$\Vert A \Vert_{G_1} \overset{?}{\geq} \Vert A \Vert_{G_2}$, if $G_1 \succeq G_2 \succeq 0 $

Question

I have a G-matrix norm which is definied as $\Vert A \Vert_G = \sqrt{\text{trace}(A^T G A )} $ with a symmetric positive semidefinite matrix $G$.

I have already found out that the relationship to the Frobenius norm is as follows: $G = M^T M$

$\Vert A \Vert_G = \Vert MA \Vert_F $

As in the title stated, I want to know, if $\Vert A \Vert_{G_1} \geq \Vert A \Vert_{G_2}$ always hold, when $G_1 \succeq G_2 \succeq 0 $?

Answer 1

I think, I solved the problem and the statement ist true.

$\Vert A \Vert_{G_1} \geq \Vert A \Vert_{G_2} \Leftrightarrow\, \Vert A \Vert_{G_1}^2 \geq \Vert A \Vert_{G_2}^2 \Leftrightarrow \Vert A \Vert_{G_1}^2 - \Vert A \Vert_{G_2}^2 \geq 0 \\ \Leftrightarrow \text{trace}(A^TG_1A) - \text{trace}(A^TG_2 A) = \text{trace}(A^T(G_1 - G_2)A) = \Vert A \Vert_{G_1-G_2}^2 \geq 0$

Last inequality is true, since $G_1-G_2 \succeq 0 $ and every matrix norm is greater than or equal to zero.