3
$\begingroup$

Suppose I have positive semidefinite matrices $A$ and $B$. Then

$$\begin{bmatrix} A & X\\ X^T & B\end{bmatrix} \succeq 0$$

for $X = A^{\frac 12} C B^{\frac 12}$, 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.

$\endgroup$
0

1 Answer 1

1
$\begingroup$

We have the following linear matrix inequality (LMI)

$$\begin{bmatrix} \mathrm A \,\, & \mathrm X\\ \mathrm X^{\top} & \mathrm B\end{bmatrix} \succeq \mathrm O$$

where $\mathrm X = \mathrm A^{\frac 12} \mathrm C \, \mathrm B^{\frac 12}$ and $\mathrm A, \mathrm B \succeq \mathrm O$. Hence,

$$\begin{bmatrix} \mathrm A^{\frac 12} \mathrm A^{\frac 12} & \mathrm A^{\frac 12} \mathrm C \, \mathrm B^{\frac 12}\\ \mathrm B^{\frac 12} \mathrm C^{\top} \mathrm A^{\frac 12} & \mathrm B^{\frac 12} \mathrm B^{\frac 12}\end{bmatrix} = \begin{bmatrix} \mathrm A^{\frac 12} & \\ & \mathrm B^{\frac 12}\end{bmatrix} \begin{bmatrix} \mathrm I & \mathrm C\\ \mathrm C^{\top} & \mathrm I\end{bmatrix} \begin{bmatrix} \mathrm A^{\frac 12} & \\ & \mathrm B^{\frac 12}\end{bmatrix} \succeq \mathrm O$$

which holds if

$$\begin{bmatrix} \mathrm I & \mathrm C\\ \mathrm C^{\top} & \mathrm I\end{bmatrix} \succeq \mathrm O$$

Using the Schur complement, the LMI above can be rewritten in the form

$$\mathrm I - \mathrm C^{\top} \mathrm C \succeq \mathrm O$$

which is equivalent to

$$\lambda_{\min} (\mathrm I - \mathrm C^{\top} \mathrm C) = 1 - \lambda_{\max} (\mathrm C^{\top} \mathrm C) = 1 - \| \mathrm C \|_2^2 \geq 0 $$

and, thus, we obtain an upper bound on the spectral norm of $\rm C$

$$\color{blue}{\| \mathrm C \|_2 \leq 1}$$

We conclude that

$$\| \mathrm C \|_2 \leq 1 \implies \begin{bmatrix} \mathrm A \,\, & \mathrm X\\ \mathrm X^{\top} & \mathrm B\end{bmatrix} \succeq \mathrm O$$

If $\rm C$ is symmetric, then its eigenvalues are real and its eigenvectors are orthogonal. Let its spectral decomposition be $\rm C = Q \Lambda Q^{\top}$. Hence,

$$\mathrm I - \mathrm C^{\top} \mathrm C = \mathrm I - \mathrm C^2 = \mathrm Q \, \left( \mathrm I - \Lambda^2 \right) \, \mathrm Q^{\top} \succeq \mathrm O$$

which is equivalent to $\mathrm I - \Lambda^2 \succeq \mathrm O$, i.e., all the eigenvalues of $\rm C$ are in $[-1,1]$.

$\endgroup$
8
  • $\begingroup$ Thank you very much for explaining the detail. Actually in my case, I have two known positive semi-definite matrices A and B and the matrix X is unknown. I also have the representation of X in terms of contraction matrix C. What I need is some kind of parametric form of matrix C which can give me the bounds of matrix X? $\endgroup$ May 15, 2017 at 2:02
  • $\begingroup$ What is the definition of "contraction matrix"? Is it symmetric? Are the eigenvalues real? Is the $(2,1)$-th block $\rm X$ or $\rm X^{\top}$, after all? $\endgroup$ May 15, 2017 at 9:18
  • $\begingroup$ The (2,1) block is X. Since, I assume that matrix X is positive semi definite, I can make the symmetric assumption on contraction matrix C. $\endgroup$ May 15, 2017 at 10:01
  • $\begingroup$ If the $(2,1)$ block is $\rm X$ then the block matrix is not symmetric. Only the symmetric part contributes to a quadratic form. $\endgroup$ May 15, 2017 at 10:03
  • 1
    $\begingroup$ I am sorry, the (1,2) block is X and (2,1) block is X^T. The joint matrix is a co variance matrix. $\endgroup$ May 15, 2017 at 11:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.