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](https://zbmath.org/?q=an:0729.15001).


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.



Bounding the norm of a contraction matrix
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 $[-1,1]$.


