Let $A$ and $V$ be positive semidefinite matrices, and let $U$ be an orthogonal matrix, partitioned columnwise:
$$U = \begin{bmatrix} U_1 & U_2 \end{bmatrix}$$
Consider the matrix
$$M = A + U_1^T V U_1 - U_1^T V U_2 (U_2^T V U_2)^+ U_2^T V U_1$$
where $X^+$ denotes the Moore-Penrose inverse of matrix $X$.
Conjecture: Assume $A + U_1^T V U_1 > 0$. Then $M > 0$.
I would like to either prove the conjecture or find a counterexample. The following examples lead me to believe that the conjecture is true, but a proof has remained elusive:
1) If $V = 0$, then we must have $A > 0$ for the assumption in the conjecture to hold. Then $M = A > 0$.
2) If $A = 0$ and $U = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$, with $V = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}$ or $V = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$, then $M = 1 > 0$.
The conjecture is false. Here is a counterexample:
$$A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} \qquad V = \begin{bmatrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{bmatrix} \qquad U_1 = \begin{bmatrix} 1 & 0 \\ 0 & 1 \\ 0 & 0 \end{bmatrix} \qquad U_2 = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$$
Then $A + U_1^T V U_1 = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix} > 0$, but $M = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}$.