# If $\|A\|_p \leq \|B\|_p$ does it follow that $\|A\|_q \leq \|B\|_q$ where $\| \cdot \|_p$ is the Schatten $p$-norm.

Let $T$ be bounded linear operators on a Hilbert space and define the Schatten $p$-norm $$\|T\|_p = \left( \sum_j s_j^p(T)\right)^{1/p}$$ where $p \geq 1$ and $s_j(T)$ are the singular values of $T$. If we know that $\|A\|_p \leq \|B\|_p$ for some bounded linear operators $A$ and $B$ and some fixed $p$, does it follow that $$\|A\|_q \leq \|B\|_q$$ for any $q \geq 1$? F0r example, does $\|A\|_2 \leq \|B\|_2 \implies \|A\|_1 \leq \|B\|_1$?

No, it does not, and you can find counterexamples in the $2 \times 2$ diagonal matrices.
EDIT: More generally, in any normed linear space $X$, if you have two norms $\|\cdot\|_1$ and $\|\cdot\|_2$ such that $\|A\|_1 \le \|B\|_1$ implies $\|A\|_2 \le \|B\|_2$, then $\|A\|_1 = \|B\|_1$ implies $\|A\|_2 = \|B\|_2$, and then $\|A\|_1/\|A\|_2$ must be the same for all $A \ne 0$, i.e. there is a constant $C$ such that $\|A\|_1 = C \|A\|_2$ for all $A \in X$.