Why is the following matrix product inequality true?

Question

Suppose that $A$ is a non-singular square matrix and $D$ is a diagonal matrix. I read a paper that used the following inequality

$$ADA^{-1} A^{-\ast}D^\ast A^\ast \succeq \lambda_\min(A^{-1}A^{-\ast}) ADD^\ast A^\ast$$

Why is this inequality true? I understand that $A^{-1} A^{-\ast} \succeq \lambda_\min(A^{-1}A^{-\ast})I$ but how exactly does this result in the above inequality?

Answer 1

We have $$ A^{-1} A^{-\ast} \succeq \lambda_\min(A^{-1}A^{-\ast})I \implies\\ A^{-1} A^{-\ast} - \lambda_\min(A^{-1}A^{-\ast})I \succeq 0 \implies \\ AD(A^{-1} A^{-\ast} - \lambda_\min(A^{-1}A^{-\ast})I)D^*A^* \succeq 0 \implies\\ AD(A^{-1} A^{-\ast})D^*A^* \succeq AD(\lambda_\min(A^{-1}A^{-\ast})I)D^*A^* \implies \\ ADA^{-1} A^{-\ast}D^\ast A^\ast \succeq \lambda_\min(A^{-1}A^{-\ast}) \cdot ADD^\ast A^\ast $$ as was desired.