Assume $langle v,s rangle + langle s,v rangle leq langle s,s rangle$If $langle v,srangle+langle s,vrangleleq langle s,srangle$ for all $sin S$, why is $vin S^perp$?$||u||leq ||u+av|| Longrightarrow langle u,vrangle=0$Proving an inequality on $sum_1leq i,j leq n langle c_i ,c_j rangle times langle l_i ,l_j rangle$Does $langle f+h,grangle=langle f,grangle+langle h,grangle$ hold for all elements $f, g, h$ of an inner product space?How do I prove that $left langle T(x),x right rangle=0,forall xin VRightarrow T=T_0$?Show $langle u,v rangle = u_1v_1+3u_2v_2$ is an inner productProve: $langle p, vrangle = |p|^2$What are the maximum and minimum values of $langle u, vrangle + langle v, wrangle + langle w, urangle$?Show that $langle x,y rangle = 0$.$V$ is a inner product space, prove $langle av, vrangle langle v, avrangle le langle av, avrangle $
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Assume $langle v,s rangle + langle s,v rangle leq langle s,s rangle$
If $langle v,srangle+langle s,vrangleleq langle s,srangle$ for all $sin S$, why is $vin S^perp$?$||u||leq ||u+av|| Longrightarrow langle u,vrangle=0$Proving an inequality on $sum_1leq i,j leq n langle c_i ,c_j rangle times langle l_i ,l_j rangle$Does $langle f+h,grangle=langle f,grangle+langle h,grangle$ hold for all elements $f, g, h$ of an inner product space?How do I prove that $left langle T(x),x right rangle=0,forall xin VRightarrow T=T_0$?Show $langle u,v rangle = u_1v_1+3u_2v_2$ is an inner productProve: $langle p, vrangle = |p|^2$What are the maximum and minimum values of $langle u, vrangle + langle v, wrangle + langle w, urangle$?Show that $langle x,y rangle = 0$.$V$ is a inner product space, prove $langle av, vrangle langle v, avrangle le langle av, avrangle $
$begingroup$
The question is
Let $V$ be a complex inner product space, and let $S$ be a subspace of $V$. Suppose that $vin V$ is a vector for which $langle s,vrangle + langle v,srangle leq langle s,srangle$ for all $sin S$. Prove that $vin S^perp$.
I am thinking about proving it by contradiction, but I am not sure what $langle s,vrangle + langle v,srangle leq langle s,srangle$ can tell me. What I am sure about right now is that $v$ must not be in $S$ since if $v$ is in $S$, then $v$ will be equal to some $s$ in $S$, then there exists such $s$ that
$langle s,vrangle + langle v,srangle = langle s,srangle + langle s,srangle = 2langle s,srangle gt langle s,srangle$.
So $v$ must be in somewhere else. What else can I know, I am so confused right now, can somebody give me some hints?
inner-product-space
edited Mar 14 at 6:40
miracle173
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
$endgroup$
add a comment |
$begingroup$
The question is
Let $V$ be a complex inner product space, and let $S$ be a subspace of $V$. Suppose that $vin V$ is a vector for which $langle s,vrangle + langle v,srangle leq langle s,srangle$ for all $sin S$. Prove that $vin S^perp$.
I am thinking about proving it by contradiction, but I am not sure what $langle s,vrangle + langle v,srangle leq langle s,srangle$ can tell me. What I am sure about right now is that $v$ must not be in $S$ since if $v$ is in $S$, then $v$ will be equal to some $s$ in $S$, then there exists such $s$ that
$langle s,vrangle + langle v,srangle = langle s,srangle + langle s,srangle = 2langle s,srangle gt langle s,srangle$.
So $v$ must be in somewhere else. What else can I know, I am so confused right now, can somebody give me some hints?
inner-product-space
edited Mar 14 at 6:40
miracle173
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
$endgroup$
$begingroup$
As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:53
1
$begingroup$
Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:55
$begingroup$
@ArturoMagidin the question does not provide it is finite dimensional or not.
$endgroup$
– PixieBlade
Mar 14 at 4:58
$begingroup$
@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
$endgroup$
– miracle173
Mar 14 at 5:08
$begingroup$
Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
$endgroup$
– anomaly
Mar 14 at 5:19
add a comment |
$begingroup$
The question is
Let $V$ be a complex inner product space, and let $S$ be a subspace of $V$. Suppose that $vin V$ is a vector for which $langle s,vrangle + langle v,srangle leq langle s,srangle$ for all $sin S$. Prove that $vin S^perp$.
I am thinking about proving it by contradiction, but I am not sure what $langle s,vrangle + langle v,srangle leq langle s,srangle$ can tell me. What I am sure about right now is that $v$ must not be in $S$ since if $v$ is in $S$, then $v$ will be equal to some $s$ in $S$, then there exists such $s$ that
$langle s,vrangle + langle v,srangle = langle s,srangle + langle s,srangle = 2langle s,srangle gt langle s,srangle$.
So $v$ must be in somewhere else. What else can I know, I am so confused right now, can somebody give me some hints?
inner-product-space
edited Mar 14 at 6:40
miracle173
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
$endgroup$
The question is
Let $V$ be a complex inner product space, and let $S$ be a subspace of $V$. Suppose that $vin V$ is a vector for which $langle s,vrangle + langle v,srangle leq langle s,srangle$ for all $sin S$. Prove that $vin S^perp$.
I am thinking about proving it by contradiction, but I am not sure what $langle s,vrangle + langle v,srangle leq langle s,srangle$ can tell me. What I am sure about right now is that $v$ must not be in $S$ since if $v$ is in $S$, then $v$ will be equal to some $s$ in $S$, then there exists such $s$ that
$langle s,vrangle + langle v,srangle = langle s,srangle + langle s,srangle = 2langle s,srangle gt langle s,srangle$.
So $v$ must be in somewhere else. What else can I know, I am so confused right now, can somebody give me some hints?
inner-product-space
inner-product-space
edited Mar 14 at 6:40
miracle173
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
edited Mar 14 at 6:40
miracle173
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
edited Mar 14 at 6:40
miracle173
7,36122247
edited Mar 14 at 6:40
miracle173
7,36122247
edited Mar 14 at 6:40
miracle173
7,36122247
7,36122247
asked Mar 14 at 4:42
PixieBladePixieBlade
113
asked Mar 14 at 4:42
PixieBladePixieBlade
113
asked Mar 14 at 4:42
PixieBladePixieBlade
113
113
$begingroup$
As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:53
1
$begingroup$
Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:55
$begingroup$
@ArturoMagidin the question does not provide it is finite dimensional or not.
$endgroup$
– PixieBlade
Mar 14 at 4:58
$begingroup$
@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
$endgroup$
– miracle173
Mar 14 at 5:08
$begingroup$
Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
$endgroup$
– anomaly
Mar 14 at 5:19
add a comment |
$begingroup$
As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:53
1
$begingroup$
Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:55
$begingroup$
@ArturoMagidin the question does not provide it is finite dimensional or not.
$endgroup$
– PixieBlade
Mar 14 at 4:58
$begingroup$
@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
$endgroup$
– miracle173
Mar 14 at 5:08
$begingroup$
Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
$endgroup$
– anomaly
Mar 14 at 5:19
$begingroup$
As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:53
$begingroup$
As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:53
1
1
$begingroup$
Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:55
$begingroup$
Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
$endgroup$
– Arturo Magidin
Mar 14 at 4:55
$begingroup$
@ArturoMagidin the question does not provide it is finite dimensional or not.
$endgroup$
– PixieBlade
Mar 14 at 4:58
$begingroup$
@ArturoMagidin the question does not provide it is finite dimensional or not.
$endgroup$
– PixieBlade
Mar 14 at 4:58
$begingroup$
@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
$endgroup$
– miracle173
Mar 14 at 5:08
$begingroup$
@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
$endgroup$
– miracle173
Mar 14 at 5:08
$begingroup$
Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
$endgroup$
– anomaly
Mar 14 at 5:19
$begingroup$
Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
$endgroup$
– anomaly
Mar 14 at 5:19
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Fix $s in S$. Replace $s$ by $epsilon s$ and divide by $epsilon$ to get $ langle v, s rangle +langle s, v rangle leq epsilon |s|^2$. Letting $epsilon to 0$ we see that Real part of $ langle v, s rangle$ is $leq 0$. Replace $s$ by $-s$ to see that the real part is $0$. Replace $s$ by $is$ to see that the imaginary part is also $0$.
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
$endgroup$
add a comment |
$begingroup$
Hint: Think about what happens as we make $langle s,srangle$ smaller and smaller.
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
$endgroup$
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
$endgroup$
– PixieBlade
Mar 14 at 5:02
$begingroup$
Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
$endgroup$
– Isaac Browne
Mar 14 at 5:05
add a comment |
$begingroup$
If $$Sne0$$
we can choose an $$s_1ne 0$$ and that means $$langle s_1,s_1ranglene 0$$
We set $$s_2=frac1sqrtlangle s_1, s_1rangles_1$$
then we have $$langle s_2, s_2 rangle=1$$
For an arbitrary $v in V$ we set
$$s=langle v,s_2rangle s_2 in S$$
and for the LHS of
$$langle s,vrangle + langle v,srangle leq langle s,srangle$$
we get
$$langle s,vrangle + langle v,srangle \=langle langle v,s_2rangle s_2,vrangle + langle v,langle v,s_2rangle s_2rangle\=langle v,s_2rangle langle s_2,vrangle+overlinelangle v,s_2rangle langle v,s_2rangle\=2overlinelangle v,s_2rangle langle v,s_2rangle$$
For the RHS we get
$$langle s ,s rangle\=langle langle v,s_2rangle s_2,langle v,s_2rangle s_2 rangle\=langle v,s_2rangle overlinelangle v,s_2ranglelangle s_2,s_2 rangle=\langle v,s_2rangle overlinelangle v,s_2rangle$$
It follows that
$$2overlinelangle v,s_2rangle langle v,s_2rangle le overlinelangle v,s_2rangle langle v,s_2rangle$$
which is a contradiction for $$overlinelangle v,s_2rangle langle v,s_2ranglene 0$$
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
$endgroup$
add a comment |
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3 Answers
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3 Answers
3
active
oldest
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active
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$begingroup$
Fix $s in S$. Replace $s$ by $epsilon s$ and divide by $epsilon$ to get $ langle v, s rangle +langle s, v rangle leq epsilon |s|^2$. Letting $epsilon to 0$ we see that Real part of $ langle v, s rangle$ is $leq 0$. Replace $s$ by $-s$ to see that the real part is $0$. Replace $s$ by $is$ to see that the imaginary part is also $0$.
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
$endgroup$
add a comment |
$begingroup$
Fix $s in S$. Replace $s$ by $epsilon s$ and divide by $epsilon$ to get $ langle v, s rangle +langle s, v rangle leq epsilon |s|^2$. Letting $epsilon to 0$ we see that Real part of $ langle v, s rangle$ is $leq 0$. Replace $s$ by $-s$ to see that the real part is $0$. Replace $s$ by $is$ to see that the imaginary part is also $0$.
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
$endgroup$
add a comment |
$begingroup$
Fix $s in S$. Replace $s$ by $epsilon s$ and divide by $epsilon$ to get $ langle v, s rangle +langle s, v rangle leq epsilon |s|^2$. Letting $epsilon to 0$ we see that Real part of $ langle v, s rangle$ is $leq 0$. Replace $s$ by $-s$ to see that the real part is $0$. Replace $s$ by $is$ to see that the imaginary part is also $0$.
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
$endgroup$
Fix $s in S$. Replace $s$ by $epsilon s$ and divide by $epsilon$ to get $ langle v, s rangle +langle s, v rangle leq epsilon |s|^2$. Letting $epsilon to 0$ we see that Real part of $ langle v, s rangle$ is $leq 0$. Replace $s$ by $-s$ to see that the real part is $0$. Replace $s$ by $is$ to see that the imaginary part is also $0$.
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
edited Mar 14 at 6:28
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
answered Mar 14 at 5:23
Kavi Rama MurthyKavi Rama Murthy
68.7k53169
68.7k53169
add a comment |
add a comment |
$begingroup$
Hint: Think about what happens as we make $langle s,srangle$ smaller and smaller.
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
$endgroup$
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
$endgroup$
– PixieBlade
Mar 14 at 5:02
$begingroup$
Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
$endgroup$
– Isaac Browne
Mar 14 at 5:05
add a comment |
$begingroup$
Hint: Think about what happens as we make $langle s,srangle$ smaller and smaller.
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
$endgroup$
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
$endgroup$
– PixieBlade
Mar 14 at 5:02
$begingroup$
Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
$endgroup$
– Isaac Browne
Mar 14 at 5:05
add a comment |
$begingroup$
Hint: Think about what happens as we make $langle s,srangle$ smaller and smaller.
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
$endgroup$
Hint: Think about what happens as we make $langle s,srangle$ smaller and smaller.
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
answered Mar 14 at 4:52
Isaac BrowneIsaac Browne
5,29751234
5,29751234
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
$endgroup$
– PixieBlade
Mar 14 at 5:02
$begingroup$
Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
$endgroup$
– Isaac Browne
Mar 14 at 5:05
add a comment |
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
$endgroup$
– PixieBlade
Mar 14 at 5:02
$begingroup$
Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
$endgroup$
– Isaac Browne
Mar 14 at 5:05
$begingroup$
Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
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– PixieBlade
Mar 14 at 5:02
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Then ⟨𝑠,𝑠⟩ will be extremely close to 0? but it still larger than ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ and since either ⟨𝑠,𝑣⟩ or ⟨𝑣,𝑠⟩ is positive definite, so ⟨𝑠,𝑣⟩+⟨𝑣,𝑠⟩ aproaches to 0 so ⟨𝑠,𝑣⟩=⟨𝑣,𝑠⟩=0? so v⊥s and s⊥v? What am I thinking...
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– PixieBlade
Mar 14 at 5:02
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Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
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– Isaac Browne
Mar 14 at 5:05
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Yep, you're on the right track! You can rigorize this by taking a vector $vecv_d$ for each direction in $S$ and then saying that this property must hold for all scalar multiples of each vector.
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– Isaac Browne
Mar 14 at 5:05
add a comment |
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If $$Sne0$$
we can choose an $$s_1ne 0$$ and that means $$langle s_1,s_1ranglene 0$$
We set $$s_2=frac1sqrtlangle s_1, s_1rangles_1$$
then we have $$langle s_2, s_2 rangle=1$$
For an arbitrary $v in V$ we set
$$s=langle v,s_2rangle s_2 in S$$
and for the LHS of
$$langle s,vrangle + langle v,srangle leq langle s,srangle$$
we get
$$langle s,vrangle + langle v,srangle \=langle langle v,s_2rangle s_2,vrangle + langle v,langle v,s_2rangle s_2rangle\=langle v,s_2rangle langle s_2,vrangle+overlinelangle v,s_2rangle langle v,s_2rangle\=2overlinelangle v,s_2rangle langle v,s_2rangle$$
For the RHS we get
$$langle s ,s rangle\=langle langle v,s_2rangle s_2,langle v,s_2rangle s_2 rangle\=langle v,s_2rangle overlinelangle v,s_2ranglelangle s_2,s_2 rangle=\langle v,s_2rangle overlinelangle v,s_2rangle$$
It follows that
$$2overlinelangle v,s_2rangle langle v,s_2rangle le overlinelangle v,s_2rangle langle v,s_2rangle$$
which is a contradiction for $$overlinelangle v,s_2rangle langle v,s_2ranglene 0$$
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
$endgroup$
add a comment |
$begingroup$
If $$Sne0$$
we can choose an $$s_1ne 0$$ and that means $$langle s_1,s_1ranglene 0$$
We set $$s_2=frac1sqrtlangle s_1, s_1rangles_1$$
then we have $$langle s_2, s_2 rangle=1$$
For an arbitrary $v in V$ we set
$$s=langle v,s_2rangle s_2 in S$$
and for the LHS of
$$langle s,vrangle + langle v,srangle leq langle s,srangle$$
we get
$$langle s,vrangle + langle v,srangle \=langle langle v,s_2rangle s_2,vrangle + langle v,langle v,s_2rangle s_2rangle\=langle v,s_2rangle langle s_2,vrangle+overlinelangle v,s_2rangle langle v,s_2rangle\=2overlinelangle v,s_2rangle langle v,s_2rangle$$
For the RHS we get
$$langle s ,s rangle\=langle langle v,s_2rangle s_2,langle v,s_2rangle s_2 rangle\=langle v,s_2rangle overlinelangle v,s_2ranglelangle s_2,s_2 rangle=\langle v,s_2rangle overlinelangle v,s_2rangle$$
It follows that
$$2overlinelangle v,s_2rangle langle v,s_2rangle le overlinelangle v,s_2rangle langle v,s_2rangle$$
which is a contradiction for $$overlinelangle v,s_2rangle langle v,s_2ranglene 0$$
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
$endgroup$
add a comment |
$begingroup$
If $$Sne0$$
we can choose an $$s_1ne 0$$ and that means $$langle s_1,s_1ranglene 0$$
We set $$s_2=frac1sqrtlangle s_1, s_1rangles_1$$
then we have $$langle s_2, s_2 rangle=1$$
For an arbitrary $v in V$ we set
$$s=langle v,s_2rangle s_2 in S$$
and for the LHS of
$$langle s,vrangle + langle v,srangle leq langle s,srangle$$
we get
$$langle s,vrangle + langle v,srangle \=langle langle v,s_2rangle s_2,vrangle + langle v,langle v,s_2rangle s_2rangle\=langle v,s_2rangle langle s_2,vrangle+overlinelangle v,s_2rangle langle v,s_2rangle\=2overlinelangle v,s_2rangle langle v,s_2rangle$$
For the RHS we get
$$langle s ,s rangle\=langle langle v,s_2rangle s_2,langle v,s_2rangle s_2 rangle\=langle v,s_2rangle overlinelangle v,s_2ranglelangle s_2,s_2 rangle=\langle v,s_2rangle overlinelangle v,s_2rangle$$
It follows that
$$2overlinelangle v,s_2rangle langle v,s_2rangle le overlinelangle v,s_2rangle langle v,s_2rangle$$
which is a contradiction for $$overlinelangle v,s_2rangle langle v,s_2ranglene 0$$
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
$endgroup$
If $$Sne0$$
we can choose an $$s_1ne 0$$ and that means $$langle s_1,s_1ranglene 0$$
We set $$s_2=frac1sqrtlangle s_1, s_1rangles_1$$
then we have $$langle s_2, s_2 rangle=1$$
For an arbitrary $v in V$ we set
$$s=langle v,s_2rangle s_2 in S$$
and for the LHS of
$$langle s,vrangle + langle v,srangle leq langle s,srangle$$
we get
$$langle s,vrangle + langle v,srangle \=langle langle v,s_2rangle s_2,vrangle + langle v,langle v,s_2rangle s_2rangle\=langle v,s_2rangle langle s_2,vrangle+overlinelangle v,s_2rangle langle v,s_2rangle\=2overlinelangle v,s_2rangle langle v,s_2rangle$$
For the RHS we get
$$langle s ,s rangle\=langle langle v,s_2rangle s_2,langle v,s_2rangle s_2 rangle\=langle v,s_2rangle overlinelangle v,s_2ranglelangle s_2,s_2 rangle=\langle v,s_2rangle overlinelangle v,s_2rangle$$
It follows that
$$2overlinelangle v,s_2rangle langle v,s_2rangle le overlinelangle v,s_2rangle langle v,s_2rangle$$
which is a contradiction for $$overlinelangle v,s_2rangle langle v,s_2ranglene 0$$
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
edited Mar 14 at 6:39
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
answered Mar 14 at 5:43
miracle173miracle173
7,36122247
7,36122247
add a comment |
add a comment |
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As a side comment, you need to assume $sneq mathbf0$ in your bit of argument. Otherwise, $2langle s,srangle$ could equal $langle s,srangle$ (both zero). Not an issue, though, since the zero vector also happens to lie in $S^“perp$.
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– Arturo Magidin
Mar 14 at 4:53
1
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Is $S$ finite dimensional? If so, write $v=s+p$, where $sin S$ and $Pin S^perp$.
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– Arturo Magidin
Mar 14 at 4:55
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@ArturoMagidin the question does not provide it is finite dimensional or not.
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– PixieBlade
Mar 14 at 4:58
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@ArturoMagidin If v is orthogonal to any finite subspace of S then v is is orthogonal to any s in S. so v is orthogonal to S.
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– miracle173
Mar 14 at 5:08
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Fix nonzero $sin S$ (the case $S = 0$ is trivial), and consider $lambda sin S$ for small $lambda$.
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– anomaly
Mar 14 at 5:19