Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $A$ be a nonsingular real square matrix. Is it true that the matrix $$\frac{1}{2}(A+A')-2(A^{-1}+(A^{-1})')^{-1}$$ is positive semidefinite?

Here, $A'$ denotes the transpose of $A$.


Let $A,B$ be positive definite matrices of the same size, is it true that $\frac{1}{2}(AB+BA)−2((AB)^{−1}+(BA)^{−1})^{−1}$ is positive semidefinite?

share|cite|improve this question
I have numerically checked and the answer is no. You can use the following MATLAB code to see it for yourself. A=rand(5);eig(0.5*(A+A')-2*inv(inv(A)+inv(A'))) – user13838 Aug 23 '11 at 18:41
A remark: replacing $A$ by $-A$ multiplies the entire expression by $-1$, so the only way it is positive semidefinite is if it, in fact, equals zero for any matrix $A$. – alexx Aug 23 '11 at 19:07
...I should have said: the only way it is positive semidefinite for every $A$ is if it equals $0$ for every $A$. – alexx Aug 23 '11 at 19:24
up vote 2 down vote accepted

Your new hypothesis is still incorrect. Counterexample: for $A={\rm diag}(8,1)$ and $B=\begin{pmatrix}1&2\\2&8\end{pmatrix}$, we have $M=\frac{1}{2}(AB+BA)−2((AB)^{−1}+(BA)^{−1})^{−1} = \frac{49}{17}\begin{pmatrix}8&9\\9&8\end{pmatrix}$, whose two eigenvalues are $\frac{-49}{17}$ and $49$ resp., so $M$ is neither positive semidefinite nor negative semidefinite.

Next time, you may do some computational experiments first to see whether your hypothesis is violated. Counterexamples can usually be easily generated. The above counterexample of mine, for instance, is modified from some random output of the following Matlab code:


D=diag(rand(1,2)); W=rand(2,2); Q=expm(W-W'); B=Q*D*Q';

M=0.5*(A*B+B*A)-2*inv(inv(A*B)+inv(B*A)); eig(M)

share|cite|improve this answer

Added: This answer the updated question.

Your updated conjecture is also false. Consider:

$$ A = \left( \begin{array}{cc} 10 & -5 \\ 9 & 3 \\ \end{array} \right) \qquad\qquad\qquad B = \left( \begin{array}{cc} 1 & -6 \\ 8 & 8 \\ \end{array} \right) $$


$$ \frac{1}{2}(A.B + B.A) - 2 ( (A.B)^{-1} + (B.A)^{-1} )^{-1} = \frac{1}{7474} \left( \begin{array}{cc} -98938 & -164451 \\ 247345 & -61502 \\ \end{array} \right) $$

The matrix above is negative-definite.

Added: This portion answers the first variant of the problem:

No, it is not true. Counterexample:

$$ A = \left( \begin{array}{ccc} 3 & 3 & -4 \\ 1 & -2 & 0 \\ -1 & -2 & -3 \\ \end{array} \right) $$


$$ \frac{1}{2}( A + A^t) - 2 ( A^{-1} + (A^{-1})^t)^{-1} = \left( \begin{array}{ccc} \frac{127}{198} & \frac{4}{33} & -\frac{163}{198} \\ \frac{4}{33} & -\frac{2}{11} & \frac{5}{33} \\ -\frac{163}{198} & \frac{5}{33} & \frac{59}{99} \\ \end{array} \right) $$

It eigenvalues are, approximately, $1.44$, $-0.387$ and zero.

Added: In order to create such a counterexample I have used Mathematica:

f[a_] := 1/2 (a + Transpose[a]) - 
  2 Inverse[# + Transpose[#] &[Inverse[a]]]

Now this generates random integer-valued matrices until a non-degenerate one is generated with the combination in question having eigenvalues of opposite signs:

  While[Det[a = RandomInteger[{-4, 4}, {2, 2}]] == 0, Null];
  If[ Intersection[Sign[Eigenvalues[f[a]]], {-1, 1}] == {-1, 1}, 

Here is a screenshot:

enter image description here

share|cite|improve this answer
How did you create that example? Is there a 2 by 2 matrix counterexample? – Sunni Aug 23 '11 at 18:38
@Sunni Yes, here is 2 by 2 counterexample $A = \left( \begin{array}{cc} 2 & -4 \\ 1 & 0 \\ \end{array} \right)$. I created it using Mathematica. I will post the code shortly – Sasha Aug 23 '11 at 18:39
My original problem is: Let $A, B$ be positive definite matrices of the same size, is it true that $\frac{1}{2}(AB+BA)-2((AB)^{-1}+(BA)^{-1})^{-1}$ is positive definite? – Sunni Aug 23 '11 at 18:47
@Sunni : Up to numerical accuracy the answer is negative semi-definite with one eigenvalue being zero. – user13838 Aug 23 '11 at 18:59
Do you mean the answer is negative? – Sunni Aug 23 '11 at 19:12

A slightly more rigorous argument is as follows: Note that the expressions can be rewritten as $$ \begin{align} 0 &\stackrel{?}{\preceq} \frac{1}{2}(A+A^T) -2((A^{T})^{-1}+A^{-1})^{-1} \tag{Q}\\ &=\frac{1}{2}(A+A^T) -2 A^T(A+A^T)^{-1}A \tag{1}\\ &= \frac{1}{2}(A+A^T) -2 A(A+A^T)^{-1}A^T\tag{1'}\\ &= \frac{1}{2}(A+A^T)(A+A^T)^{-1}(A+A^T)-2 A^T(A+A^T)^{-1}A\\ &= \frac{1}{2}(A-A^T)(A+A^T)^{-1}(A-A^T)\\ &= -\frac{1}{2}(A-A^T)^T(A+A^T)^{-1}(A-A^T) \tag{2} \end{align} $$ Here (1) and (1') are equal and depending which side you are factoring the $A^{-1}$. And then we just recognize a "completing the square" step.

Here, note that (2) is nothing but a congruence transformation i.e. $P^TSP$ with skew symmetric $P=-P^T$, hence the sign of $S$ is preserved. Hence, whatever can be said about the (in)definiteness of $-(A+A^T)$, same holds for $(Q)$. Therefore, the answer is negative, (Q) does not hold in general. Replacing $A$ with $AB$ does not require any modification.

One important detail is that even $A,B\succ 0$, this does not imply that $AB+BA\succ 0$ since $AB$ is not necessarily symmetric. (user1551 already constructed one example.)

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.