91Ó°ÊÓ

The 2-norm and Frobenius norm for a given matrix A are defined as: The 2-norm (\( \|A\|_{2} \)): it is the largest singular value of A The Frobenius norm (\( \|A\|_{F} \)): it is the square root of the sum of the squares of all entries of A

Since both norms are non-negative, we can square both sides of the inequality to simplify the comparison: \( \|A\|_{2}^2 \leq \|A\|_{F}^2 \) Now, we know that the square of the 2-norm is equal to the largest singular value (\(\sigma_{1}^{2}\)) of A, and the square of the Frobenius norm is equal to the sum of the squares of all entries of A (\(\sum_{i,j} a_{ij}^2\)): \( \sigma_{1}^2 \leq \sum_{i,j} a_{ij}^2 \) By the definition of singular values, we know that all singular values are non-negative and the largest singular value is greater than or equal to the others: \( \sigma_{1}^2 \geq \sigma_{2}^2 \geq ... \geq \sigma_{k}^2 \geq 0 \) Where k = min(m, n), Now summing up all the squared singular values: \( \sigma_{1}^2 + \sigma_{2}^2 + ... + \sigma_{k}^2 \) Since \( \sigma_{1}^2 \leq \sum_{i,j} a_{ij}^2 \), we can conclude that: \( \|A\|_{2}^2 \leq \|A\|_{F}^2 \) Which implies that: \( \|A\|_{2} \leq \|A\|_{F} \) So, we proved the first part of the exercise. Now we'll move to the second part. (b) Conditions when the 2-norm and the Frobenius norm of A are equal

To have the equality between the 2-norm and the Frobenius norm: \( \|A\|_{2} = \|A\|_{F} \)

By analyzing the previous proof, we see that the equality will only hold when: \( \sigma_{1}^2 = \sum_{i,j} a_{ij}^2 \) Which implies that: \( \sigma_{2}^2 = \sigma_{3}^2 = ... = \sigma_{k}^2 = 0 \) This means that all singular values except the largest (\(\sigma_{1}\)) must be 0. Such scenario occurs when matrix A has rank 1. So, the 2-norm and the Frobenius norm of matrix A will be equal if matrix A has rank 1.