# Bounding the norm of a contraction matrix

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. ## 1 Answer 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$[-1,1]$. • 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? Commented May 15, 2017 at 2:02 • 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? Commented May 15, 2017 at 9:18 • 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. Commented May 15, 2017 at 10:01 • If the$(2,1)$block is$\rm X\$ then the block matrix is not symmetric. Only the symmetric part contributes to a quadratic form. Commented May 15, 2017 at 10:03
• I am sorry, the (1,2) block is X and (2,1) block is X^T. The joint matrix is a co variance matrix. Commented May 15, 2017 at 11:55