Does $A, B in S_n^{+}$ imply $A + B - |A - B| in S_n^{+}$? (where $S_n^{+}$ is the set of positive semidefinite symmetric matrices)

If $M$ is a (say real) matrix, we can define its modulus $|M|$ as the square root of $M' M$.

Now if $A$, $B$ are positive semidefinite symmetric matrices, is $A + B - | A - B |$ positive semidefinite?

If $A$ and $B$ commute, they can be diagonalized simultaneously, in which case the answer is yes. What is the situation if $A$ and $B$ do not commute?

Thanks for your help.

Answer

HINT:

You are asking whether the "formal" $\min$ of two positive matrices is still positive, where
$$\min(A,B) = \frac{1}{2}( A+B - |A-B|)$$

No, it may be non positive semidefinite.

Let's explain first the what is the meaning behind the fact that it may be non positive. Notice that $\min(A,B)$ is $\preceq$ both $A$ and $B$ for all $A$, $B$ hermitian. Assume now that indeed it is true that whenever $A$, $B \succeq 0$ we also have $\min(A,B) \succeq 0$. Then it would follow that whenever $A$, $B \succeq C$ we also have $\min(A,B) \succeq C$. That would mean that $\min(A,B)$ is a true infimmum for $A$ and $B$. But, alas, the hermitian matrices with the order $\succeq$ is not a lattice... That is the meaning of the statement.

Now, how to get a counterexample. It will work already for $2\times 2$ real symmetric matrices. Take $A$ positive and $B$ positive semidefinite with a $0$ eigenvalue ( that is, determinant $0$). Make sure that $A$ and $B$ do not commute. Now calculate $C = \min(A,B)$. Since $C\preceq B$ its smallest eigenvalue has to be $\le 0$, the smallest eigenvalue of $B$. If $A$ and $B$ do not commute you probably will get a $C$ with determinant $\ne 0$ so it will not be positive semidefinite. From here, add a little positive to $B$, and still have $C\not\succeq 0$.

I recommend calculating the square roots of matrices with Mathematica or Wolframalpha ( command : MatrixPower[ D, 1/2], for $D= (A-B)^2$)