The number of negative eigenvalues of the difference between two positive semidefinite matrices

Question

Let $A := B - C$, where both $B$ and $C$ are $p\times p$ (symmetric) positive semidefinite matrices and the rank of $C$ is $r < p$. What is the maximum number of negative eigenvalues of $A$?

It seems the Weyl theorem helps, but I cannot see how it works.

Answer 1

Hints. (First approach.) Any two positive semidefinite matrices are simultaneously diagonalisable by congruence. So, by Sylvester's law of inertia, you may assume that $B$ and $C$ are nonnegative diagonal matrices.

(Second approach.) Use Weyl's inequality $\lambda_{r+1}^\uparrow(B-C) \ge \lambda_\min(B) + \lambda_{r+1}^\uparrow(-C)$.

(Third approach.) If $A$ has $k$ negative eigenvalues, its eigenvectors for these eigenvalues form a $k$-dimensional subspace $V$. Hence $C$ is positive definite on $V$. So, by Courant-Fischer minimax principle, the $k$-th largest eigenvalue of $C$ is positive.

Answer 2

Assume that $D = -C$ and so $A = B+D$.

Suppose that $a_1 \geq a_2 \geq \cdots \geq a_p$, $b_1 \geq b_2 \geq \cdots\geq b_p$, and $d_1\geq d_2 \geq \cdots\geq d_p$ denote the eigenvalues of $A$, $B$, and $D$ respectively.

Since $B$ is positive semidefinite, its eigenvalues are nonnegative, i.e., $b_i\geq 0$. Since the rank of $D$ is $r$, we have

$d_i < 0\quad\forall i = 1,\cdots,p-r$

$d_i = 0\quad\forall i = p-r+1,\cdots,p$

According to Weyl theorem, we can obtain

$d_i + b_p \leq a_i \leq d_i + b_1$.

It follows from the above inequalities that

$a_i \geq 0 \quad\forall i = 1,\cdots,p-r$.

Therefore, there are at most $r$ negative eigenvalues in $A$.