Recipe
Find a basis for an annihilator W^{\circ}
Use when: you're given a spanning set for a subspace W and need an explicit basis for its annihilator W^{\circ}
Goal: a set of explicit functionals that both annihilate W and span all of W^{\circ}
Taught in Annihilators
Quick recipe
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1
Write \varphi(x,y)=ax+by in \mathbb R^2 (one coefficient per coordinate), with every coefficient still unknown.
Example
For W=\operatorname{span}\{(2,1)\} in \mathbb R^2, write \varphi(x,y)=ax+by.
Why: an annihilator functional is a linear combination of coordinates just like any other functional, so the unknowns have to be named before they can be solved for.
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2
Substitute each spanning vector of W into \varphi's general form and set the result equal to zero, one equation per spanning vector.
Example
\varphi(2,1)=2a+b=0, the single equation coming from W's one spanning vector.
Why: vanishing on a spanning set forces vanishing on all of W, since every vector in W is a combination of that spanning set.
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3
Solve for the dependent coefficients in terms of whichever ones are left free; the number of free coefficients equals \dim(W^{\circ}).
Example
From 2a+b=0, b=-2a, leaving a free. So every \varphi\in W^{\circ} has the form \varphi(x,y)=a(x-2y).
Why: a homogeneous system's free coefficients directly count how many independent functionals span the solution set.
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4
For each free coefficient in turn, set it to 1 and every other free coefficient to 0, then substitute back to get one basis functional.
Example
Setting a=1 above gives the single basis functional \psi(x,y)=x-2y.
- exactly one free coefficient — W^{\circ} is one-dimensional, spanned by the single functional that comes from setting it to 1.
- more than one free coefficient, such as one vector spanning W inside \mathbb R^3 — build one basis functional per free coefficient, each by isolating a different one at a time.
Why: setting exactly one free coefficient to 1 at a time is what turns an infinite family of solutions into a finite, genuinely independent basis.
Common mistakes
- Imposing \varphi(\mathbf w)=0 on only one vector of W when the spanning set has more than one vector.
- Reporting the coefficients a and b themselves as the answer, instead of the functional \varphi(x,y)=\cdots they define.
- Missing a second basis functional when W turns out smaller than expected, leaving more than one free coefficient.
- Forgetting that W^{\circ} lives in V^{*}, not back inside V itself.