Positive operator is symmetric?
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If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?
However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?
Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.
Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed
begin{align*}
0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
end{align*}
This implies that
$$langle Tx,yrangle = -langle Ty,xrangle.$$
Exchanging $x$ with $ix$ yields
$$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
why
$$langle Tx,yrangle = langle Ty,xrangle$$
all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.
Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:
We have that
$$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.
matrices functional-analysis operator-theory
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
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add a comment |
$begingroup$
If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?
However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?
Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.
Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed
begin{align*}
0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
end{align*}
This implies that
$$langle Tx,yrangle = -langle Ty,xrangle.$$
Exchanging $x$ with $ix$ yields
$$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
why
$$langle Tx,yrangle = langle Ty,xrangle$$
all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.
Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:
We have that
$$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.
matrices functional-analysis operator-theory
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
$endgroup$
add a comment |
$begingroup$
If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?
However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?
Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.
Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed
begin{align*}
0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
end{align*}
This implies that
$$langle Tx,yrangle = -langle Ty,xrangle.$$
Exchanging $x$ with $ix$ yields
$$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
why
$$langle Tx,yrangle = langle Ty,xrangle$$
all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.
Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:
We have that
$$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.
matrices functional-analysis operator-theory
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
$endgroup$
If I understand correctly then for an operator $mathcal{A}$ defined on a Hilbert space $mathcal{H}$, $langle mathcal{A}x,xranglegeq 0$ does not necessarily imply that $mathcal{A}$ is Hermitian: $mathcal{A} = mathcal{A}^ast$. See for instance Is a positive operator symmetric?
However I was shown the following proof of the supposedly erroneous statement and was wondering if there is something wrong in it for I can't find it myself?
Lemma 1: $langle Tx,xrangle = 0$ for every $xin mathcal{H}$ implies that $T equiv 0$.
Proof: We show that $langle Tx,yrangle = 0$ for every pair $x,yin mathcal{H}.$ Indeed
begin{align*}
0 = langle T(x+y),x+yrangle - langle T(x-y),x-yrangle & = 2langle Tx,yrangle+2langle Ty,xrangle.
end{align*}
This implies that
$$langle Tx,yrangle = -langle Ty,xrangle.$$
Exchanging $x$ with $ix$ yields
$$0 = ilangle Tx,yrangle -ilangle Ty,xrangle$$
why
$$langle Tx,yrangle = langle Ty,xrangle$$
all in all we find that $langle Tx,yrangle = pm langle Ty,xrangle$ which implies that both are $0$.
Proof that $langle mathcal{A}x,xranglegeq 0$ implies that $mathcal{A}^ast = mathcal{A}$:
We have that
$$mathbb{R}nilangle mathcal{A}x,xrangle = langle x,mathcal{A}^ast xrangle = overline{langle mathcal{A}^ast x,xrangle } = langle mathcal{A}^ast x, xrangleRightarrow langle (mathcal{A}-mathcal{A}^ast)x,xrangle = 0$$
for every $x$ thus by the lemma $mathcal{A}-mathcal{A}^astequiv 0$.
matrices functional-analysis operator-theory
matrices functional-analysis operator-theory
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
asked Jan 3 at 11:50
Olof RubinOlof Rubin
1,131316
1,131316
add a comment |
add a comment |
1 Answer
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In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
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Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
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– Olof Rubin
Jan 3 at 11:57
add a comment |
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In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
add a comment |
$begingroup$
In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
add a comment |
$begingroup$
In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
$endgroup$
In the case complex scalars $langle Tx,x rangle=0$ for all $x$ implies $T=0$ and $langle Tx,x rangle geq 0$ for all $x$ implies $T=T^{*}$ (as you have shown). This is not true for real scalars. Rotation by $90$ degrees on $mathbb R^{2}$ is a counter example.
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
edited Jan 3 at 11:58
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
answered Jan 3 at 11:55
Kavi Rama MurthyKavi Rama Murthy
56.6k42159
56.6k42159
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
add a comment |
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
$begingroup$
Ok very interesting thank you! I see now that the counterexample in the linked post assumes that we only allow real scalars.
$endgroup$
– Olof Rubin
Jan 3 at 11:57
add a comment |
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