Prove the norm of operator derived from orthogonal projections is less or equal to 1
$begingroup$
Suppose $H$ is a Hilbert space. We have two orthogonal projections $P_1: Hrightarrow A_1$ and $P_2: Hrightarrow A_2$. Here $A_1$ and $A_2$ are two closed subspace of $H$. Prove that
$$
lVert P_1 - P_2rVert le 1
$$
Any hints may help. Thank you.
linear-algebra functional-analysis
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
$endgroup$
add a comment |
$begingroup$
Suppose $H$ is a Hilbert space. We have two orthogonal projections $P_1: Hrightarrow A_1$ and $P_2: Hrightarrow A_2$. Here $A_1$ and $A_2$ are two closed subspace of $H$. Prove that
$$
lVert P_1 - P_2rVert le 1
$$
Any hints may help. Thank you.
linear-algebra functional-analysis
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
$endgroup$
add a comment |
$begingroup$
Suppose $H$ is a Hilbert space. We have two orthogonal projections $P_1: Hrightarrow A_1$ and $P_2: Hrightarrow A_2$. Here $A_1$ and $A_2$ are two closed subspace of $H$. Prove that
$$
lVert P_1 - P_2rVert le 1
$$
Any hints may help. Thank you.
linear-algebra functional-analysis
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
$endgroup$
Suppose $H$ is a Hilbert space. We have two orthogonal projections $P_1: Hrightarrow A_1$ and $P_2: Hrightarrow A_2$. Here $A_1$ and $A_2$ are two closed subspace of $H$. Prove that
$$
lVert P_1 - P_2rVert le 1
$$
Any hints may help. Thank you.
linear-algebra functional-analysis
linear-algebra functional-analysis
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
asked Dec 3 '18 at 21:55
SyuizenSyuizen
934411
934411
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
HINT:
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $|P-frac{1}{2}I|=frac{1}{2}$ or $|2P-I|=1$, where $I$ is the identity operator on $H$. To show this, note that $2P-I$ is a unitary.
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
$endgroup$
add a comment |
$begingroup$
You can write $A_1 = B_1oplus (A_1cap A_2)$ and $A_2=B_2oplus(A_1cap A_2)$ where $oplus$ is an orthogonal decomposition. Then
$$
P_{A_1}=P_{B_1}+P_{A_1cap A_2}\
P_{A_2}=P_{B_2}+P_{A_1cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1perp B_2$, it is easy to see that $|P_{B_1}-P_{B_2}|le 1$ because
$$
|P_{B_1}x-P_{B_2}x|^2=|P_{B_1}x|^2+|P_{B_1}x|^2=|P_{B_1oplus B_2}x|^2le |x|^2.
$$
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
add a comment |
$begingroup$
For any $xin H$ with $|x|=1$, we have for an orthogonal projection $P$ $$langle Px,xrangle=langle P^*Px,xrangle=langle Px,Pxrangle=|Px|^2leq |Px|^2+|(I-P)x|^2=|x|^2=1.$$
So
$$
langle (P_1-P_2)x,xrangle=langle P_1x,xrangle-langle P_2x,xranglein[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$.
For any selfadjoint operator $T$, we have $|T|=sup{|langle Tx,xrangle: |x|=1}$. So
$$
|P_1-P_2|=sup{|langle (P_1-P_2)x,xrangle|: |x|=1}leq1.
$$
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
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add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
HINT:
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $|P-frac{1}{2}I|=frac{1}{2}$ or $|2P-I|=1$, where $I$ is the identity operator on $H$. To show this, note that $2P-I$ is a unitary.
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
$endgroup$
add a comment |
$begingroup$
HINT:
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $|P-frac{1}{2}I|=frac{1}{2}$ or $|2P-I|=1$, where $I$ is the identity operator on $H$. To show this, note that $2P-I$ is a unitary.
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
$endgroup$
add a comment |
$begingroup$
HINT:
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $|P-frac{1}{2}I|=frac{1}{2}$ or $|2P-I|=1$, where $I$ is the identity operator on $H$. To show this, note that $2P-I$ is a unitary.
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
$endgroup$
HINT:
Note that it suffices to show that for any orthogonal projection $P$ on $H$, then $|P-frac{1}{2}I|=frac{1}{2}$ or $|2P-I|=1$, where $I$ is the identity operator on $H$. To show this, note that $2P-I$ is a unitary.
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
answered Dec 3 '18 at 23:02
AweyganAweygan
14.3k21441
14.3k21441
add a comment |
add a comment |
$begingroup$
You can write $A_1 = B_1oplus (A_1cap A_2)$ and $A_2=B_2oplus(A_1cap A_2)$ where $oplus$ is an orthogonal decomposition. Then
$$
P_{A_1}=P_{B_1}+P_{A_1cap A_2}\
P_{A_2}=P_{B_2}+P_{A_1cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1perp B_2$, it is easy to see that $|P_{B_1}-P_{B_2}|le 1$ because
$$
|P_{B_1}x-P_{B_2}x|^2=|P_{B_1}x|^2+|P_{B_1}x|^2=|P_{B_1oplus B_2}x|^2le |x|^2.
$$
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
add a comment |
$begingroup$
You can write $A_1 = B_1oplus (A_1cap A_2)$ and $A_2=B_2oplus(A_1cap A_2)$ where $oplus$ is an orthogonal decomposition. Then
$$
P_{A_1}=P_{B_1}+P_{A_1cap A_2}\
P_{A_2}=P_{B_2}+P_{A_1cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1perp B_2$, it is easy to see that $|P_{B_1}-P_{B_2}|le 1$ because
$$
|P_{B_1}x-P_{B_2}x|^2=|P_{B_1}x|^2+|P_{B_1}x|^2=|P_{B_1oplus B_2}x|^2le |x|^2.
$$
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
add a comment |
$begingroup$
You can write $A_1 = B_1oplus (A_1cap A_2)$ and $A_2=B_2oplus(A_1cap A_2)$ where $oplus$ is an orthogonal decomposition. Then
$$
P_{A_1}=P_{B_1}+P_{A_1cap A_2}\
P_{A_2}=P_{B_2}+P_{A_1cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1perp B_2$, it is easy to see that $|P_{B_1}-P_{B_2}|le 1$ because
$$
|P_{B_1}x-P_{B_2}x|^2=|P_{B_1}x|^2+|P_{B_1}x|^2=|P_{B_1oplus B_2}x|^2le |x|^2.
$$
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
$endgroup$
You can write $A_1 = B_1oplus (A_1cap A_2)$ and $A_2=B_2oplus(A_1cap A_2)$ where $oplus$ is an orthogonal decomposition. Then
$$
P_{A_1}=P_{B_1}+P_{A_1cap A_2}\
P_{A_2}=P_{B_2}+P_{A_1cap A_2},
$$
where the $P_{X}$ is the orthogonal projection onto $X$. So
$$
P_{A_1}-P_{A_2}=P_{B_1}-P_{B_2}.
$$
Because $B_1perp B_2$, it is easy to see that $|P_{B_1}-P_{B_2}|le 1$ because
$$
|P_{B_1}x-P_{B_2}x|^2=|P_{B_1}x|^2+|P_{B_1}x|^2=|P_{B_1oplus B_2}x|^2le |x|^2.
$$
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
answered Dec 3 '18 at 23:03
DisintegratingByPartsDisintegratingByParts
59.5k42580
59.5k42580
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
add a comment |
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
1
1
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
$begingroup$
Why does $B_1perp B_2$ hold?
$endgroup$
– Syuizen
Dec 4 '18 at 14:27
add a comment |
$begingroup$
For any $xin H$ with $|x|=1$, we have for an orthogonal projection $P$ $$langle Px,xrangle=langle P^*Px,xrangle=langle Px,Pxrangle=|Px|^2leq |Px|^2+|(I-P)x|^2=|x|^2=1.$$
So
$$
langle (P_1-P_2)x,xrangle=langle P_1x,xrangle-langle P_2x,xranglein[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$.
For any selfadjoint operator $T$, we have $|T|=sup{|langle Tx,xrangle: |x|=1}$. So
$$
|P_1-P_2|=sup{|langle (P_1-P_2)x,xrangle|: |x|=1}leq1.
$$
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
For any $xin H$ with $|x|=1$, we have for an orthogonal projection $P$ $$langle Px,xrangle=langle P^*Px,xrangle=langle Px,Pxrangle=|Px|^2leq |Px|^2+|(I-P)x|^2=|x|^2=1.$$
So
$$
langle (P_1-P_2)x,xrangle=langle P_1x,xrangle-langle P_2x,xranglein[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$.
For any selfadjoint operator $T$, we have $|T|=sup{|langle Tx,xrangle: |x|=1}$. So
$$
|P_1-P_2|=sup{|langle (P_1-P_2)x,xrangle|: |x|=1}leq1.
$$
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
add a comment |
$begingroup$
For any $xin H$ with $|x|=1$, we have for an orthogonal projection $P$ $$langle Px,xrangle=langle P^*Px,xrangle=langle Px,Pxrangle=|Px|^2leq |Px|^2+|(I-P)x|^2=|x|^2=1.$$
So
$$
langle (P_1-P_2)x,xrangle=langle P_1x,xrangle-langle P_2x,xranglein[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$.
For any selfadjoint operator $T$, we have $|T|=sup{|langle Tx,xrangle: |x|=1}$. So
$$
|P_1-P_2|=sup{|langle (P_1-P_2)x,xrangle|: |x|=1}leq1.
$$
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
For any $xin H$ with $|x|=1$, we have for an orthogonal projection $P$ $$langle Px,xrangle=langle P^*Px,xrangle=langle Px,Pxrangle=|Px|^2leq |Px|^2+|(I-P)x|^2=|x|^2=1.$$
So
$$
langle (P_1-P_2)x,xrangle=langle P_1x,xrangle-langle P_2x,xranglein[-1,1],
$$
since is it a difference of two numbers each in $[0,1]$.
For any selfadjoint operator $T$, we have $|T|=sup{|langle Tx,xrangle: |x|=1}$. So
$$
|P_1-P_2|=sup{|langle (P_1-P_2)x,xrangle|: |x|=1}leq1.
$$
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
answered Dec 4 '18 at 1:59
Martin ArgeramiMartin Argerami
127k1182183
127k1182183
add a comment |
add a comment |
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