Inner Product Algebra

In this article, we give some useful algebraic tricks for inner products that will be useful in deriving range proofs (and encoding circuits as inner products) later. Each rule will be accompanied by a simple proof.

Notation

Variables in bold, like a\mathbf{a}, denote a vector. Variables not in bold, like vv, denote a scalar. The operator ∘\circ is the Hadamard product (elementwise multiplication) of two vectors, i.e. [a1,…,an]∘[b1,…,bn]=[a1b1,…,anbn][a_1, \dots, a_n]\circ[b_1, \dots, b_n] = [a_1b_1, \dots, a_nb_n]. We use the shorthand “lhs” and “rhs” to refer to the “left-hand side” and “right-hand side” of an equation, respectively. A “summand” is an element of an addition, e.g. if a+b=ca + b = c, then aa and bb would be called summands. The 1\mathbf{1} vector is a vector of all ones, i.e. [1,1,…,1][1, 1, \dots, 1]. All vectors are implied to be of the same length nn unless otherwise stated.

Rule 1: An inner product where one of the vectors is a sum of vectors can be expanded

Suppose we’re calculating an inner product where one of the vectors is a sum of two vectors - for example ⟨a+b,c⟩\langle\mathbf{a} + \mathbf{b}, \mathbf{c}\rangle. We can split this up into the sum of two inner products:
⟨a+b,c⟩=⟨a,c⟩+⟨b,c⟩\langle \mathbf{a} + \mathbf{b}, \mathbf{c} \rangle = \langle \mathbf{a}, \mathbf{c}\rangle + \langle \mathbf{b}, \mathbf{c} \rangle

Proof:
The lhs can be written as

∑i=1n(ai+bi)ci\sum_{i=1}^n(a_i+b_i)c_i

The rhs can be written as

∑i=1naici+∑i=1ncibi=∑i=1n(aici+cibi)=∑i=1n(ai+bi)ci\begin{align*} \sum_{i=1}^na_ic_i+\sum_{i=1}^nc_ib_i &=\sum_{i=1}^n(a_ic_i+c_ib_i) \\ &=\sum_{i=1}^n(a_i+b_i)c_i \end{align*}

Rule 2: Inner products with common terms can be combined

The two inner products on the lhs below have a common vector of c\mathbf{c}. Therefore, they can be combined:

⟨a,c⟩+⟨b,c⟩=⟨a+b,c⟩\langle \mathbf{a}, \mathbf{c}\rangle + \langle \mathbf{b}, \mathbf{c} \rangle = \langle \mathbf{a} + \mathbf{b}, \mathbf{c} \rangle

This is really Rule 1 with the lhs and the rhs swapped.

The proof is the same as Rule 1.

Rule 3: Moving vectors to the other side of the inner product

An inner product can be re-written as the 1\mathbf{1} vector with the Hadamard product of the original vectors:

⟨a,b⟩=⟨1,a∘b⟩\langle \mathbf{a}, \mathbf{b} \rangle= \langle \mathbf{1}, \mathbf{a\circ b} \rangle

Proof:

⟨a,b⟩=∑i=1naibi⟨1,a∘b⟩=∑i=1n1∗(aibi)∑i=1naibi=∑i=1n1∗(aibi)\begin{align*} \langle \mathbf{a}, \mathbf{b} \rangle&=\sum_{i=1}^na_ib_i \\ \langle \mathbf{1}, \mathbf{a\circ b} \rangle&=\sum_{i=1}^n1*(a_ib_i)\\ \sum_{i=1}^na_ib_i &= \sum_{i=1}^n1*(a_ib_i)\\ \end{align*}

Rule 4: We can add vectors to one of the terms of the inner product to force two inner products to have common terms

Suppose we’re adding an inner product ⟨x,b+c⟩\langle\mathbf{x}, \mathbf{b}+\mathbf{c}\rangle and an inner product ⟨y,b⟩\langle\mathbf{y}, \mathbf{b}\rangle, and the sum of the inner products is vv. Note that they have different components, so we can’t add them with Rule 2. Nevertheless, the following equality

⟨x,b+c⟩+⟨y,b⟩=v\langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{y}, \mathbf{b}\rangle = v

can be written as

⟨x+y,b+c⟩=v+⟨y,c⟩\langle \mathbf{x} + \mathbf{y}, \mathbf{b} + \mathbf{c}\rangle = v + \langle\mathbf{y},\mathbf{c}\rangle

In the above scenario, we can add ⟨y,c⟩\langle\mathbf{y},\mathbf{c}\rangle to both sides.

⟨x,b+c⟩+⟨y,b⟩+⟨y,c⟩=v+⟨y,c⟩⟨x,b+c⟩+⟨y,b⟩+⟨y,c⟩=v+⟨y,c⟩\begin{align*} \langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{y}, \mathbf{b}\rangle + \boxed{\langle\mathbf{y},\mathbf{c}\rangle}&= v + \boxed{\langle\mathbf{y},\mathbf{c}\rangle}\\ \langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{y}, \mathbf{b}\rangle + \langle\mathbf{y},\mathbf{c}\rangle&= v + \langle\mathbf{y},\mathbf{c}\rangle \end{align*}

We now have common y\mathbf{y} terms we can combine using Rule 2:

⟨x,b+c⟩+⟨y,b⟩+⟨y,c⟩=v+⟨y,c⟩⟨x,b+c⟩+⟨y,b+c⟩=v+⟨y,c⟩⟨x,b+c⟩+⟨y,b+c⟩=v+⟨y,c⟩\begin{align*} \langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{\fbox{y}}, \mathbf{b}\rangle + \langle\mathbf{\fbox{y}},\mathbf{c}\rangle&= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{\fbox{y}}, \mathbf{b} + \mathbf{c}\rangle &= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{y}, \mathbf{b} + \mathbf{c}\rangle &= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \end{align*}

Now that we have forced the two inner products to have common term ⟨b+c⟩\langle \mathbf{b} + \mathbf{c} \rangle on the lhs, we can combine them into one vector using Rule 2 again:

⟨x,b+c⟩+⟨y,b+c⟩=v+⟨y,c⟩⟨x+y,b+c⟩=v+⟨y,c⟩⟨x+y,b+c⟩=v+⟨y,c⟩\begin{align*} \langle \mathbf{x}, \boxed{\mathbf{b} + \mathbf{c}}\rangle + \langle \mathbf{y}, \boxed{\mathbf{b} + \mathbf{c}}\rangle &= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \langle \mathbf{x} + \mathbf{y}, \boxed{\mathbf{b} + \mathbf{c}}\rangle &= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \langle \mathbf{x} + \mathbf{y}, \mathbf{b} + \mathbf{c}\rangle &= v + \langle\mathbf{y},\mathbf{c}\rangle\\ \end{align*}

Therefore,

⟨x,b+c⟩+⟨y,b⟩=v\langle \mathbf{x}, \mathbf{b} + \mathbf{c}\rangle + \langle \mathbf{y}, \mathbf{b}\rangle = v

can be rewritten as

⟨x+y,b+c⟩=v+⟨y,c⟩\langle \mathbf{x} + \mathbf{y}, \mathbf{b} + \mathbf{c}\rangle = v + \langle\mathbf{y},\mathbf{c}\rangle

Rule 5: Adding two inner products with unrelated vectors

We can add ⟨a1,b1⟩+⟨a2,b2⟩\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle (which have no vectors in common) and obtain:

⟨a1,b1⟩+⟨a2,b2⟩=⟨a1+a2,b1+b2⟩−⟨a1,b2⟩−⟨a2,b1⟩\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle=\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_1+\mathbf{b}_2\rangle-\langle\mathbf{a_1},\mathbf{b_2}\rangle-\langle\mathbf{a_2},\mathbf{b_1}\rangle

Proof:

⟨a1,b1⟩+⟨a2,b2⟩=⟨a1,b1⟩+⟨a2,b2⟩⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩=⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩add ⟨a1,b2⟩ to both sides⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩=⟨a1,b1⟩+⟨a1+a2,b2⟩combine b2 terms⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩+⟨a2,b1⟩=⟨a1,b1⟩+⟨a1+a2,b2⟩+⟨a2,b1⟩add ⟨a2,b1⟩ to both sides⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩+⟨a2,b1⟩=⟨a1+a2,b1⟩+⟨a1+a2,b2⟩combine b1 terms⟨a1,b1⟩+⟨a2,b2⟩+⟨a1,b2⟩+⟨a2,b1⟩=⟨a1+a2,b1+b2⟩combine right-hand side⟨a1,b1⟩+⟨a2,b2⟩=⟨a1+a2,b1+b2⟩−⟨a1,b2⟩−⟨a2,b1⟩subtract ⟨a1,b2⟩+⟨a2,b1⟩\begin{align*} \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle&=\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle&=\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle&&\text{add }\langle\mathbf{a}_1,\mathbf{b}_2\rangle \text{ to both sides}\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle&=\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_2\rangle&&\text{combine }\mathbf{b}_2 \text{ terms}\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle+\langle\mathbf{a}_2,\mathbf{b}_1\rangle&=\langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_2,\mathbf{b}_1\rangle&&\text{add }\langle\mathbf{a}_2,\mathbf{b}_1\rangle\text{ to both sides}\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle+\langle\mathbf{a}_2,\mathbf{b}_1\rangle&=\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_1\rangle+\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_2\rangle&&\text{combine }\mathbf{b}_1\text{ terms}\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle+\langle\mathbf{a}_1,\mathbf{b}_2\rangle+\langle\mathbf{a}_2,\mathbf{b}_1\rangle&=\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_1+\mathbf{b}_2\rangle&&\text{combine right-hand side}\\ \langle\mathbf{a}_1,\mathbf{b}_1\rangle+\langle\mathbf{a}_2,\mathbf{b}_2\rangle&=\langle\mathbf{a}_1+\mathbf{a}_2,\mathbf{b}_1+\mathbf{b}_2\rangle-\langle\mathbf{a}_1,\mathbf{b}_2\rangle-\langle\mathbf{a}_2,\mathbf{b}_1\rangle&&\text{subtract }\langle\mathbf{a}_1,\mathbf{b}_2\rangle+\langle\mathbf{a}_2,\mathbf{b}_1\rangle \end{align*}

The proof illustrates that it may be handy sometimes to be creative about finding inner products to add to both sides of the equation.

Rule 6: Scalars can be brought inside and outside of an inner product

z⋅⟨a,b⟩=⟨z⋅a,b⟩=⟨a,z⋅b⟩z\cdot\langle\mathbf{a},\mathbf{b}\rangle = \langle z\cdot\mathbf{a},\mathbf{b}\rangle = \langle\mathbf{a},z\cdot\mathbf{b}\rangle

The proof for this statement is left as an exercise for the reader. As a hint, constant terms can be brought in and out of a summation.

This tutorial is part of our series on ZK Bulletproofs.

Ready to Get Started?Join Thousands of Users Today

Start your free trial now and experience the difference. No credit card required.

© 2025 Better-Start. All rights reserved.

Product

Resources

Company

Legal