Given symmetric semidefinite matrix A and B, prove AB = 0 if and only if tr(AB)=0 [closed]
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Given $Ain S^{n}_{+}$ means that $A$ and $B$ are symmetric semidefinite matrix. Can we prove that $operatorname{tr}(AB)=0$ if and only if $AB=0$ ?
linear-algebra trace
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
asked Nov 16 at 1:41
Yu Lin
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closed as off-topic by LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs Nov 16 at 23:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
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favorite
Given $Ain S^{n}_{+}$ means that $A$ and $B$ are symmetric semidefinite matrix. Can we prove that $operatorname{tr}(AB)=0$ if and only if $AB=0$ ?
linear-algebra trace
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
asked Nov 16 at 1:41
Yu Lin
62
closed as off-topic by LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs Nov 16 at 23:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given $Ain S^{n}_{+}$ means that $A$ and $B$ are symmetric semidefinite matrix. Can we prove that $operatorname{tr}(AB)=0$ if and only if $AB=0$ ?
linear-algebra trace
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
asked Nov 16 at 1:41
Yu Lin
62
Given $Ain S^{n}_{+}$ means that $A$ and $B$ are symmetric semidefinite matrix. Can we prove that $operatorname{tr}(AB)=0$ if and only if $AB=0$ ?
linear-algebra trace
linear-algebra trace
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
asked Nov 16 at 1:41
Yu Lin
62
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
asked Nov 16 at 1:41
Yu Lin
62
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
edited Nov 16 at 15:22
Davide Giraudo
124k16149253
124k16149253
asked Nov 16 at 1:41
Yu Lin
62
asked Nov 16 at 1:41
Yu Lin
62
asked Nov 16 at 1:41
Yu Lin
62
62
closed as off-topic by LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs Nov 16 at 23:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs Nov 16 at 23:39
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – LinAlg, amWhy, user10354138, Trevor Gunn, Gibbs
If this question can be reworded to fit the rules in the help center, please edit the question.
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3 Answers
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accepted
Hint: $operatorname{tr}(AB)=operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=left(A^{1/2}B^{1/2}right)^astleft(A^{1/2}B^{1/2}right)$ is positive semidefinite.
answered Nov 16 at 1:46
user1551
70.3k566125
add a comment |
up vote
0
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See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg
A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.
answered Nov 16 at 2:06
John Nash
6218
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
add a comment |
up vote
0
down vote
You have to know the following results:
$tr(AB)=tr(BA)$ - trace commute in blocks- Any symmetric matrix is diagonalizable
- A symmetric matrix is positive-semidefinite $iff$ all its eigenvalues are $geq 0$
Well, $AB=0 implies tr(AB)=0$ its trivial.
Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!
answered Nov 16 at 2:22
Robson
725221
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
Hint: $operatorname{tr}(AB)=operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=left(A^{1/2}B^{1/2}right)^astleft(A^{1/2}B^{1/2}right)$ is positive semidefinite.
answered Nov 16 at 1:46
user1551
70.3k566125
add a comment |
up vote
2
down vote
accepted
Hint: $operatorname{tr}(AB)=operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=left(A^{1/2}B^{1/2}right)^astleft(A^{1/2}B^{1/2}right)$ is positive semidefinite.
answered Nov 16 at 1:46
user1551
70.3k566125
add a comment |
up vote
2
down vote
accepted
up vote
2
down vote
accepted
Hint: $operatorname{tr}(AB)=operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=left(A^{1/2}B^{1/2}right)^astleft(A^{1/2}B^{1/2}right)$ is positive semidefinite.
answered Nov 16 at 1:46
user1551
70.3k566125
Hint: $operatorname{tr}(AB)=operatorname{tr}(B^{1/2}AB^{1/2})$ and $B^{1/2}AB^{1/2}=left(A^{1/2}B^{1/2}right)^astleft(A^{1/2}B^{1/2}right)$ is positive semidefinite.
answered Nov 16 at 1:46
user1551
70.3k566125
edited Nov 16 at 2:08
answered Nov 16 at 1:46
user1551
70.3k566125
answered Nov 16 at 1:46
user1551
70.3k566125
answered Nov 16 at 1:46
user1551
70.3k566125
70.3k566125
add a comment |
add a comment |
up vote
0
down vote
See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg
A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.
answered Nov 16 at 2:06
John Nash
6218
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
add a comment |
up vote
0
down vote
See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg
A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.
answered Nov 16 at 2:06
John Nash
6218
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
add a comment |
up vote
0
down vote
up vote
0
down vote
See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg
A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.
answered Nov 16 at 2:06
John Nash
6218
See it if any confusion please ask https://i.stack.imgur.com/TxtTb.jpg
A is diagonal matrix of the form A = diag(a1, . . . , an), a1 ≥ . . . ≥ an ≥ 0.
answered Nov 16 at 2:06
John Nash
6218
answered Nov 16 at 2:06
John Nash
6218
answered Nov 16 at 2:06
John Nash
6218
answered Nov 16 at 2:06
John Nash
6218
6218
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
add a comment |
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
I am a bit confused, why A is a diagonal matrix?
– Yu Lin
Nov 16 at 2:15
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
This is very elementary see your question and proof. Think it yourself.
– John Nash
Nov 16 at 2:32
add a comment |
up vote
0
down vote
You have to know the following results:
$tr(AB)=tr(BA)$ - trace commute in blocks- Any symmetric matrix is diagonalizable
- A symmetric matrix is positive-semidefinite $iff$ all its eigenvalues are $geq 0$
Well, $AB=0 implies tr(AB)=0$ its trivial.
Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!
answered Nov 16 at 2:22
Robson
725221
add a comment |
up vote
0
down vote
You have to know the following results:
$tr(AB)=tr(BA)$ - trace commute in blocks- Any symmetric matrix is diagonalizable
- A symmetric matrix is positive-semidefinite $iff$ all its eigenvalues are $geq 0$
Well, $AB=0 implies tr(AB)=0$ its trivial.
Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!
answered Nov 16 at 2:22
Robson
725221
add a comment |
up vote
0
down vote
up vote
0
down vote
You have to know the following results:
$tr(AB)=tr(BA)$ - trace commute in blocks- Any symmetric matrix is diagonalizable
- A symmetric matrix is positive-semidefinite $iff$ all its eigenvalues are $geq 0$
Well, $AB=0 implies tr(AB)=0$ its trivial.
Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!
answered Nov 16 at 2:22
Robson
725221
You have to know the following results:
$tr(AB)=tr(BA)$ - trace commute in blocks- Any symmetric matrix is diagonalizable
- A symmetric matrix is positive-semidefinite $iff$ all its eigenvalues are $geq 0$
Well, $AB=0 implies tr(AB)=0$ its trivial.
Then for the converse, you can use 1.2.3. to understand why the hint given by @user1551 helps you!
answered Nov 16 at 2:22
Robson
725221
answered Nov 16 at 2:22
Robson
725221
answered Nov 16 at 2:22
Robson
725221
answered Nov 16 at 2:22
Robson
725221
725221
add a comment |
add a comment |
