RMS Norm

Characterizing properties of the Root-Mean-Square Norm for vectors

Posted Jun 5, 2025 Updated Mar 15, 2026

By Ji-Ha Kim

10 min read

RMS Norm

Root-Mean-Square (RMS) Norm for Vectors

While various norms are ubiquitous in mathematics and engineering, deep learning practice often benefits from a dimension-invariant scale for vectors. The RMS norm provides this by normalizing the Euclidean norm by the square root of the dimension.

Definition. RMS Norm (for Vectors)

For
$x \in \mathbb{R}^n$
(
$n \ge 1$
), the root-mean-square norm is
$\Vert x \Vert_{\mathrm{RMS}} \;=\; \frac{\Vert x \Vert_2}{\sqrt{n}} \;=\; \sqrt{\frac{1}{n}\sum_{i=1}^n x_i^2}$
where
$\Vert x \Vert_2 = \sqrt{\sum_{i=1}^n x_i^2}$
is the standard Euclidean norm.

Properties and Rationale for Vector RMS Norm

Dimension neutrality. If the coordinates

\[ x_i \]

\[ x \]

are i.i.d. random variables with zero mean and unit variance (e.g.,

x_i \sim \mathcal{N}(0,1)

), then

\mathbb{E}[\Vert x \Vert_{\mathrm{RMS}}^2] = 1

, and

\mathbb{E}[\Vert x \Vert_{\mathrm{RMS}}] \approx 1

for large

\[ n \]

. This property makes the notion of “unit-size” more consistent for vectors of varying dimensions, such as network activations from layers of different widths.

Rotational and Orthogonal Invariance. The RMS norm is a positive scalar multiple of the

\ell_2

-norm. The

\ell_2

-norm is invariant under rotations (

\[ SO(n) \]

) and, more generally, under all orthogonal transformations (

\[ O(n) \]

), which include reflections. The RMS norm inherits these crucial geometric symmetries.

Theorem 1. Rotationally Invariant Functions

A function
$f: \mathbb{R}^n \to \mathbb{R}$
(
$n \ge 2$
) is rotationally invariant (i.e.,
$\[ f(Rx) = f(x) \]$
for all rotation matrices
$R \in SO(n)$
and all
$x \in \mathbb{R}^n$
) if and only if there exists a function
$g: \mathbb{R}_{\ge 0} \to \mathbb{R}$
such that:
$f(x) = g(\Vert x \Vert_2) \quad \forall x \in \mathbb{R}^n$

Note on the case
$\[ n=1 \]$

For
$\[ n=1 \]$
, the space is
$\mathbb{R}$
. The special orthogonal group
$\[ SO(1) \]$
contains only the identity matrix
$\[ [1] \]$
. Thus, any function
$f: \mathbb{R} \to \mathbb{R}$
is trivially rotationally invariant (i.e.,
$f(1 \cdot x_1) = f(x_1)$
). Such a function can be written as
$f(x_1) = g(\Vert x_1 \Vert_2) = g(\vert x_1 \vert)$
if and only if
$\[ f \]$
is an even function (i.e.,
$\[ f(x_1)=f(-x_1) \]$
).

Proof of Theorem 1.

(

\Leftarrow

) **Sufficiency (for all

n \ge 1

):** Assume

f(x) = g(\Vert x \Vert_2)

for some function

g: \mathbb{R}_{\ge 0} \to \mathbb{R}

. For any rotation matrix

R \in SO(n)

, rotations preserve the Euclidean norm:

\Vert Rx \Vert_2 = \Vert x \Vert_2

. Then,

f(Rx) = g(\Vert Rx \Vert_2) = g(\Vert x \Vert_2) = f(x)

. Thus,

\[ f \]

is rotationally invariant.

(

\Rightarrow

) Necessity:

**Case

n \ge 2

:** Assume

f: \mathbb{R}^n \to \mathbb{R}

is rotationally invariant. * If

x = \mathbf{0}

, define

g(0) = f(\mathbf{0})

. Then

f(\mathbf{0}) = g(\Vert \mathbf{0} \Vert_2)

. * If

x \ne \mathbf{0}

, let

r_0 = \Vert x \Vert_2 > 0

. For any

y \in \mathbb{R}^n

with

\Vert y \Vert_2 = r_0

, there exists

R \in SO(n)

such that

\[ y = Rx \]

(since

\[ SO(n) \]

acts transitively on spheres for

n \ge 2

). By rotational invariance,

\[ f(y) = f(Rx) = f(x) \]

. Thus,

\[ f(x) \]

depends only on

\Vert x \Vert_2

. Define

\[ g(r_0) = f(x_0) \]

for any fixed

\[ x_0 \]

with

\Vert x_0 \Vert_2 = r_0

(e.g.,

x_0 = (r_0, 0, \ldots, 0)^\top

). Then

f(x) = g(\Vert x \Vert_2)

. (Using

\[ r_0 \]

to avoid confusion with function

\[ g(r) \]

**Case

\[ n=1 \]

:** (Covered in the note within the theorem statement.) For completeness, if

f(x_1) = g(\vert x_1 \vert)

, then

f(-x_1) = g(\vert -x_1 \vert) = g(\vert x_1 \vert) = f(x_1)

, so

\[ f \]

must be an even function. Conversely, if

\[ f \]

is an even function, define

\[ g(r) = f(r) \]

for

r \ge 0

. Then for any

x_1 \in \mathbb{R}

g(\vert x_1 \vert) = f(\vert x_1 \vert)

. Since

\[ f \]

is even,

f(\vert x_1 \vert) = f(x_1)

x_1 \ge 0

and

f(\vert x_1 \vert) = f(-x_1) = f(x_1)

\[ x_1 < 0 \]

. So

f(x_1) = g(\vert x_1 \vert)

Corollary 1.1. Rotationally Invariant Norms

If a function
$\Vert \cdot \Vert : \mathbb{R}^n \to \mathbb{R}$
is a norm and is rotationally invariant, then it must be a positive scalar multiple of the Euclidean norm:
$\Vert x \Vert = c \Vert x \Vert_2 \quad \forall x \in \mathbb{R}^n, \text{ for some constant } c > 0$

Proof of Corollary 1.1.

Let

\Vert \cdot \Vert

be a rotationally invariant norm.

By Theorem 1:
- For

n \ge 2

, since

\Vert \cdot \Vert

is rotationally invariant,

\Vert x \Vert = g(\Vert x \Vert_2)

for some

g: \mathbb{R}_{\ge 0} \to \mathbb{R}

. * For

\[ n=1 \]

, a norm

\Vert x_1 \Vert

is an even function (

\Vert -x_1 \Vert = \vert -1 \vert \Vert x_1 \Vert = \Vert x_1 \Vert

). By the

\[ n=1 \]

case of Theorem 1,

\Vert x_1 \Vert = g(\vert x_1 \vert) = g(\Vert x_1 \Vert_2)

. Thus, for any

n \ge 1

\Vert x \Vert = g(\Vert x \Vert_2)

By absolute homogeneity of norms,

\Vert \alpha x \Vert = \vert \alpha \vert \Vert x \Vert

. So,

g(\Vert \alpha x \Vert_2) = \vert \alpha \vert g(\Vert x \Vert_2)

. Since

\Vert \alpha x \Vert_2 = \vert \alpha \vert \Vert x \Vert_2

, we have

g(\vert \alpha \vert \Vert x \Vert_2) = \vert \alpha \vert g(\Vert x \Vert_2)

. Let

r = \Vert x \Vert_2 \ge 0

and

\lambda = \vert \alpha \vert \ge 0

. Then

g(\lambda r) = \lambda g(r)

for all

\lambda, r \ge 0

. This is Cauchy’s functional equation for homogeneous functions on

\mathbb{R}_{\ge 0}

\[ r > 0 \]

, set

\[ r=1 \]

to get

g(\lambda) = \lambda g(1)

. Let

\[ c = g(1) \]

. Then

g(\lambda) = c\lambda

for

\lambda > 0

. Since

\Vert \mathbf{0} \Vert = 0

, we have

g(\Vert \mathbf{0} \Vert_2) = g(0)=0

. The relation

g(\lambda)=c\lambda

also gives

\[ g(0)=0 \]

, so it holds for all

\lambda \ge 0

. Thus,

\Vert x \Vert = g(\Vert x \Vert_2) = c \Vert x \Vert_2

Since

\Vert \cdot \Vert

is a norm, for

x \ne \mathbf{0}

\Vert x \Vert > 0

. Thus

c \Vert x \Vert_2 > 0

, which implies

\[ c > 0 \]

. (For instance,

c=g(1)=\Vert e_1 \Vert > 0

e_1 \ne \mathbf{0}

Theorem 2. Orthogonal Invariance of Euclidean-Derived Norms

The Euclidean norm (
$\ell_2$
-norm) is orthogonally invariant: for any orthogonal matrix
$Q \in O(n)$
(satisfying
$Q^\top Q = I$
) and any
$x \in \mathbb{R}^n$
,
$\Vert Qx \Vert_2 = \Vert x \Vert_2$
Consequently, any norm of the form
$\Vert x \Vert = c \Vert x \Vert_2$
with
$\[ c > 0 \]$
is also orthogonally invariant. This implies that any rotationally invariant norm is also orthogonally invariant.

Proof of Theorem 2.

The Euclidean norm squared is

\Vert x \Vert_2^2 = x^\top x

. For

Q \in O(n)

\Vert Qx \Vert_2^2 = (Qx)^\top (Qx) = x^\top Q^\top Q x = x^\top I x = x^\top x = \Vert x \Vert_2^2

. Since norms are non-negative, taking the square root gives

\Vert Qx \Vert_2 = \Vert x \Vert_2

If a norm is of the form

\Vert x \Vert = c \Vert x \Vert_2

for some

\[ c > 0 \]

, then

\Vert Qx \Vert = c \Vert Qx \Vert_2 = c \Vert x \Vert_2 = \Vert x \Vert

. So, such norms are orthogonally invariant. By Corollary 1.1, any rotationally invariant norm (

\[ SO(n) \]

-invariant norm) must be of the form

c \Vert x \Vert_2

for some

\[ c>0 \]

. Therefore, any rotationally invariant norm is also orthogonally invariant (

\[ O(n) \]

-invariant).

Corollary 2.1. Uniqueness of RMS Norm Family

The RMS norm,
$\Vert x \Vert_{\mathrm{RMS}} = \frac{1}{\sqrt{n}}\Vert x \Vert_2$
, is a positive scalar multiple of the
$\ell_2$
-norm. Therefore, it is rotationally and orthogonally invariant.

Furthermore, consider a family of norms
$\{\mathcal{N}_n(\cdot)\}_{n \ge 1}$
, where each
$\mathcal{N}_n: \mathbb{R}^n \to \mathbb{R}$
is a norm on
$\mathbb{R}^n$
. If this family satisfies:

Rotational Invariance: Each

$\mathcal{N}_n(\cdot)$
is rotationally invariant.

Dimensional Normalization: For a class of random vectors

$X^{(n)} \in \mathbb{R}^n$
(whose components
$\[ X_i \]$
are i.i.d. with zero mean and unit variance, ensuring
$\mathbb{E}[\Vert X^{(n)} \Vert_{\mathrm{RMS}}] \approx 1$
), the expected value
$\mathbb{E}[\mathcal{N}_n(X^{(n)})]$
is a constant
$\[ K > 0 \]$
independent of
$\[ n \]$
. (More precisely, assume
$\mathbb{E}[\Vert X^{(n)} \Vert_2] = \sqrt{n}$
for this class of vectors.)

Then, each norm
$\mathcal{N}_n(x)$
must be of the form
$K \cdot \Vert x \Vert_{\mathrm{RMS}}$
. If
$\[ K=1 \]$
, the RMS norm family is the unique family of norms satisfying these conditions.

Proof of Corollary 2.1.

The RMS norm is

(1/\sqrt{n})\Vert x \Vert_2

. Since

1/\sqrt{n}>0

, it’s a rotationally invariant norm by Corollary 1.1, and thus orthogonally invariant by Theorem 2.

For the second part:

Rotational Invariance: By Corollary 1.1, each

\mathcal{N}_n(x) = c_n \Vert x \Vert_2

for some constant

\[ c_n > 0 \]

Dimensional Normalization: We are given

\mathbb{E}[\mathcal{N}_n(X^{(n)})] = K

for all

\[ n \]

. Substituting the form from (1):

\mathbb{E}[c_n \Vert X^{(n)} \Vert_2] = c_n \mathbb{E}[\Vert X^{(n)} \Vert_2] = K

. The condition on the random vectors implies

\mathbb{E}[\Vert X^{(n)} \Vert_2] = \sqrt{n}

. (This is derived from the motivating property

\mathbb{E}[\Vert X^{(n)} \Vert_{\mathrm{RMS}}] = 1

, which means

\mathbb{E}[\Vert X^{(n)} \Vert_2 / \sqrt{n}] = 1

.) Plugging this into the equation for

\[ K \]

c_n \sqrt{n} = K

. So,

c_n = K / \sqrt{n}

Form of the Norm: Therefore,

\mathcal{N}_n(x) = (K/\sqrt{n}) \Vert x \Vert_2 = K \cdot (\Vert x \Vert_2 / \sqrt{n}) = K \cdot \Vert x \Vert_{\mathrm{RMS}}

. If

\[ K=1 \]

, then

\mathcal{N}_n(x) = \Vert x \Vert_{\mathrm{RMS}}

Tip. When to use the vector RMS norm

Employ the vector
$\Vert \cdot \Vert_{\mathrm{RMS}}$
when a scale for vectors is needed that is simultaneously rotationally symmetric (thus orthogonally symmetric) and normalized for vector dimension. This is useful, for example, when comparing activations from neural network layers of different widths or designing width-robust regularizers.

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Root-Mean-Square (RMS) Norm for Vectors

Properties and Rationale for Vector RMS Norm

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