Mixing
Norm estimate for a product of two orthogonal projectors
$begingroup$
Let $H$ denote a Hilbert space.
Consider two orthogonal projectors $,P,Qinmathscr L(H),$ such that $H=operatorname{Im}Poplusoperatorname{Im}Q,,$ that is
- both $,operatorname{Im}Q,$ and $,operatorname{Im}P,$ are closed subspaces of $H$,
$operatorname{Im}P+operatorname{Im}Q=H$,
${0}=operatorname{Im}Pcapoperatorname{Im}Q,$.
It is not assumed that $operatorname{Im}Pperpoperatorname{Im}Q$, or equivalently $P+Q=mathbb 1$.
Is it true then that $|PQ|<1$ ?
Note that $|PQ|=|QP|$ as the involution is isometric.
This is a follow-up question to
A "Crookedness criterion" for a pair of orthogonal projectors? .
Its answer shows that it is necessary to assume that
$,operatorname{Im}P+operatorname{Im}Q,$ is closed in $H$.
functional-analysis hilbert-spaces norm projection
asked Jan 29 at 15:38
HannoHanno
2,484628
$endgroup$
add a comment |
$begingroup$
Let $H$ denote a Hilbert space.
Consider two orthogonal projectors $,P,Qinmathscr L(H),$ such that $H=operatorname{Im}Poplusoperatorname{Im}Q,,$ that is
- both $,operatorname{Im}Q,$ and $,operatorname{Im}P,$ are closed subspaces of $H$,
$operatorname{Im}P+operatorname{Im}Q=H$,
${0}=operatorname{Im}Pcapoperatorname{Im}Q,$.
It is not assumed that $operatorname{Im}Pperpoperatorname{Im}Q$, or equivalently $P+Q=mathbb 1$.
Is it true then that $|PQ|<1$ ?
Note that $|PQ|=|QP|$ as the involution is isometric.
This is a follow-up question to
A "Crookedness criterion" for a pair of orthogonal projectors? .
Its answer shows that it is necessary to assume that
$,operatorname{Im}P+operatorname{Im}Q,$ is closed in $H$.
functional-analysis hilbert-spaces norm projection
asked Jan 29 at 15:38
HannoHanno
2,484628
$endgroup$
add a comment |
$begingroup$
Let $H$ denote a Hilbert space.
Consider two orthogonal projectors $,P,Qinmathscr L(H),$ such that $H=operatorname{Im}Poplusoperatorname{Im}Q,,$ that is
- both $,operatorname{Im}Q,$ and $,operatorname{Im}P,$ are closed subspaces of $H$,
$operatorname{Im}P+operatorname{Im}Q=H$,
${0}=operatorname{Im}Pcapoperatorname{Im}Q,$.
It is not assumed that $operatorname{Im}Pperpoperatorname{Im}Q$, or equivalently $P+Q=mathbb 1$.
Is it true then that $|PQ|<1$ ?
Note that $|PQ|=|QP|$ as the involution is isometric.
This is a follow-up question to
A "Crookedness criterion" for a pair of orthogonal projectors? .
Its answer shows that it is necessary to assume that
$,operatorname{Im}P+operatorname{Im}Q,$ is closed in $H$.
functional-analysis hilbert-spaces norm projection
asked Jan 29 at 15:38
HannoHanno
2,484628
$endgroup$
Let $H$ denote a Hilbert space.
Consider two orthogonal projectors $,P,Qinmathscr L(H),$ such that $H=operatorname{Im}Poplusoperatorname{Im}Q,,$ that is
- both $,operatorname{Im}Q,$ and $,operatorname{Im}P,$ are closed subspaces of $H$,
$operatorname{Im}P+operatorname{Im}Q=H$,
${0}=operatorname{Im}Pcapoperatorname{Im}Q,$.
It is not assumed that $operatorname{Im}Pperpoperatorname{Im}Q$, or equivalently $P+Q=mathbb 1$.
Is it true then that $|PQ|<1$ ?
Note that $|PQ|=|QP|$ as the involution is isometric.
This is a follow-up question to
A "Crookedness criterion" for a pair of orthogonal projectors? .
Its answer shows that it is necessary to assume that
$,operatorname{Im}P+operatorname{Im}Q,$ is closed in $H$.
functional-analysis hilbert-spaces norm projection
functional-analysis hilbert-spaces norm projection
asked Jan 29 at 15:38
HannoHanno
2,484628
asked Jan 29 at 15:38
HannoHanno
2,484628
asked Jan 29 at 15:38
HannoHanno
2,484628
asked Jan 29 at 15:38
HannoHanno
2,484628
asked Jan 29 at 15:38
HannoHanno
2,484628
2,484628
add a comment |
add a comment |
1 Answer
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$begingroup$
The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=operatorname{Im}Poplusoperatorname{Im}Q,,$
let $Zinmathscr L(H),$ denote the projector onto the second summand. In general, this projection onto $operatorname{Im}Q,$ along $operatorname{Im}P,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $operatorname{Im}Q,$ nor $operatorname{Im}P,$ equals ${0}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$|PQ|:leqslant:sqrt{1-|Z|^{-2}};.$$
Proof:$ $ If $xinoperatorname{Im}Q,$ is a unit vector, then $,Zx=x,$ and $,ZPx=0$. Thus,
$$1=|Zx-ZPx|:leqslant:|Z|: |(1-P)x|\[2ex]
Longrightarrowquadfrac1{|Z|^2} :leqslant: biglangle(1-P)xmid (1-P)xbigrangle
= biglangle(1-P)xmid xbigrangle
= 1-|Px|^2qquadtext{or}quad|Px|;leqslant; sqrt{1-|Z|^{-2}}$$
The claim now follows from
$$|PQ|:=:sup_{uin H,,|u|leq 1}|PQu|
:=:sup_{xinoperatorname{Im}Q,,|x|leqslant 1}|Px|:.$$
Remark:$ $ Every non-zero projector satisfies $|Z|geqslant1$.
And $|Z|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
answered Jan 31 at 17:36
HannoHanno
2,484628
$endgroup$
add a comment |
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$begingroup$
The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=operatorname{Im}Poplusoperatorname{Im}Q,,$
let $Zinmathscr L(H),$ denote the projector onto the second summand. In general, this projection onto $operatorname{Im}Q,$ along $operatorname{Im}P,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $operatorname{Im}Q,$ nor $operatorname{Im}P,$ equals ${0}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$|PQ|:leqslant:sqrt{1-|Z|^{-2}};.$$
Proof:$ $ If $xinoperatorname{Im}Q,$ is a unit vector, then $,Zx=x,$ and $,ZPx=0$. Thus,
$$1=|Zx-ZPx|:leqslant:|Z|: |(1-P)x|\[2ex]
Longrightarrowquadfrac1{|Z|^2} :leqslant: biglangle(1-P)xmid (1-P)xbigrangle
= biglangle(1-P)xmid xbigrangle
= 1-|Px|^2qquadtext{or}quad|Px|;leqslant; sqrt{1-|Z|^{-2}}$$
The claim now follows from
$$|PQ|:=:sup_{uin H,,|u|leq 1}|PQu|
:=:sup_{xinoperatorname{Im}Q,,|x|leqslant 1}|Px|:.$$
Remark:$ $ Every non-zero projector satisfies $|Z|geqslant1$.
And $|Z|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
answered Jan 31 at 17:36
HannoHanno
2,484628
$endgroup$
add a comment |
$begingroup$
The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=operatorname{Im}Poplusoperatorname{Im}Q,,$
let $Zinmathscr L(H),$ denote the projector onto the second summand. In general, this projection onto $operatorname{Im}Q,$ along $operatorname{Im}P,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $operatorname{Im}Q,$ nor $operatorname{Im}P,$ equals ${0}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$|PQ|:leqslant:sqrt{1-|Z|^{-2}};.$$
Proof:$ $ If $xinoperatorname{Im}Q,$ is a unit vector, then $,Zx=x,$ and $,ZPx=0$. Thus,
$$1=|Zx-ZPx|:leqslant:|Z|: |(1-P)x|\[2ex]
Longrightarrowquadfrac1{|Z|^2} :leqslant: biglangle(1-P)xmid (1-P)xbigrangle
= biglangle(1-P)xmid xbigrangle
= 1-|Px|^2qquadtext{or}quad|Px|;leqslant; sqrt{1-|Z|^{-2}}$$
The claim now follows from
$$|PQ|:=:sup_{uin H,,|u|leq 1}|PQu|
:=:sup_{xinoperatorname{Im}Q,,|x|leqslant 1}|Px|:.$$
Remark:$ $ Every non-zero projector satisfies $|Z|geqslant1$.
And $|Z|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
answered Jan 31 at 17:36
HannoHanno
2,484628
$endgroup$
add a comment |
$begingroup$
The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=operatorname{Im}Poplusoperatorname{Im}Q,,$
let $Zinmathscr L(H),$ denote the projector onto the second summand. In general, this projection onto $operatorname{Im}Q,$ along $operatorname{Im}P,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $operatorname{Im}Q,$ nor $operatorname{Im}P,$ equals ${0}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$|PQ|:leqslant:sqrt{1-|Z|^{-2}};.$$
Proof:$ $ If $xinoperatorname{Im}Q,$ is a unit vector, then $,Zx=x,$ and $,ZPx=0$. Thus,
$$1=|Zx-ZPx|:leqslant:|Z|: |(1-P)x|\[2ex]
Longrightarrowquadfrac1{|Z|^2} :leqslant: biglangle(1-P)xmid (1-P)xbigrangle
= biglangle(1-P)xmid xbigrangle
= 1-|Px|^2qquadtext{or}quad|Px|;leqslant; sqrt{1-|Z|^{-2}}$$
The claim now follows from
$$|PQ|:=:sup_{uin H,,|u|leq 1}|PQu|
:=:sup_{xinoperatorname{Im}Q,,|x|leqslant 1}|Px|:.$$
Remark:$ $ Every non-zero projector satisfies $|Z|geqslant1$.
And $|Z|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
answered Jan 31 at 17:36
HannoHanno
2,484628
$endgroup$
The norm estimate does indeed hold.
I was told that S. Afriat proved it in the mid-1950s, the reference is in the footnote. Actually, I haven't seen the paper, but received a cue, and here's the reasoning.
Given the direct sum decomposition $H=operatorname{Im}Poplusoperatorname{Im}Q,,$
let $Zinmathscr L(H),$ denote the projector onto the second summand. In general, this projection onto $operatorname{Im}Q,$ along $operatorname{Im}P,$ is oblique. And along that the hypothesis that the direct sum decomposition is not trivial is made, i.e., neither $operatorname{Im}Q,$ nor $operatorname{Im}P,$ equals ${0}$, corresponding to the cases $Z=0$ or $Z=1$, respectively. Then one has
$$|PQ|:leqslant:sqrt{1-|Z|^{-2}};.$$
Proof:$ $ If $xinoperatorname{Im}Q,$ is a unit vector, then $,Zx=x,$ and $,ZPx=0$. Thus,
$$1=|Zx-ZPx|:leqslant:|Z|: |(1-P)x|\[2ex]
Longrightarrowquadfrac1{|Z|^2} :leqslant: biglangle(1-P)xmid (1-P)xbigrangle
= biglangle(1-P)xmid xbigrangle
= 1-|Px|^2qquadtext{or}quad|Px|;leqslant; sqrt{1-|Z|^{-2}}$$
The claim now follows from
$$|PQ|:=:sup_{uin H,,|u|leq 1}|PQu|
:=:sup_{xinoperatorname{Im}Q,,|x|leqslant 1}|Px|:.$$
Remark:$ $ Every non-zero projector satisfies $|Z|geqslant1$.
And $|Z|=1$ holds if and only if $Z$ is an orthogonal projector.
Sydney N. Afriat "Orthogonal and oblique projectors and the characteristics of pairs of vector spaces", Proc. Cambridge Philos. Soc. 53, 1957
answered Jan 31 at 17:36
HannoHanno
2,484628
edited Feb 5 at 11:51
answered Jan 31 at 17:36
HannoHanno
2,484628
answered Jan 31 at 17:36
HannoHanno
2,484628
answered Jan 31 at 17:36
HannoHanno
2,484628
2,484628
add a comment |
add a comment |
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