linear-algebra - linear algebra - Maximum bound of Contraction Matrix - answerstu - answerstu

linear algebra - Maximum bound of Contraction Matrix

Suppose I have positive semi-definite matrices A and B. Then $[A$ $X;X$ $B]$$>$$=0$ for $X$=$A$^$0.5$$C$$B$^$0.5$, where $C$ is the contraction matrix with maximum eigenvalue less than 1.

Horn, Roger A.; Johnson, Charles R., Topics in matrix analysis, Cambridge etc.: Cambridge University Press. viii, 607 p. \sterling 45.00; \$ 59.50 (1991). ZBL0729.15001.

I have some questions:

(1) Is it possible for $C$ to have negative eigenvalues?

(2) Are their any properties of $C$ other than eigenvalue<1? (Please, suggest a book or something).

(3) Is it possible to compute the maximum bounds of the matrix $C$ (Where any random contraction is inside that maximum bound)?

I shall be very thankful for any guidance and suggestion. Thanks

1 Answer

  1. Alva- Reply


    Before addressing your questions, it should be clarified that in the result of Horn and Johnson, the matrices $A$ and $B$ do not need to have the same dimensions, and therefore $C$ does not need to be square. So, the restriction in their result is really that the largest singular value of $C$ be bounded above by $1$ rather than $C$'s eigenvalues (which wouldn’t apply in the non-square case). Note also that the bound is not strict: the largest singular value of $C$ can equal one.

    The bound that $\sigma_1(C)\le 1$ is in fact, a maximal bound for $C$ and is the best bound obtainable for this result (answering your question #3): Horn and Johnson establish the equivalence (not just a one-way implication) of this bound and the positive semidefiniteness of

    $$\begin{bmatrix} A & X \\ X^* & B \end{bmatrix}$$

    in their proof.

    In Horn and Johnson's proof, they also note that when $A$ and $B$ are positive definite, one can simply solve for $C$ as $C=A^{-\frac{1}{2}}X B^{-\frac{1}{2}}.$ For $A$ and $B$ positive definite, this form of $C$ is the only constraint we have besides the requirement that its largest singular value be bounded by 1 (answering your question# 2). In other words, pick any $X$ such that $\sigma_1(C)\le 1$ where $C=A^{-\frac{1}{2}}X B^{-\frac{1}{2}}$ and you have yourself a positive definite matrix

    $$\begin{bmatrix} A & X \\ X^* & B \end{bmatrix}.$$

    The extension to positive semidefinite matrices follows from a limiting argument (as further noted by Horn and Johnson).

    For question #1, we can answer it in the case where $A$ and $B$ have the same dimensions by building a small example: letting $A = B = 1$ and $C=-\frac{1}{2}$ yields $X=-\frac{1}{2}$ and the positive definite matrix

    $$\begin{bmatrix} 1 & -\frac{1}{2} \\ -\frac{1}{2} & 1 \end{bmatrix}.$$

    So, yes, $C$ can have negative eigenvalues.

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