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Exercise 296Topology › Topology of Mn(ℝ) › Norms and metric structureL2Added 7 Nov 2026

Entries versus Frobenius norm

#Frobenius norm#max norm#equivalence of norms

Statement

Show that max⁡i,j∣aij∣≤∥A∥F≤nmax⁡i,j∣aij∣\max_{i,j}|a_{ij}|\le\|A\|_F\le n\max_{i,j}|a_{ij}| for every A∈Mn(R)A\in M_n(\mathbb{R}).

Hint

Compare a single term aij2a_{ij}^2 with the sum of all n2n^2 terms.

For the right inequality, bound every aij2a_{ij}^2 by the same number.

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Solution

Idea. A sum of n2n^2 non-negative numbers is at least its largest term and at most n2n^2 times its largest term.

Step 1: left inequality. Fix (k,l)(k,l). Since all terms are non-negative,
akl2≤∑i,jaij2=∥A∥F2.a_{kl}^2\le\sum_{i,j}a_{ij}^2=\|A\|_F^2.
Taking square roots, ∣akl∣≤∥A∥F|a_{kl}|\le\|A\|_F, and then take the maximum over (k,l)(k,l).

Step 2: right inequality. Let m=max⁡i,j∣aij∣m=\max_{i,j}|a_{ij}|. Then each aij2≤m2a_{ij}^2\le m^2 and there are n2n^2 terms, so
∥A∥F2≤n2m2,∥A∥F≤nm.\|A\|_F^2\le n^2m^2,\qquad\|A\|_F\le nm.

Equality cases. The left inequality is an equality iff AA has at most one non-zero entry (e.g. A=E11A=E_{11}). The right one is an equality iff ∣aij∣=m|a_{ij}|=m for all i,ji,j (e.g. the all-ones matrix, for which ∥J∥F=n\|J\|_F=n and m=1m=1).

Remark. These inequalities show directly that the Frobenius norm and the max-entry norm are equivalent. They are the simplest instance of the fact (proved later) that all norms on Mn(R)M_n(\mathbb{R}) are equivalent. In particular a sequence of matrices converges iff each entry converges.