Recipe

Construct a dual basis by solving directly

Use when: you have a specific basis \{b_1,\dots,b_n\} of V and need the functionals \varphi_1,\dots,\varphi_n satisfying \varphi_i(b_j)=\delta_{ij}

Goal: every dual basis functional, each written as an explicit formula in coordinates

Taught in Dual bases

1. 1

   Write \varphi_i(x,y)=ax+by (one more unknown coefficient per extra dimension of V), with the coefficients still unknown.

   Example

   For V=\mathbb R^2 with basis \{b_1,b_2\}=\{(2,1),(1,3)\}, write \varphi_1(x,y)=ax+by.

   Why: the dual basis functionals are unknown linear combinations of coordinates, so writing the general form first turns the problem into solving for coefficients.

2. 2

   Substitute each basis vector b_j into \varphi_i's general form, setting the result to 1 when j=i and to 0 for every other j.

   Example

   \varphi_1(2,1)=2a+b=1 and \varphi_1(1,3)=a+3b=0, one equation per basis vector.

   Why: this substitution is exactly the defining property of a dual basis, turned into a solvable system of equations.

3. 3

   Solve the system from Step 2 using ordinary elimination or substitution, the same way any two-equation, two-unknown system is solved.

   Example

   From a+3b=0, a=-3b. Substituting into 2a+b=1 gives -6b+b=1, so b=-\tfrac15 and a=\tfrac35.

   Why: the coefficients are exactly what makes the defining property hold, and nothing shortcuts actually solving for them.

4. 4

   Redo Steps 1 through 3 once for each index i from 1 to n, changing only which basis vector gets the value 1 on the right side.

   Example

   For \varphi_2: 2c+d=0 and c+3d=1, giving c=-\tfrac15, d=\tfrac25, so \varphi_2(x,y)=-\tfrac15x+\tfrac25y.

   Why: a dual basis has exactly as many functionals as V has dimensions, and each one needs its own system solved.

5. 5

   Plug every basis vector into every functional and confirm the pattern of 1s and 0s matches exactly.

   • every pairing matches the Kronecker delta pattern — the dual basis is confirmed correct.
   • any pairing is off by even one entry — an arithmetic slip happened in Step 3, and that system needs to be resolved.

   ⚡ Shortcut

   For the standard basis \{e_1,\dots,e_n\} of \mathbb F^n, skip Steps 1 through 4 entirely: the dual basis is just the coordinate functionals \varphi_i(x_1,\dots,x_n)=x_i, with no system to solve.

   Why: the defining property is the only thing that makes these functionals a genuine dual basis, so checking it catches an arithmetic slip before it spreads.