Return to Answer

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(C^{1/2}A^{1/2} ) \leq \\ \leq \sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) = \tr(C^{1/2}A^{1/2} ) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}).$$

Here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(C^{1/2}A^{1/2} ) \leq \sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) .$$

By the weak majorization property of the singular values of the product of matrices (e.g. see Theorem 3 of this paper of Horn), we have

$$ \sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}). $$

Now, the result at hand.

Note that here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(A^{1/2} C^{1/2}) \leq \\ \leq \sum_{i=1}^n \lambda_i(A^{1/2} C^{1/2}) = \tr(A^{1/2} C^{1/2}) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}).$$

Here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(C^{1/2}A^{1/2} ) \leq \\ \leq \sum_{i=1}^n \lambda_i(C^{1/2}A^{1/2} ) = \tr(C^{1/2}A^{1/2} ) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}).$$

Here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(A^{1/2} C^{1/2}) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}). $$

If $\left(\begin{array}{} A & B \\ B^T & C \end{array}\right)\succeq 0$ then there exists a contraction $K$ (i.e. $\lambda_n(K)=\|K\| \leq 1$) such that $B = A^{1/2} K C^{1/2}$ (e.g. see Theorem IX.5.9 of Bhatia's book: Matrix Analysis). By Von-Neumann's trace inequality, we have $\newcommand{tr}{\mathrm{tr}}$ $$ \tr(B) = \tr(A^{1/2} K C^{1/2}) \leq\sum_{i=1}^n \lambda_i(K)\lambda_i(A^{1/2} C^{1/2}) \leq \\ \leq \sum_{i=1}^n \lambda_i(A^{1/2} C^{1/2}) = \tr(A^{1/2} C^{1/2}) \leq \sum_{i=1}^n \lambda_i(A^{1/2})\lambda_i(C^{1/2}).$$

Here $\lambda_i(\cdot)$ stands for the singular values, which is equal to the eigenvalues for positive semidefinite matrices.