---
tag: K33BQX70N0
kind: definition
name: Frobenius norm
msc: [15A60]
defines: JT8Q204E4R
uses: [54T6Y6DSY4, 0VD9ZSWWJG, 0XC1RDBY7A, 0DXKE10PHF, 09VMDSRXWJ]
sources:
  - url: https://en.wikipedia.org/wiki/Matrix_norm
    locator: Frobenius norm
---

## Definition

Let $\mathbb{K}$ be $\R$ or $\C$, let $m, n \ge 1$, and let $A = (a_{ij})$ be an $m \times n$ [[matrix|0VD9ZSWWJG]] over $\mathbb{K}$.
The *Frobenius norm* of $A$ is
$$\norm{A}_F = \sqrt{\sum_{i=1}^{m} \sum_{j=1}^{n} \abs{a_{ij}}^2}.$$

## Remarks

Equivalently, $\norm{A}_F = \sqrt{\tr(A^* A)}$, where $A^*$ is the [[conjugate transpose|0DXKE10PHF]] and $\tr$ is the [[trace|0XC1RDBY7A]].
If $\sigma_1, \ldots, \sigma_r$ are the nonzero [[singular values|09VMDSRXWJ]] of $A$, then $\norm{A}_F^2 = \sigma_1^2 + \cdots + \sigma_r^2$.
Example: for $A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix}$, $\norm{A}_F = \sqrt{30}$.
The Frobenius norm is a submultiplicative [[matrix norm|54T6Y6DSY4]], by the Cauchy–Schwarz inequality, and $\norm{UAV}_F = \norm{A}_F$ for unitary $U$ and $V$.
It is not an operator norm when $n \ge 2$: every operator norm gives $\norm{I_n} = 1$, but $\norm{I_n}_F = \sqrt{n}$.