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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

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

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 $[-1,1]$.

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 $[-1,1]$.

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

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

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 $[-1,1]$.

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

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

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 $[-1,1]$.

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

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

Let us consider the special case where $\rm C$ is symmetric and, thus, its eigenvalues are real. 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 holds if

$$\mathrm I - \Lambda^2 \succeq \mathrm O$$

Since $\rm C$ is a contraction matrix, its spectral radius is less than $1$. Thus, $\mathrm I - \Lambda^2 \succeq \mathrm O$ does indeed hold. We conclude that

$$\rho (\mathrm C) < 1 \text{ and } \mathrm C = \mathrm C^{\top} \implies \begin{bmatrix} \mathrm A \,\, & \mathrm X\\ \mathrm X^{\top} & \mathrm B\end{bmatrix} \succeq \mathrm O$$

The general (non-symmetric) case is left as an exercise for the reader.

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

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

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 $[-1,1]$.