(2.15) and $\Lambda\le2\sqrt{\operatorname{rank}(\mathcal P_A(M))}\|M\|_\infty\le2\sqrt{2\operatorname{rank}(A)}\|M\|_\infty\le\sqrt{2\operatorname{rank}(A)}\lambda$

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(2.15) and $\Lambda\le2\sqrt{\operatorname{rank}(\mathcal P_A(M))}\|M\|_\infty\le2\sqrt{2\operatorname{rank}(A)}\|M\|_\infty\le\sqrt{2\operatorname{rank}(A)}\lambda$ · Prove2Me