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
14.1k1633
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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 |
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2 Answers
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2 Answers
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$begingroup$
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
14.1k1633
$endgroup$
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
14.1k1633
$endgroup$
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
14.1k1633
$endgroup$
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
14.1k1633
edited Dec 17 '18 at 15:33
answered Dec 17 '18 at 14:19
SongSong
14.1k1633
answered Dec 17 '18 at 14:19
SongSong
14.1k1633
answered Dec 17 '18 at 14:19
SongSong
14.1k1633
14.1k1633
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 |
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$begingroup$
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