Quaternion Norm

The norm of a quaternion is its Euclidean magnitude in four-dimensional component space. It is closely connected to quaternion multiplication because multiplying a quaternion by its conjugate produces the square of that norm. The norm is multiplicative, provides the normalization used for rotation quaternions, and supplies the denominator in the quaternion inverse.

For

$\displaystyle q=q_w+q_x\mathbf i+q_y\mathbf j+q_z\mathbf k =q_w+\mathbf q,$

(1)

the modern quaternion norm is

$\displaystyle \boxed{ \lVert q\rVert =\sqrt{q_w^2+q_x^2+q_y^2+q_z^2}. }$

(2)

Derivation from the conjugate

The conjugate is

$\displaystyle q^*=q_w-\mathbf q.$

Using the scalar–vector Hamilton product,

$\begin{equation*}\begin{aligned} qq^* &=(q_w+\mathbf q)(q_w-\mathbf q)\ &=q_w^... ...mathbf q+q_w\mathbf q +\mathbf q\times(-\mathbf q). \end{aligned}\end{equation*}$

Because

$\displaystyle \mathbf q\times\mathbf q=\mathbf0,$

all vector terms vanish. Therefore

$\displaystyle \boxed{ qq^*=q^*q =q_w^2+\mathbf q\cdot\mathbf q =q_w^2+q_x^2+q_y^2+q_z^2. }$

(4)

It follows that

$\displaystyle \boxed{\lVert q\rVert=\sqrt{qq^*}=\sqrt{q^*q}.}$

(5)

The squared norm is therefore

$\displaystyle \boxed{\lVert q\rVert^2=qq^*=q^*q.}$

(6)

Basic properties

The norm is nonnegative:

$\displaystyle \lVert q\rVert\ge0.$

(7)

It vanishes only for the zero quaternion:

$\displaystyle \boxed{ \lVert q\rVert=0 \quad\Longleftrightarrow\quad q=0. }$

(8)

conjugation does not change the norm:

$\displaystyle \boxed{\lVert q^*\rVert=\lVert q\rVert.}$

(9)

For a real scalar $a$

$\displaystyle \boxed{\lVert aq\rVert=\vert a\vert\,\lVert q\rVert.}$

(10)

As an ordinary Euclidean norm on $\mathbb{R}^4$ , it also satisfies the triangle inequality

$\displaystyle \boxed{\lVert p+q\rVert\le\lVert p\rVert+\lVert q\rVert.}$

(11)

The norm of a product

A particularly important property is multiplicativity:

$\displaystyle \boxed{\lVert pq\rVert=\lVert p\rVert\,\lVert q\rVert.}$

(12)

The cleanest proof uses the conjugate product rule

$\displaystyle (pq)^*=q^*p^*.$

Then

$\begin{equation*}\begin{aligned} \lVert pq\rVert^2 &=(pq)(pq)^*\ &=(pq)(q^*p^*)\ &=p(qq^*)p^*. \end{aligned}\end{equation*}$

Since $qq^*=\lVert q\rVert^2$ is a real scalar, it commutes with all quaternions, so

$\begin{equation*}\begin{aligned} \lVert pq\rVert^2 &=\lVert q\rVert^2pp^*\ &=\lVert p\rVert^2\lVert q\rVert^2. \end{aligned}\end{equation*}$

Taking the nonnegative square root gives equation (12).

This property is sometimes called the composition property. It is one of the defining features that makes the real quaternions a normed division algebra.

Historical terminology: tensor and norm

Historical quaternion sources use terminology that differs from modern engineering practice. Hamilton and many later quaternionists used the tensor $Tq$ for the positive magnitude corresponding to the modern $\lVert q\rVert$ . The historical norm was commonly the squared quantity

$\displaystyle \boxed{Nq=qKq=(Tq)^2,}$

(15)

where

denotes the conjugate.

Thus an older statement such as “the norm of $q$ ” may mean what a modern engineering text writes as $\lVert q\rVert^2$ . PhysicsLibrary uses the modern convention:

norm $\displaystyle =\lVert q\rVert,$ squared norm $\displaystyle =\lVert q\rVert^2.$

Unit quaternions and normalization

A unit quaternion is a quaternion of norm one:

$\displaystyle \boxed{\lVert q\rVert=1.}$

(16)

Every nonzero quaternion can be normalized:

$\displaystyle \boxed{ \widehat q=\frac{q}{\lVert q\rVert}. }$

(17)

Indeed,

$\displaystyle \lVert\widehat q\rVert =\frac{\lVert q\rVert}{\lVert q\rVert}=1.$

(18)

Unit quaternions form the subset used to represent proper three-dimensional rotations. Their geometric interpretation is developed later in the PhysicsLibrary quaternion sequence.

Normalization is also important numerically. A theoretically unit quaternion propagated with finite-precision arithmetic may drift slightly away from unit norm, so many attitude algorithms periodically renormalize it. A zero or numerically near-zero quaternion must not be normalized.

Worked examples

Example 1: direct norm

For

$\displaystyle q=2+\mathbf i-2\mathbf j+2\mathbf k,$

$\displaystyle \lVert q\rVert^2=4+1+4+4=13,$

$\displaystyle \boxed{\lVert q\rVert=\sqrt{13}.}$

Example 2: norm composition

Let

$\displaystyle p=1+\mathbf i, \qquad q=1+\mathbf j.$

Then

$\displaystyle \lVert p\rVert=\sqrt2, \qquad \lVert q\rVert=\sqrt2.$

Their product is

$\displaystyle pq=1+\mathbf i+\mathbf j+\mathbf k,$

$\displaystyle \lVert pq\rVert=2.$

This agrees with

$\displaystyle \lVert p\rVert\lVert q\rVert=\sqrt2\sqrt2=2.$

Example 3: normalize a quaternion

Suppose

$\displaystyle \widetilde q=(2,2,0,0).$

Then

$\displaystyle \lVert\widetilde q\rVert=2\sqrt2,$

and

$\displaystyle \boxed{ \widehat q=\frac{1}{\sqrt2}(1,1,0,0). }$

A direct check gives $\lVert\widehat q\rVert=1$ .

Common pitfalls

Confusing norm and squared norm. Modern notation distinguishes $\lVert q\rVert$ from $\lVert q\rVert^2$ .
Reading historical terminology as if it were modern. Hamilton's commonly corresponds to the modern squared norm.
Forgetting all four components. The scalar component contributes to the norm just like each vector component.
Normalizing the zero quaternion. The expression $q/\lVert q\rVert$ is defined only when $q\neq0$ .
Assuming normalization changes the represented direction only. Normalization rescales the entire four-component quaternion. In rotation applications this is valid because the attitude representation is specifically restricted to unit quaternions.

Exercises

Find $\lVert q\rVert^2$ and $\lVert q\rVert$ for

$\displaystyle q=3-2\mathbf i+\mathbf j+4\mathbf k.$
Prove from that $\lVert q^*\rVert=\lVert q\rVert$ .
Prove that $\lVert aq\rVert=\vert a\vert\lVert q\rVert$ for real .
Prove

$\displaystyle \lVert pq\rVert=\lVert p\rVert\lVert q\rVert$
without expanding the four components of .
Normalize

$\displaystyle q=(2,-2,2,-2).$
Let $\mathbf v=2\mathbf i-\mathbf j+2\mathbf k$ . Compute its quaternion norm and compare it with the ordinary Euclidean length of the corresponding three-vector.
If $\lVert p\rVert=3$ and $\lVert q\rVert=4$ , what is $\lVert pq\rVert$ ?
Explain the difference between Hamilton's historical and the modern quantity $\lVert q\rVert$ .

Solutions

1. Direct norm

$\displaystyle \lVert q\rVert^2=3^2+(-2)^2+1^2+4^2=30,$

$\displaystyle \boxed{\lVert q\rVert=\sqrt{30}.}$

2. Norm of the conjugate

Since

$\displaystyle \lVert q^*\rVert^2=q^*(q^*)^*=q^*q=qq^*=\lVert q\rVert^2.$

Both norms are nonnegative, hence

$\displaystyle \boxed{\lVert q^*\rVert=\lVert q\rVert.}$

3. Real scaling

$\displaystyle \lVert aq\rVert^2=(aq)(aq)^*=a^2qq^*=a^2\lVert q\rVert^2.$

Taking the nonnegative square root gives

$\displaystyle \boxed{\lVert aq\rVert=\vert a\vert\lVert q\rVert.}$

4. Multiplicative norm

$\begin{equation*} \begin{aligned} \lVert pq\rVert^2 &=(pq)(pq)^*\ &=pqq^*p^*\\... ...\rVert^2pp^*\ &=\lVert p\rVert^2\lVert q\rVert^2. \end{aligned}\end{equation*}$

Taking square roots yields

$\displaystyle \boxed{\lVert pq\rVert=\lVert p\rVert\lVert q\rVert.}$

5. Normalization

For

$\displaystyle \lVert q\rVert=4.$

Therefore

$\displaystyle \boxed{ \widehat q=\left(\frac12,-\frac12,\frac12,-\frac12\right). }$

6. Pure quaternion norm

$\displaystyle \lVert\mathbf v\rVert=\sqrt{2^2+(-1)^2+2^2}=3.$

This is exactly the Euclidean length of the corresponding vector in $\mathbb{R}^3$ .

7. Product norm

By multiplicativity,

$\displaystyle \boxed{\lVert pq\rVert=3\cdot4=12.}$

8. Historical terminology

Hamilton's $Nq$

commonly denotes the product $qKq=(Tq)^2$

, corresponding to modern $\lVert q\rVert^2$ . Hamilton's tensor $Tq$

corresponds to the modern positive norm $\lVert q\rVert$ .

What comes next

For every nonzero quaternion, multiplication by the conjugate produces the nonzero scalar $\lVert q\rVert^2$ . The next PhysicsLibrary article uses this fact to derive the multiplicative inverse.

Sources and historical notes

Hamilton, Tait, Joly, and Hathaway all connect quaternion magnitude with the product of a quaternion and its conjugate. Their historical terminology uses tensor and norm differently from much modern engineering notation, which is why the distinction is made explicit here. The modern notation, derivations, examples, exercises, and solutions in this article are newly written.

Bibliography

1: W. R. Hamilton, Elements of Quaternions, 2nd ed., Vol. I, edited by C. J. Joly, Longmans, Green, and Co., London, 1899. Public-domain historical source. Internet Archive copy
2: P. G. Tait, An Elementary Treatise on Quaternions, 3rd ed., Cambridge University Press, Cambridge, 1890. Public-domain historical source.
3: C. J. Joly, A Manual of Quaternions, Macmillan and Co., London, 1905. Public-domain historical source.
4: A. S. Hathaway, A Primer of Quaternions, 1896. Public-domain text; Project Gutenberg edition and L^ATEX source available. Project Gutenberg edition

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