Does $A,B>0$ imply that the projection onto the positive subspace of $A-B$ is smaller than $A+B$ for positive semidefinite matrices?

Question

I am trying to find a counterexample for the following matrix inequality which I suspect not to hold: $$P^+(A-B)\leq A+B$$ where $A,B$ are positive semidefinite matrices and $P^+(X)$ the projector onto the positive subspace of $X$.

Answer 1

Let $A=\frac{1}{10}\begin{bmatrix} 2&4\\4&8\end{bmatrix},$ $B=\frac{1}{10}\begin{bmatrix}9&-3\\-3&1\end{bmatrix}.$ Then $A-B=\frac{1}{10}\begin{bmatrix}-7&7\\7&7\end{bmatrix},$ which has eigenvalues $\pm \sqrt{\frac{49}{50}},$ with corresponding eigenvectors $\begin{bmatrix}1\\1\pm\sqrt{2}\end{bmatrix}.$ Thus $P^{+}(A-B)=\frac{\sqrt{49/50}}{6+2\sqrt{2}}\begin{bmatrix}1&1+\sqrt{2}\\1+\sqrt{2}&5+2\sqrt{2}\end{bmatrix}.$ Letting $x=\frac{1}{\sqrt{6+2\sqrt{2}}}\begin{bmatrix}1\\1+\sqrt{2}\end{bmatrix},$ we see that $x^{*}P^{+}(A-B)x=\sqrt{\frac{49}{50}}\approx 0.9899,$ while $x^{*}(A+B)x=\frac{4+2\sqrt{2}}{6+2\sqrt{2}}\approx 0.7735.$ Therefore, $P^{+}(A-B)\not\leq A+B$ in this case.

Admittedly, $A$ and $B$ were both positive semidefinite here, but a slight modification yields a counterexample with $A'$ and $B'$ positive definite. Letting $\varepsilon>0,$ observe that $A+\varepsilon I,$ $B+\varepsilon I$ are positive definite, $(A+\varepsilon I)-(B+\varepsilon I)=A-B,$ and $x^{*}(A+B+2\varepsilon I)x=x^{*}(A+B)x+2\varepsilon,$ so for $\varepsilon<0.1,$ we still have $x^{*}(A'+B')x<x^{*}P^{+}(A'-B')x.$

Several factors are important to making the example above work. They both have one large eigenvalue and one $0$ (or close to $0$) eigenvalue, and the eigenvectors corresponding to the large eigenvalues are nearly orthogonal ($\frac{\langle [1,2],[3,-1]\rangle}{\sqrt{5}\sqrt{10}}=\frac{1}{5\sqrt{2}}$). Since $A$ and $B$ are roughly the same size, this means that the eigenvectors of $A-B$ are changed essentially as much as possible from the large eigenvector of either $A$ or $B$. This means that the eigenvector of $A-B$ corresponding to the positive eigenvalue is relatively far from either of the large eigenvectors of $A$ or $B$, which means that when we compute $x^{*}P^{+}(A-B)x$ for the largest eigenvector $x$, scaled to unit length, we obtain the largest eigenvalue of $P^{+}(A-B),$ but when we compute $x^{*}Ax$ (or $x^{*}Bx$), we are getting an average of the largest eigenvalue, $1$, and the smallest eigenvalue, $0$ (or $\varepsilon,$ if we consider the second example). It then turns out that the sum of these two ends up being less than the positive eigenvalue of $A-B$ (here is where a bit of tinkering might be necessary to make sure this happens as desired.

In higher dimensions, I would suggest considering matrices that are again of the form $uu^{T}+\varepsilon I$ for a unit vector $u,$ and with the vectors $u$ close to being orthogonal (note that if they are orthogonal, the matrices $A$ and $B$ will be simultaneously diagonalizable, and the counterexample will be impossible). Then $A-B=uu^{T}-vv^{T}$ will have one positive and one negative eigenvalue, and the positive unit eigenvector should satisfy $x^{*}P^{+}(A-B)x>x^{*}(A+B)x$ when $\varepsilon$ is sufficiently small (this obviously requires proof).