For a vector subspace ,
In finite dimension, .
An annihilator of a vector subspace collects the linear functionals that ignore every input from that vector subspace. For , a linear functional vanishes on exactly when . Hence is the one-dimensional span of . The zero level set in the linear functional diagram is precisely .
For a vector subspace of a finite-dimensional vector space , its annihilator of a vector subspace determines exactly: . Indeed, a basis of can be extended to one of . If , include among the added basis vectors and define a linear functional taking value one at and zero on all the other basis vectors. This functional vanishes on but not at , excluding every outside vector from the intersection.
It suffices to intersect the kernels of any basis of : every other annihilating functional is their linear combination. The dual basis for a basis of extending a basis of shows . If , the empty intersection is . This translates a subspace description by spanning vectors into an exact description by independent linear equations.
For a linear map , an invariant subspace has an annihilator of a vector subspace invariant under the dual map . Indeed, for and ,
For a finite-dimensional vector space, the converse also holds. If , every vanishes on for each . The dual basis description of shows that the common zero set of its linear functionals is exactly , so . Thus
The dimension identity exchanges a one-dimensional invariant subspace with an invariant hyperplane in the dual space.

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