$P,Q$ are Hermitian operators with only non-negative eigenvalues, then tr($PQ$) = 0 $implies$ that $PQ = 0$.
$begingroup$
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
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
add a comment |
$begingroup$
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
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
$endgroup$
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
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
edited Jan 19 at 6:36
Andrews
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
edited Jan 19 at 6:36
Andrews
4081317
edited Jan 19 at 6:36
Andrews
4081317
edited Jan 19 at 6:36
Andrews
4081317
4081317
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
asked Jan 19 at 5:26
Jiashen TangJiashen Tang
82
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
add a comment |
$begingroup$
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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
Your Answer
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1 Answer
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1 Answer
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votes
$begingroup$
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.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
$endgroup$
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.
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
answered Jan 19 at 16:17
Martin ArgeramiMartin Argerami
127k1182183
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
add a comment |
$begingroup$
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
add a comment |
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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