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Let $A, B$ be two positive definite $2 times 2$ matrices. Prove or disprove: $AB+BA$ is positive definite.

Positive definiteness and eigenvalue inequalities of two symmetric matricesInequality of Positive-definite matrix.Invertibility of block matrices, with the property of being symmetric, positive definite, and of full rank:Prove that a matrix is positive definitePositive Definite or Negative Definite MatrixIs a complex symmetric matrix with positive definite real part diagonalizable?Geometric interpretation of positive semi-definiteness of sum of matricesWhy is the product of two matrices the product of their partitions?How can I find two symmetric positive definite roots of a symmetric positive definite matrix?How to construct negative definite $AB$ if $A$ and $B$ are positive definite?

4

1

$begingroup$

I know that $AB+BA$ is not necessarily positive definite, as this question has been asked before on here. What I don't understand is how one would go about constructing counter-examples. Previous answers to this question just state the counter-examples, such as $A = beginbmatrix 6 & 0 \
0 & 1 \ endbmatrix$, $B = beginbmatrix 2 & 1 \
1 & 1 \ endbmatrix$. Is there some intuition behind how these matrices are chosen, or is it more a matter of just fiddling with $2 times 2$ matrices until we find a counter-example?

Thanks in advance.

linear-algebra matrices positive-definite

share|cite|improve this question

asked Mar 15 at 1:32

Mike DMike D

1084

$endgroup$

add a comment | 

4

1

$begingroup$

I know that $AB+BA$ is not necessarily positive definite, as this question has been asked before on here. What I don't understand is how one would go about constructing counter-examples. Previous answers to this question just state the counter-examples, such as $A = beginbmatrix 6 & 0 \
0 & 1 \ endbmatrix$, $B = beginbmatrix 2 & 1 \
1 & 1 \ endbmatrix$. Is there some intuition behind how these matrices are chosen, or is it more a matter of just fiddling with $2 times 2$ matrices until we find a counter-example?

Thanks in advance.

linear-algebra matrices positive-definite

share|cite|improve this question

asked Mar 15 at 1:32

Mike DMike D

1084

$endgroup$

add a comment | 

$begingroup$

I know that $AB+BA$ is not necessarily positive definite, as this question has been asked before on here. What I don't understand is how one would go about constructing counter-examples. Previous answers to this question just state the counter-examples, such as $A = beginbmatrix 6 & 0 \
0 & 1 \ endbmatrix$, $B = beginbmatrix 2 & 1 \
1 & 1 \ endbmatrix$. Is there some intuition behind how these matrices are chosen, or is it more a matter of just fiddling with $2 times 2$ matrices until we find a counter-example?

Thanks in advance.

linear-algebra matrices positive-definite

share|cite|improve this question

asked Mar 15 at 1:32

Mike DMike D

1084

$endgroup$

share|cite|improve this question

asked Mar 15 at 1:32

Mike DMike D

1084

share|cite|improve this question

asked Mar 15 at 1:32

Mike DMike D

1084

asked Mar 15 at 1:32

Mike DMike D

1084

asked Mar 15 at 1:32

Mike DMike D

1084

1 Answer
1

active

oldest

votes

2

$begingroup$

Here is one way to think about the problem. If $AB+BA$ is always positive definite, by passing $A$ to the limit, $AB+BA$ should be positive semidefinite too when $A$ is a rank-one positive semidefinite matrix.

So, with such an $A$, if you can construct a counterexample such that $Bsucc0$ but $AB+BA$ is indefinite, then by a continuity argument, you can get a counterexample with a positive definite $A$.

E.g. $AB+BA$ is indefinite when $A=pmatrix1&0\ 0&0$ and $B=pmatrix2&1\ 1&1$. It follows that $AB+BA$ is indefinite when $A=pmatrix1\ &t$ and $t>0$ is sufficiently small (the value of $B$ here is actually unimportant; as long as $B$ is not diagonal, we get a counterexample). Multiply $A$ by $frac1t$, we may pick $A=pmatrixfrac1t\ &1$ as well. Now we love integer matrices. So, let's set $frac1t=n$ for some positive integer $n$. It remains to find an $n$ that really works. Since
$$
det(AB+BA)=detpmatrix4n&n+1\ n+1&2=8n-(n+1)^2=8-(n-3)^2,
$$
any $n>3+sqrt8$ with do the job. The least integer such $n$ is $6$, which is the value used in the answer you have mentioned.

share|cite|improve this answer

answered Mar 15 at 7:08

user1551user1551

73.8k566129

$endgroup$

add a comment | 

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2

$begingroup$

Here is one way to think about the problem. If $AB+BA$ is always positive definite, by passing $A$ to the limit, $AB+BA$ should be positive semidefinite too when $A$ is a rank-one positive semidefinite matrix.

So, with such an $A$, if you can construct a counterexample such that $Bsucc0$ but $AB+BA$ is indefinite, then by a continuity argument, you can get a counterexample with a positive definite $A$.

E.g. $AB+BA$ is indefinite when $A=pmatrix1&0\ 0&0$ and $B=pmatrix2&1\ 1&1$. It follows that $AB+BA$ is indefinite when $A=pmatrix1\ &t$ and $t>0$ is sufficiently small (the value of $B$ here is actually unimportant; as long as $B$ is not diagonal, we get a counterexample). Multiply $A$ by $frac1t$, we may pick $A=pmatrixfrac1t\ &1$ as well. Now we love integer matrices. So, let's set $frac1t=n$ for some positive integer $n$. It remains to find an $n$ that really works. Since
$$
det(AB+BA)=detpmatrix4n&n+1\ n+1&2=8n-(n+1)^2=8-(n-3)^2,
$$
any $n>3+sqrt8$ with do the job. The least integer such $n$ is $6$, which is the value used in the answer you have mentioned.

share|cite|improve this answer

answered Mar 15 at 7:08

user1551user1551

73.8k566129

$endgroup$

add a comment | 

2

$begingroup$

Here is one way to think about the problem. If $AB+BA$ is always positive definite, by passing $A$ to the limit, $AB+BA$ should be positive semidefinite too when $A$ is a rank-one positive semidefinite matrix.

So, with such an $A$, if you can construct a counterexample such that $Bsucc0$ but $AB+BA$ is indefinite, then by a continuity argument, you can get a counterexample with a positive definite $A$.

E.g. $AB+BA$ is indefinite when $A=pmatrix1&0\ 0&0$ and $B=pmatrix2&1\ 1&1$. It follows that $AB+BA$ is indefinite when $A=pmatrix1\ &t$ and $t>0$ is sufficiently small (the value of $B$ here is actually unimportant; as long as $B$ is not diagonal, we get a counterexample). Multiply $A$ by $frac1t$, we may pick $A=pmatrixfrac1t\ &1$ as well. Now we love integer matrices. So, let's set $frac1t=n$ for some positive integer $n$. It remains to find an $n$ that really works. Since
$$
det(AB+BA)=detpmatrix4n&n+1\ n+1&2=8n-(n+1)^2=8-(n-3)^2,
$$
any $n>3+sqrt8$ with do the job. The least integer such $n$ is $6$, which is the value used in the answer you have mentioned.

share|cite|improve this answer

answered Mar 15 at 7:08

user1551user1551

73.8k566129

$endgroup$

add a comment | 

$begingroup$

Here is one way to think about the problem. If $AB+BA$ is always positive definite, by passing $A$ to the limit, $AB+BA$ should be positive semidefinite too when $A$ is a rank-one positive semidefinite matrix.

So, with such an $A$, if you can construct a counterexample such that $Bsucc0$ but $AB+BA$ is indefinite, then by a continuity argument, you can get a counterexample with a positive definite $A$.

E.g. $AB+BA$ is indefinite when $A=pmatrix1&0\ 0&0$ and $B=pmatrix2&1\ 1&1$. It follows that $AB+BA$ is indefinite when $A=pmatrix1\ &t$ and $t>0$ is sufficiently small (the value of $B$ here is actually unimportant; as long as $B$ is not diagonal, we get a counterexample). Multiply $A$ by $frac1t$, we may pick $A=pmatrixfrac1t\ &1$ as well. Now we love integer matrices. So, let's set $frac1t=n$ for some positive integer $n$. It remains to find an $n$ that really works. Since
$$
det(AB+BA)=detpmatrix4n&n+1\ n+1&2=8n-(n+1)^2=8-(n-3)^2,
$$
any $n>3+sqrt8$ with do the job. The least integer such $n$ is $6$, which is the value used in the answer you have mentioned.

share|cite|improve this answer

answered Mar 15 at 7:08

user1551user1551

73.8k566129

$endgroup$

share|cite|improve this answer

answered Mar 15 at 7:08

user1551user1551

73.8k566129

answered Mar 15 at 7:08

user1551user1551

73.8k566129

answered Mar 15 at 7:08

user1551user1551

73.8k566129