How to Prove the Dot Product Is Distributive

The dot product distributes over vector addition in either argument. The clearest proof uses the coordinate definition: add the corresponding vector components, multiply by the matching component of the other vector, and then distribute the ordinary multiplication over addition.

This proof applies to real vectors in the same Euclidean space, such as vectors in $\mathbb{R}^2$ or $\mathbb{R}^3$. It also shows exactly why the identity holds, rather than relying on a diagram or a geometric formula.

The coordinate definition

Let $\mathbf{u}=(u_1,\ldots,u_n)$ and $\mathbf{v}=(v_1,\ldots,v_n)$ be vectors in $\mathbb{R}^n$. Their dot product is defined by

$$\mathbf{u}\cdot\mathbf{v}=\sum_{i=1}^{n}u_i v_i.$$

In words, multiply corresponding coordinates and add the results. For example, if $\mathbf{u}=(2,-1,3)$ and $\mathbf{v}=(4,5,0)$, then

$$\mathbf{u}\cdot\mathbf{v}=2(4)+(-1)(5)+3(0)=3.$$

To prove distributivity, take three vectors $\mathbf{u},\mathbf{v},\mathbf{w}\in\mathbb{R}^n$. The sum $\mathbf{u}+\mathbf{v}$ is formed coordinate by coordinate, so its $i$th coordinate is $u_i+v_i$.

Proof of the distributive laws

For distributivity in the first argument, expand the definition and use the usual distributive law for real numbers in each coordinate:

$$\begin{aligned} (\mathbf{u}+\mathbf{v})\cdot\mathbf{w} &=\sum_{i=1}^{n}(u_i+v_i)w_i\\ &=\sum_{i=1}^{n}(u_iw_i+v_iw_i)\\ &=\sum_{i=1}^{n}u_iw_i+\sum_{i=1}^{n}v_iw_i\\ &=\mathbf{u}\cdot\mathbf{w}+\mathbf{v}\cdot\mathbf{w}. \end{aligned}$$

The third equality uses the fact that a finite sum of term-by-term sums can be separated into two sums. That is precisely what allows the two dot products to appear.

The dot product is also distributive in its second argument. The same calculation gives

$$\begin{aligned} \mathbf{u}\cdot(\mathbf{v}+\mathbf{w}) &=\sum_{i=1}^{n}u_i(v_i+w_i)\\ &=\sum_{i=1}^{n}(u_iv_i+u_iw_i)\\ &=\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{w}. \end{aligned}$$

Thus both identities hold:

$$(\mathbf{u}+\mathbf{v})\cdot\mathbf{w} =\mathbf{u}\cdot\mathbf{w}+\mathbf{v}\cdot\mathbf{w}, \qquad \mathbf{u}\cdot(\mathbf{v}+\mathbf{w}) =\mathbf{u}\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{w}.$$

Worked example

Choose $\mathbf{u}=(1,2,-1)$, $\mathbf{v}=(3,-1,4)$, and $\mathbf{w}=(2,0,5)$. First add the vectors in the left-hand expression:

$$\mathbf{u}+\mathbf{v}=(4,1,3), \qquad (\mathbf{u}+\mathbf{v})\cdot\mathbf{w}=4(2)+1(0)+3(5)=23.$$

Now evaluate the right-hand side separately:

$$\mathbf{u}\cdot\mathbf{w}=1(2)+2(0)+(-1)(5)=-3,$$ $$\mathbf{v}\cdot\mathbf{w}=3(2)+(-1)(0)+4(5)=26.$$

Adding these dot products gives $-3+26=23$, the same result as $(\mathbf{u}+\mathbf{v})\cdot\mathbf{w}$. The example checks the identity for these vectors; the coordinate proof establishes it for all vectors in $\mathbb{R}^n$.

Why the result makes sense geometrically

If $\mathbf{w}$ is fixed, the number $\mathbf{u}\cdot\mathbf{w}$ depends linearly on $\mathbf{u}$. In particular, adding two input vectors adds their dot products with $\mathbf{w}$. This is the behavior of a linear function from vectors to real numbers, often called a linear functional.

For nonzero vectors $\mathbf{u}$ and $\mathbf{w}$, the geometric formula $\mathbf{u}\cdot\mathbf{w}=\|\mathbf{u}\|\,\|\mathbf{w}\|\cos\theta$ also describes the dot product, where $\theta$ is the angle between the vectors. But it is usually not the simplest way to prove distributivity: when you replace $\mathbf{u}$ with $\mathbf{u}+\mathbf{v}$, both its length and its angle with $\mathbf{w}$ may change. The coordinate definition avoids that difficulty.

Common mistakes

• Adding the wrong quantities: The dot product is a scalar, not a vector. The identity says to add two scalar dot products, not to add coordinates after taking the dot product.
• Changing the grouping: $(\mathbf{u}+\mathbf{v})\cdot\mathbf{w}$ equals $\mathbf{u}\cdot\mathbf{w}+\mathbf{v}\cdot\mathbf{w}$; it does not generally equal $\mathbf{u}\cdot\mathbf{v}+\mathbf{w}$.
• Using incompatible dimensions: The vectors being added must have the same number of components, and the dot products must be defined between vectors of the same dimension.
• Treating the dot product as ordinary multiplication: The proof uses ordinary distributivity on each coordinate product, but the dot product itself is a sum of products, not a single product.

Scalar multiplication

For real scalars $c$, the same coordinate definition also gives

$$(c\mathbf{u})\cdot\mathbf{v}=c(\mathbf{u}\cdot\mathbf{v}), \qquad \mathbf{u}\cdot(c\mathbf{v})=c(\mathbf{u}\cdot\mathbf{v}).$$

Together with additivity in both arguments, these rules say that the real dot product is bilinear. For complex vectors, the usual inner product includes complex conjugation, so scalar multiplication behaves differently in one argument; the addition identities still hold.