The Frobenius norm is at most r√ r as much as the spectral radius, and this is probably tight (see the section on equivalence of norms in Wikipedia). Note that the Schatten 2 2 -norm is equal …

Norm type, specified as 2 (default), a different positive integer scalar, Inf, or -Inf. The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as …

The Frobenius norm is the diagonal of that box, and the determinant is the volume. The usual norm defined as sup∥x∥=1 ∥Ax∥ corresponds to the longest side of the box.

Norm type, specified as 2 (default), a positive real scalar, Inf, or -Inf. The valid values of p and what they return depend on whether the first input to norm is a matrix or vector, as shown in …

The Frobenius norm (i.e. the sum of singular values) is a matrix norm (it fulfills the norm axioms), but not an operator norm, since no vector norm exists so that the above definition for the …

Jan 7, 2021 · I've been trying to prove that the Frobenius and 1 norm are equivalent, satisfying the following inequalities (A ∈ Fm×n A ∈ F m × n, with m ≥ n m ≥ n):

Your proof is correct, but it can be conceptually simplified as follows. Note that both the operator norm and Frobenius norm are unitarily invariant. Therefore, by singular value decomposition, …

Here is yet another way to do it that I think provides a lot of intuition. First, notice that applying a unitary matrix to M on either the left or right side will not change the Frobinous norm. This is …

Jan 4, 2019 · I want to show that it is unitarily invariant, so that for unitary U we have that ‖UA‖F = ‖A‖F = ‖AU‖F

Dec 30, 2020 · Considering the Frobenius norm ∥.∥F ‖. ‖ F, how can we prove that it's submultiplicative? N.b: I noticed there is a user who asked a similar question but proof looked …