$P,Q$ are Hermitian operators with only non-negative eigenvalues, then tr($PQ$) = 0 $implies$ that $PQ = 0$.












1












$begingroup$


So far I can prove:



suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.



Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...



Now there are at least two possibilities to make tr() = $0$.



i) all $p_i$ = $0$, then of course $PQ = 0$



ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...



iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???



I think they should be a very easy problem... But I am just stuck somehow..






linear-algebra operator-theory







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  • $begingroup$
    Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
    $endgroup$
    – dantopa
    Jan 19 at 5:30






  • 1




    $begingroup$
    Try to use square roots $sqrt{P}, sqrt{Q}$.
    $endgroup$
    – Song
    Jan 19 at 5:40
















1












$begingroup$


So far I can prove:



suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.



Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...



Now there are at least two possibilities to make tr() = $0$.



i) all $p_i$ = $0$, then of course $PQ = 0$



ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...



iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???



I think they should be a very easy problem... But I am just stuck somehow..






linear-algebra operator-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
    $endgroup$
    – dantopa
    Jan 19 at 5:30






  • 1




    $begingroup$
    Try to use square roots $sqrt{P}, sqrt{Q}$.
    $endgroup$
    – Song
    Jan 19 at 5:40














1












1








1





$begingroup$


So far I can prove:



suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.



Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...



Now there are at least two possibilities to make tr() = $0$.



i) all $p_i$ = $0$, then of course $PQ = 0$



ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...



iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???



I think they should be a very easy problem... But I am just stuck somehow..






linear-algebra operator-theory







share|cite|improve this question











$endgroup$




So far I can prove:



suppose there is a basis set ${|i>}$ that diagonalize $P$ with eigenvalue $p_i$, then tr($PQ$) = $sum_i<i|PQ|i>$ = $sum_ip_i<i|Q|i>$.



Since $P, Q$ are Hermitian with non-negative eigenvalues, both $p_i$ and $<i|Q|i>$ is non-negative...



Now there are at least two possibilities to make tr() = $0$.



i) all $p_i$ = $0$, then of course $PQ = 0$



ii) all $<i|Q|i>$ = $0$, then I can't prove $PQ = 0$...



iii) $p_i$, $<i|Q|i>$ alternates to be $0$, $PQ$ = ???



I think they should be a very easy problem... But I am just stuck somehow..






linear-algebra operator-theory




linear-algebra operator-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 19 at 6:36









Andrews

4081317




4081317










asked Jan 19 at 5:26









Jiashen TangJiashen Tang

82




82












  • $begingroup$
    Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
    $endgroup$
    – dantopa
    Jan 19 at 5:30






  • 1




    $begingroup$
    Try to use square roots $sqrt{P}, sqrt{Q}$.
    $endgroup$
    – Song
    Jan 19 at 5:40


















  • $begingroup$
    Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
    $endgroup$
    – dantopa
    Jan 19 at 5:30






  • 1




    $begingroup$
    Try to use square roots $sqrt{P}, sqrt{Q}$.
    $endgroup$
    – Song
    Jan 19 at 5:40
















$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30




$begingroup$
Welcome to Mathematics Stack Exchange! Take the short tour to see how how to get the most from your time here.
$endgroup$
– dantopa
Jan 19 at 5:30




1




1




$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40




$begingroup$
Try to use square roots $sqrt{P}, sqrt{Q}$.
$endgroup$
– Song
Jan 19 at 5:40










1 Answer
1






active

oldest

votes


















0












$begingroup$

Here is a fairly direct argument.



You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
    $endgroup$
    – Jiashen Tang
    Jan 19 at 19:18










  • $begingroup$
    "Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
    $endgroup$
    – Martin Argerami
    Jan 19 at 21:36











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









0












$begingroup$

Here is a fairly direct argument.



You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
    $endgroup$
    – Jiashen Tang
    Jan 19 at 19:18










  • $begingroup$
    "Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
    $endgroup$
    – Martin Argerami
    Jan 19 at 21:36
















0












$begingroup$

Here is a fairly direct argument.



You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
    $endgroup$
    – Jiashen Tang
    Jan 19 at 19:18










  • $begingroup$
    "Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
    $endgroup$
    – Martin Argerami
    Jan 19 at 21:36














0












0








0





$begingroup$

Here is a fairly direct argument.



You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.






share|cite|improve this answer









$endgroup$



Here is a fairly direct argument.



You have
$$
0=operatorname{tr}(PQ)=operatorname{tr}(P^{1/2}P^{1/2}Q)=operatorname{tr}(P^{1/2}QP^{1/2}).
$$
As $Qgeq0$, you have $P^{1/2}QP^{1/2}=P^{1/2}Q(P^{1/2})^*geq0$. And the trace is faithful, so $P^{1/2}QP^{1/2}=0$. Then
$$
0=P^{1/2}QP^{1/2}=P^{1/2}Q^{1/2}Q^{1/2}P^{1/2}=(Q^{1/2}P^{1/2})^*Q^{1/2}P^{1/2}=0.
$$
So $Q^{1/2}P^{1/2}=0$. Now multiply by $Q^{1/2}$ on the left and by $P^{1/2}$ on the right.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 19 at 16:17









Martin ArgeramiMartin Argerami

127k1182183




127k1182183












  • $begingroup$
    Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
    $endgroup$
    – Jiashen Tang
    Jan 19 at 19:18










  • $begingroup$
    "Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
    $endgroup$
    – Martin Argerami
    Jan 19 at 21:36


















  • $begingroup$
    Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
    $endgroup$
    – Jiashen Tang
    Jan 19 at 19:18










  • $begingroup$
    "Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
    $endgroup$
    – Martin Argerami
    Jan 19 at 21:36
















$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18




$begingroup$
Thanks Martin! Two points I don't understand. i) $Q ≥ 0$, does it mean every entry of $Q$ matrix $≥ 0$? ii) what does 'trace is faithful' mean and why that implies $sqrt{P}Qsqrt{P}$ = $0$. I'm a physics student, I am not quite familiar with linear algebra.
$endgroup$
– Jiashen Tang
Jan 19 at 19:18












$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36




$begingroup$
"Nonnegative entries" is not a property of an operator; when talking about operators, "positive" means "positive semi-definite". That the trace is faithful means that it is injective on positive operators; this is an easy consequence of the fact that the trace is the sum of the eigenvalues.
$endgroup$
– Martin Argerami
Jan 19 at 21:36


















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