Recipe

Find coordinates relative to an orthonormal basis

Use when: you have a basis you know, or can check, is orthonormal, and need a vector's coordinates without solving a linear system

Goal: every coordinate c_i=\langle\mathbf v,\mathbf e_i\rangle, confirmed by reconstructing \mathbf v from them

Taught in Orthonormal bases and Fourier coefficients

1. 1

   Check that each basis vector has \langle\mathbf e_i,\mathbf e_i\rangle=1 and every pair has \langle\mathbf e_i,\mathbf e_j\rangle=0 for i\ne j.

   Example

   For \mathbf e_1=(\tfrac35,\tfrac45), \mathbf e_2=(-\tfrac45,\tfrac35): \langle\mathbf e_1,\mathbf e_1\rangle=\tfrac9{25}+\tfrac{16}{25}=1, \langle\mathbf e_2,\mathbf e_2\rangle=1, and \langle\mathbf e_1,\mathbf e_2\rangle=-\tfrac{12}{25}+\tfrac{12}{25}=0.

   • the basis checks out orthonormal — the coefficient formula in Step 2 applies directly.
   • the basis is NOT orthonormal — fall back to solving the ordinary change-of-basis linear system instead.

   Why: every step after this one relies on orthonormality to make the cross terms vanish, and skipping this check risks a shortcut that silently gives the wrong answer.

2. 2

   For each basis vector \mathbf e_i, compute c_i=\langle\mathbf v,\mathbf e_i\rangle directly, with no system to solve.

   Example

   For \mathbf v=(1,7): c_1=\langle\mathbf v,\mathbf e_1\rangle=\tfrac35(1)+\tfrac45(7)=\tfrac{31}5, and c_2=\langle\mathbf v,\mathbf e_2\rangle=-\tfrac45(1)+\tfrac35(7)=\tfrac{17}5.

   Why: orthonormality makes every term but j=i vanish in \langle\mathbf v,\mathbf e_i\rangle=\sum_j c_j\langle\mathbf e_j,\mathbf e_i\rangle, isolating c_i by itself.

3. 3

   Compute c_1\mathbf e_1+c_2\mathbf e_2+\cdots and confirm it equals the original vector \mathbf v.

   Example

   \tfrac{31}5(\tfrac35,\tfrac45)+\tfrac{17}5(-\tfrac45,\tfrac35)=(\tfrac{93}{25},\tfrac{124}{25})+(-\tfrac{68}{25},\tfrac{51}{25})=(1,7), matching \mathbf v exactly.

   Why: this catches an arithmetic slip in Step 2 before the coefficients get used any further, the same way substituting a root back checks a factored equation.