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

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.

• Note that if $\xi$ is an eigenvalue of the pencil $\mathbf A-\lambda\mathbf B$, then $\frac1{\xi}$ is an eigenvalue of $\mathbf B-\lambda\mathbf A$. Nov 29, 2010 at 2:56

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 $S:=A_+^{-1/2}$.
• Thanks. In fact, $B_{13}$ must be 0. Let $P_1$ be a matrix consisting of columns that form a basis of the range of $A$. Let $P_2$ be a matrix consisting of columns that form a basis of the intersection of the kernel of $A$ and the range of $B$. Let $P_3$ be a matrix consisting of columns that form a basis of the intersection of the kernel of $A$ and the kernel of $B$. Then columns of $P=(P_1, P_2, P_3)$ form a basis of $\mathbb{R}^n$ in which $A$ and $B$ took the form you showed above. In particular, $B_{13} = P_1^T B P_3 = 0$ because $B P_3 = 0$. Nov 30, 2010 at 17:35