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Suppose $A$ and $B$ are two $n\times n$ real symmetric matrices. $A$ is positive semidefinite. Then for what values of real number $k$, matrix $(kA-B)$ is positive semidefinite (we write as $kA-B\succeq0$)?

If $A$ is positive definite, we may find an $n\times n$ nonsingular matrix $D$ such that $A=D^T D$. As a result, $kA-B\succeq0$ is equivalent to $$kI\succeq (D^{-1})^TBD^{-1},$$ 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 constraint 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.

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  • $\begingroup$ Note that if $\xi$ is an eigenvalue of the pencil $\mathbf A-\lambda\mathbf B$, then $\frac1{\xi}$ is an eigenvalue of $\mathbf B-\lambda\mathbf A$. $\endgroup$ – J. M. is not a mathematician Nov 29 '10 at 2:56
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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 $S:=A_+^{-1/2}$.

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  • $\begingroup$ 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 $B P_3 = 0$. $\endgroup$ – daizhuo Nov 30 '10 at 17:35

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