For what values of $k$ is matrix $k A - B$ positive semidefinite?

Question

Suppose $A$ and $B$ are two $n \times n$ real symmetric matrices, and $A$ is positive semidefinite. For what values of $k \in \mathbb R$ is matrix $kA-B$ positive semidefinite (we write as $kA-B \succeq 0$)?

If $A$ is positive definite, we may find an $n \times n$ nonsingular matrix $D$ such that $A = D^T D$. As a result, $kA-B \succeq 0$ is equivalent to $$kI \succeq (D^{-1})^T B D^{-1}$$ or $$k \geq \lambda_{\max}((D^{-1})^TBD^{-1}))$$

But how to deal with the situation when $A$ is singular but still positive semidefinite?

I know for certain that in this case we must impose additional constraints on matrix $B$. In particular, let the columns of matrix $N$ consist of a basis of the null space of $A$, then we must have that $N^T B N \preceq 0$ (i.e., $N^T B N$ is negative semidefinite). But what is the lower bound on $k$?

Thanks.

Answer 1

You can make a orthogonal change of coordinates as follows. First choose an orthogonal basis of the range of $A$, and one of the kernel. This makes a basis of $\mathbb R^n$ in which $A$ is block-diagonal with a zero block: $$A=\begin{pmatrix} A_+ & 0 \\\\ 0 & 0 \end{pmatrix},\qquad B=\begin{pmatrix} B_1 & B_2 \\\\ B_2^T & B_3 \end{pmatrix}.$$ Then a necessary condition is that $B_3\le0$. Now change the basis of $\ker A$ by choosing orthogonal bases of the kernel and range of $B_3$. Then $A$ is unchanged, whereas $B$ becomes $$B==\begin{pmatrix} B_1 & B_{12} & B_{13} \\\\ B_{12}^T & B_{22} & 0 \\\\ B_{13}^T & 0 & 0 \end{pmatrix}.$$ By assumption, we have $B_{22}< 0$. A new necessary condition appears, that $B_{13}=0$. Now the necessary and sufficient condition over $k$ is that $$k\ge\lambda_{max}(S(B_{1}-B_{12}B_{22}^{-T}B_{12}^T)S),$$ where $S:=A_+^{-1/2}$.