessanator hint, and note that $Ucap U^{perp}={0},$ the zero vector. If $v=0,$ then we are done trivially because $0in U^{perp}.$ But now, suppose, by way of establishing a contradiction, that $vin V,$ and $vnot=0$. Then, by assumption, we can write $v=u+u^{perp},$ where $uin U$ and $u^{perp}in U^{perp}$. It must be that
$$langle v,urangle+langle u,vrangle=langle u+u^{perp},urangle+langle u,u+u^{perp}rangle=langle u,urangle+langle u,urangle =2langle u,urangle.$$
If $u=0,$ we are done and $vin U^{perp}.$ If $unot=0,$ then we have a case where $langle v,urangle+langle u,vrangle > langle u,urangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
$endgroup$
We take the Chessanator hint, and note that $Ucap U^{perp}={0},$ the zero vector. If $v=0,$ then we are done trivially because $0in U^{perp}.$ But now, suppose, by way of establishing a contradiction, that $vin V,$ and $vnot=0$. Then, by assumption, we can write $v=u+u^{perp},$ where $uin U$ and $u^{perp}in U^{perp}$. It must be that
$$langle v,urangle+langle u,vrangle=langle u+u^{perp},urangle+langle u,u+u^{perp}rangle=langle u,urangle+langle u,urangle =2langle u,urangle.$$
If $u=0,$ we are done and $vin U^{perp}.$ If $unot=0,$ then we have a case where $langle v,urangle+langle u,vrangle > langle u,urangle,$ which cannot be, by assumption. Therefore, $u=0,$ and the theorem is proved.
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
edited Jan 31 at 16:45
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
answered Jan 31 at 16:14
Adrian KeisterAdrian Keister
5,26971933
5,26971933
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
|
show 3 more comments
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
$V=Uoplus U^perp$,how does that mean that if $vnotin Uimplies vin U^perp$
$endgroup$
– user596656
Jan 31 at 16:24
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
I already made it clear that $Ucap U^{perp}={0}$. There's only one vector in both $U$ and $U^{perp}$. And, in the case where $vnot=0,$ if it's in $V$ but not $U,$ it must be in $U^{perp}.$
$endgroup$
– Adrian Keister
Jan 31 at 16:34
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
Join_PhD has a point. I would propose the following fix. We only know that $vin V$ so we know that $v=v_{1}+v_{2}$ with $v_{1}in U$ and $v_{2}in U^{perp}$. Then $langle v_{1},vrangle+langle v,v_{1}rangle=2langle v_{1},v_{1}rangle>langle v_{1},v_{1}rangle$.
$endgroup$
– Floris Claassens
Jan 31 at 16:37
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
$begingroup$
I think with that change this proof is definitely more elegant than my own.
$endgroup$
– Floris Claassens
Jan 31 at 16:38
1
1
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
$begingroup$
@Adrian Keister, yes that looks correct.
$endgroup$
– Floris Claassens
Jan 31 at 16:49
|
show 3 more comments
$begingroup$
Suppose there is some $uin U$ such that $langle u,vrangleneq 0$. Pick $lambdainmathbb{C}$ and take $w=lambda u$, such that $langle w,vrangle>0$ and $|w|=1$.
Now let $r>0$ and consider
$$langle v,rwrangle+langle rw,vrangleleq |rw|^{2}=r^{2}Leftrightarrow r^{2}-2rlangle v,wranglegeq0$$
which only holds for $rleq0$ and $rgeq 2langle v,wrangle$. So for $r=langle v,wrangle$ we find
$$langle v,rwrangle+langle rw,vrangle=2langle v,wrangle^{2}geq langle v,wrangle^{2}=|rw|^{2}$$
which is a contradiction.
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
$endgroup$
add a comment |
$begingroup$
Suppose there is some $uin U$ such that $langle u,vrangleneq 0$. Pick $lambdainmathbb{C}$ and take $w=lambda u$, such that $langle w,vrangle>0$ and $|w|=1$.
Now let $r>0$ and consider
$$langle v,rwrangle+langle rw,vrangleleq |rw|^{2}=r^{2}Leftrightarrow r^{2}-2rlangle v,wranglegeq0$$
which only holds for $rleq0$ and $rgeq 2langle v,wrangle$. So for $r=langle v,wrangle$ we find
$$langle v,rwrangle+langle rw,vrangle=2langle v,wrangle^{2}geq langle v,wrangle^{2}=|rw|^{2}$$
which is a contradiction.
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
$endgroup$
add a comment |
$begingroup$
Suppose there is some $uin U$ such that $langle u,vrangleneq 0$. Pick $lambdainmathbb{C}$ and take $w=lambda u$, such that $langle w,vrangle>0$ and $|w|=1$.
Now let $r>0$ and consider
$$langle v,rwrangle+langle rw,vrangleleq |rw|^{2}=r^{2}Leftrightarrow r^{2}-2rlangle v,wranglegeq0$$
which only holds for $rleq0$ and $rgeq 2langle v,wrangle$. So for $r=langle v,wrangle$ we find
$$langle v,rwrangle+langle rw,vrangle=2langle v,wrangle^{2}geq langle v,wrangle^{2}=|rw|^{2}$$
which is a contradiction.
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
$endgroup$
Suppose there is some $uin U$ such that $langle u,vrangleneq 0$. Pick $lambdainmathbb{C}$ and take $w=lambda u$, such that $langle w,vrangle>0$ and $|w|=1$.
Now let $r>0$ and consider
$$langle v,rwrangle+langle rw,vrangleleq |rw|^{2}=r^{2}Leftrightarrow r^{2}-2rlangle v,wranglegeq0$$
which only holds for $rleq0$ and $rgeq 2langle v,wrangle$. So for $r=langle v,wrangle$ we find
$$langle v,rwrangle+langle rw,vrangle=2langle v,wrangle^{2}geq langle v,wrangle^{2}=|rw|^{2}$$
which is a contradiction.
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
answered Jan 31 at 16:30
Floris ClaassensFloris Claassens
1,32829
1,32829
add a comment |
add a comment |
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$begingroup$
So, $V$ are complex inner product vector?
$endgroup$
– Thomas Andrews
Jan 31 at 15:45
1
$begingroup$
Hint: look at the orthogonal projection of $v$ onto $U$ (i.e. using the fact that $V = U oplus U^{perp}$).
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– Chessanator
Jan 31 at 15:59
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@Chessanator;how is that a hint
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– user596656
Jan 31 at 16:01