= Invariant subspaces and annihilators
{title2=$T(U)\subseteq U\Longrightarrow T'(U^0)\subseteq U^0$}
For a <linear map> $T:V\to V$, an <invariant subspace> $U$ has an <annihilator of a vector subspace> $U^0$ invariant under the <dual map> $T'$. Indeed, for $\varphi\in U^0$ and $u\in U$,
$$
(T'\varphi)(u)=\varphi(Tu)=0.
$$
For a <finite-dimensional vector space>, the converse also holds. If $T'(U^0)\subseteq U^0$, every $\varphi\in U^0$ vanishes on $Tu$ for each $u\in U$. The <dual basis> description of $U^0$ shows that the common zero set of its <linear functionals> is exactly $U$, so $Tu\in U$. Thus
$$
\boxed{U\text{ is invariant under }T\iff U^0\text{ is invariant under }T'}.
$$
The <dimension> identity $\dim U^0=\dim V-\dim U$ exchanges a one-dimensional <invariant subspace> with an invariant <hyperplane> in the <dual space>.
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