How Do You Expand an Inner Product of Finite Sums?

If you need to multiply out an inner product whose arguments are finite sums, distribute across both arguments. Every term in the first sum pairs with every term in the second. This is the inner-product analogue of multiplying two finite sums of numbers.

The finite-sum rule

Let $x_1,\ldots,x_m$ and $y_1,\ldots,y_n$ belong to an inner product space. Then

$$\left\langle \sum_{i=1}^m x_i,\,\sum_{j=1}^n y_j\right\rangle =\sum_{i=1}^m\sum_{j=1}^n \langle x_i,y_j\rangle.$$

For example, with two terms in each sum,

$$\langle x_1+x_2,\,y_1+y_2\rangle =\langle x_1,y_1\rangle+\langle x_1,y_2\rangle +\langle x_2,y_1\rangle+\langle x_2,y_2\rangle.$$

This identity follows by applying additivity in one argument and then additivity in the other. In a complex inner product space, the inner product is sesquilinear: it is additive in both arguments, but scalar multiplication in one argument introduces complex conjugation. That distinction matters when scalars are present; it does not change the displayed rule for sums of vectors.

Worked example in $\mathbb{R}^2$

Use the usual dot product and take

$$x_1=(1,2),\quad x_2=(3,-1),\quad y_1=(2,0),\quad y_2=(-1,4).$$

Expanding the inner product of the sums gives four pairings:

$$\begin{aligned} \langle x_1+x_2, y_1+y_2\rangle &=\langle x_1,y_1\rangle+\langle x_1,y_2\rangle +\langle x_2,y_1\rangle+\langle x_2,y_2\rangle\\ &=(1\cdot2+2\cdot0)+(1\cdot(-1)+2\cdot4)\\ &\quad +(3\cdot2+(-1)\cdot0)+(3\cdot(-1)+(-1)\cdot4)\\ &=2+7+6-7=8. \end{aligned}$$

You can check directly: $x_1+x_2=(4,1)$ and $y_1+y_2=(1,4)$, whose dot product is $4\cdot1+1\cdot4=8$. Both methods agree.

Applying the rule to a squared norm

A common special case is the squared norm of a sum. Since $\|x\|^2=\langle x,x\rangle$,

$$\left\|\sum_{i=1}^m x_i\right\|^2 =\sum_{i=1}^m\sum_{j=1}^m\langle x_i,x_j\rangle.$$

In a real inner product space, symmetry gives the equivalent expression

$$\left\|\sum_{i=1}^m x_i\right\|^2 =\sum_{i=1}^m\|x_i\|^2+2\sum_{1\le i<j\le m}\langle x_i,x_j\rangle.$$

In a complex inner product space, use conjugate symmetry, $\langle x_j,x_i\rangle=\overline{\langle x_i,x_j\rangle}$. Each pair of cross terms therefore adds to twice a real part:

$$\left\|\sum_{i=1}^m x_i\right\|^2 =\sum_{i=1}^m\|x_i\|^2+2\operatorname{Re}\left(\sum_{1\le i<j\le m}\langle x_i,x_j\rangle\right).$$

For two vectors this is $\|x+y\|^2=\|x\|^2+\|y\|^2+2\operatorname{Re}\langle x,y\rangle$. In a real space, the real-part operator can be omitted.

Common mistakes

• Keeping only matching indices. The expression $\langle\sum_i x_i,\sum_j y_j\rangle$ includes every pair $(i,j)$, not just terms with $i=j$.
• Treating a complex inner product as bilinear. Sums distribute in both arguments, but pulling out a complex scalar conjugates it in whichever argument is conjugate-linear under your convention. Check whether your text defines the inner product to be linear in the first or second argument.
• Dropping cross terms in a squared norm. In general, $\|x+y\|^2$ is not just $\|x\|^2+\|y\|^2$. The cross terms vanish when $x$ and $y$ are orthogonal; otherwise they contribute.

The finite-sum expansion is valid in any inner product space, including an infinite-dimensional Hilbert space. It uses only the inner-product axioms, not completeness or finite dimensionality.