Inequality involving projectors on a Hilbert space isomorphic to $mathbb{C}^n$
$begingroup$
Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $mathbb{H} cong mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, ${P_1, P_2} = 0$. Let $x$ be a unit $L_2$-norm vector.
Show that $(x^TP_1x)^2 + (x^TP_2x)^2 leq 1$
I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.
I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.
linear-algebra projection projection-matrices
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
$endgroup$
add a comment |
$begingroup$
Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $mathbb{H} cong mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, ${P_1, P_2} = 0$. Let $x$ be a unit $L_2$-norm vector.
Show that $(x^TP_1x)^2 + (x^TP_2x)^2 leq 1$
I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.
I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.
linear-algebra projection projection-matrices
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
$endgroup$
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56
add a comment |
$begingroup$
Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $mathbb{H} cong mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, ${P_1, P_2} = 0$. Let $x$ be a unit $L_2$-norm vector.
Show that $(x^TP_1x)^2 + (x^TP_2x)^2 leq 1$
I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.
I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.
linear-algebra projection projection-matrices
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
$endgroup$
Suppose we have Hermitian operators $P_1$ and $P_2$ on Hilbert space $mathbb{H} cong mathbb{C}^n$, such that: $P_1^2=P_1$, $P_2^2=P_2$, ${P_1, P_2} = 0$. Let $x$ be a unit $L_2$-norm vector.
Show that $(x^TP_1x)^2 + (x^TP_2x)^2 leq 1$
I can only see how one can show that the same inequality holds for $RHS=2$. That can be easily seen because of the fact that a projection of a vector is always "shorter" than the vector. But I'm confused about the $RHS=1$ case.
I think I should somehow use the restriction on the anticommutator, but I can't figure that out. It is easy to prove that in this case $P_1+P_2$ is also a projector but I can't see how this helps here.
linear-algebra projection projection-matrices
linear-algebra projection projection-matrices
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
edited Dec 17 '18 at 14:18
mechanodroid
27.8k62447
27.8k62447
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
asked Dec 17 '18 at 14:00
Aleksandr BerezutskiiAleksandr Berezutskii
103
103
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56
add a comment |
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In fact, the correct statement should be$$
x^TP_1x + x^TP_2x leq 1,quadcdots(*).
$$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let
$$
x= P_1x + P_2 x + P_3x,$$
where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $ineq j$. Thus, we have
$$
1=x^Tx =sum_{i=1}^3 x^TP_ix +sum_{ineq j}x^TP_iP_jx= sum_{i=1}^3 x^TP_ix geq x^TP_1x + x^TP_2x,
$$ as desired.
$textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQneq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pvneq 0$. By the assumption that $PQ=-QP$, we have
$$
Pv = -QPv.
$$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
$endgroup$
add a comment |
$begingroup$
The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.
Notice that for an orthogonal projection $P$ and $x in Bbb{H}$ we have
$$|Px|^2 = langle Px,Pxrangle = langle P^*Px,xrangle = langle P^2x,xrangle = langle Px,xrangle$$
Hence for $x in Bbb{H}$, $|x| = 1$ we have
begin{align}
LHS &= langle P_1x,xrangle^2 + langle P_2x,xrangle^2 \
&= |P_1x|^4 + |P_2x|^4 \
&le |P_1x|^4 + 2|P_1x|^2|P_2x|^2 + |P_2x|^4 \
&= (|P_1x|^2+|P_2x|^2)^2
end{align}
Since $P_1x perp P_2x$ we have
$$LHS le (|P_1x|^2+|P_2x|^2)^2 = |P_1x + P_2x|^4 = |(P_1+P_2)x|^4 le |x|^4 = 1$$
because $P_1 + P_2$ is an orthogonal projection and hence of norm $le 1$.
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043978%2finequality-involving-projectors-on-a-hilbert-space-isomorphic-to-mathbbcn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
In fact, the correct statement should be$$
x^TP_1x + x^TP_2x leq 1,quadcdots(*).
$$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let
$$
x= P_1x + P_2 x + P_3x,$$
where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $ineq j$. Thus, we have
$$
1=x^Tx =sum_{i=1}^3 x^TP_ix +sum_{ineq j}x^TP_iP_jx= sum_{i=1}^3 x^TP_ix geq x^TP_1x + x^TP_2x,
$$ as desired.
$textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQneq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pvneq 0$. By the assumption that $PQ=-QP$, we have
$$
Pv = -QPv.
$$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
$endgroup$
add a comment |
$begingroup$
In fact, the correct statement should be$$
x^TP_1x + x^TP_2x leq 1,quadcdots(*).
$$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let
$$
x= P_1x + P_2 x + P_3x,$$
where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $ineq j$. Thus, we have
$$
1=x^Tx =sum_{i=1}^3 x^TP_ix +sum_{ineq j}x^TP_iP_jx= sum_{i=1}^3 x^TP_ix geq x^TP_1x + x^TP_2x,
$$ as desired.
$textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQneq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pvneq 0$. By the assumption that $PQ=-QP$, we have
$$
Pv = -QPv.
$$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
$endgroup$
add a comment |
$begingroup$
In fact, the correct statement should be$$
x^TP_1x + x^TP_2x leq 1,quadcdots(*).
$$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let
$$
x= P_1x + P_2 x + P_3x,$$
where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $ineq j$. Thus, we have
$$
1=x^Tx =sum_{i=1}^3 x^TP_ix +sum_{ineq j}x^TP_iP_jx= sum_{i=1}^3 x^TP_ix geq x^TP_1x + x^TP_2x,
$$ as desired.
$textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQneq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pvneq 0$. By the assumption that $PQ=-QP$, we have
$$
Pv = -QPv.
$$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
$endgroup$
In fact, the correct statement should be$$
x^TP_1x + x^TP_2x leq 1,quadcdots(*).
$$ (But your version is still true as we can see by squaring $(*)$.) This is called Bessel's inequality. To prove this, let
$$
x= P_1x + P_2 x + P_3x,$$
where $P_3 = I-P_1-P_2$. We can see that $P_3$ is also a projection and $P_iP_j = O$ for $ineq j$. Thus, we have
$$
1=x^Tx =sum_{i=1}^3 x^TP_ix +sum_{ineq j}x^TP_iP_jx= sum_{i=1}^3 x^TP_ix geq x^TP_1x + x^TP_2x,
$$ as desired.
$textbf{EDIT:}$ I added short proof that if $P,Q$ are orthogonal projections such that $PQ+QP=0$, then $PQ=QP=0$.
Assume to the contrary that $PQneq 0$. Then there is $v$ such that $v=Qv$ and $PQv=Pvneq 0$. By the assumption that $PQ=-QP$, we have
$$
Pv = -QPv.
$$ This implies that $Q$ has $-1$ as an eigenvalue. This leads to a contradiction.
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
edited Dec 17 '18 at 15:33
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
answered Dec 17 '18 at 14:19
SongSong
14.8k1635
14.8k1635
add a comment |
add a comment |
$begingroup$
The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.
Notice that for an orthogonal projection $P$ and $x in Bbb{H}$ we have
$$|Px|^2 = langle Px,Pxrangle = langle P^*Px,xrangle = langle P^2x,xrangle = langle Px,xrangle$$
Hence for $x in Bbb{H}$, $|x| = 1$ we have
begin{align}
LHS &= langle P_1x,xrangle^2 + langle P_2x,xrangle^2 \
&= |P_1x|^4 + |P_2x|^4 \
&le |P_1x|^4 + 2|P_1x|^2|P_2x|^2 + |P_2x|^4 \
&= (|P_1x|^2+|P_2x|^2)^2
end{align}
Since $P_1x perp P_2x$ we have
$$LHS le (|P_1x|^2+|P_2x|^2)^2 = |P_1x + P_2x|^4 = |(P_1+P_2)x|^4 le |x|^4 = 1$$
because $P_1 + P_2$ is an orthogonal projection and hence of norm $le 1$.
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.
Notice that for an orthogonal projection $P$ and $x in Bbb{H}$ we have
$$|Px|^2 = langle Px,Pxrangle = langle P^*Px,xrangle = langle P^2x,xrangle = langle Px,xrangle$$
Hence for $x in Bbb{H}$, $|x| = 1$ we have
begin{align}
LHS &= langle P_1x,xrangle^2 + langle P_2x,xrangle^2 \
&= |P_1x|^4 + |P_2x|^4 \
&le |P_1x|^4 + 2|P_1x|^2|P_2x|^2 + |P_2x|^4 \
&= (|P_1x|^2+|P_2x|^2)^2
end{align}
Since $P_1x perp P_2x$ we have
$$LHS le (|P_1x|^2+|P_2x|^2)^2 = |P_1x + P_2x|^4 = |(P_1+P_2)x|^4 le |x|^4 = 1$$
because $P_1 + P_2$ is an orthogonal projection and hence of norm $le 1$.
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
$endgroup$
add a comment |
$begingroup$
The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.
Notice that for an orthogonal projection $P$ and $x in Bbb{H}$ we have
$$|Px|^2 = langle Px,Pxrangle = langle P^*Px,xrangle = langle P^2x,xrangle = langle Px,xrangle$$
Hence for $x in Bbb{H}$, $|x| = 1$ we have
begin{align}
LHS &= langle P_1x,xrangle^2 + langle P_2x,xrangle^2 \
&= |P_1x|^4 + |P_2x|^4 \
&le |P_1x|^4 + 2|P_1x|^2|P_2x|^2 + |P_2x|^4 \
&= (|P_1x|^2+|P_2x|^2)^2
end{align}
Since $P_1x perp P_2x$ we have
$$LHS le (|P_1x|^2+|P_2x|^2)^2 = |P_1x + P_2x|^4 = |(P_1+P_2)x|^4 le |x|^4 = 1$$
because $P_1 + P_2$ is an orthogonal projection and hence of norm $le 1$.
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
$endgroup$
The condition $P_1P_2 + P_2P_1 = 0$ holds if and only if $P_1$ and $P_2$ project onto orthogonal subspaces. Also it is equivalent to $P_1 + P_2$ being an orthogonal projection.
Notice that for an orthogonal projection $P$ and $x in Bbb{H}$ we have
$$|Px|^2 = langle Px,Pxrangle = langle P^*Px,xrangle = langle P^2x,xrangle = langle Px,xrangle$$
Hence for $x in Bbb{H}$, $|x| = 1$ we have
begin{align}
LHS &= langle P_1x,xrangle^2 + langle P_2x,xrangle^2 \
&= |P_1x|^4 + |P_2x|^4 \
&le |P_1x|^4 + 2|P_1x|^2|P_2x|^2 + |P_2x|^4 \
&= (|P_1x|^2+|P_2x|^2)^2
end{align}
Since $P_1x perp P_2x$ we have
$$LHS le (|P_1x|^2+|P_2x|^2)^2 = |P_1x + P_2x|^4 = |(P_1+P_2)x|^4 le |x|^4 = 1$$
because $P_1 + P_2$ is an orthogonal projection and hence of norm $le 1$.
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
answered Dec 17 '18 at 14:17
mechanodroidmechanodroid
27.8k62447
27.8k62447
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3043978%2finequality-involving-projectors-on-a-hilbert-space-isomorphic-to-mathbbcn%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
What is the meaning of ${P_1,P_2}=0. $ ?
$endgroup$
– Fred
Dec 17 '18 at 14:10
$begingroup$
That's notation for an anticommutator, i.e. ${P_1, P_2}=P_1 P_2 + P_2 P_1$
$endgroup$
– Aleksandr Berezutskii
Dec 17 '18 at 14:56