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Suppose $U$ is an invariant subspace for $T$. For $\phi \in U^0$ and $u\in U$, the dual map satisfiesbecause $Tu\in U$ and the annihilator of a vector subspace consists of the linear functionals vanishing there. ThusThis is the correspondence between invariant subspaces and annihilators.

For a vector subspace $U\subseteq V$, its annihilator of a vector subspace isLet $\dim U=k$ and $\dim V=n$. Extend a basis $u_1,\dots ,u_k$ of $U$ to a basis $u_1,\dots ,u_n$ of $V$, and let ${\epsilon }^1,\dots ,{\epsilon }^n$ be the dual basis. Every linear functional has the unique expression $\phi ={\sum }_{j=1}^n\phi (u_j){\epsilon }^j$. It belongs to $U^0$ precisely when its first $k$ coefficients vanish. ConsequentlyThe listed dual basis elements are linearly independent, so counting them proves the formula, including $U=0$ and $U=V$.

Let $\dim V=n$ and $\dim W=k$. Extend a basis $w_1,\dots ,w_k$ of the vector subspace $W$ to $w_1,\dots ,w_n$ of $V$, and let $f_1,\dots ,f_n$ be its dual basis. Every linear functional is uniquely ${\sum }_i{c}_i{f}_i$. It vanishes on $W$ exactly when $c_1=\cdots =c_k=0$. Thus the annihilator of a vector subspace has basis $f_{k+1},\dots ,f_n$, and

To apply this to a matrix, regard its rows as elements of $({\mathbb{R}}^n{)}^{\ast }$ and let $R$ be their linear span. Elementary invertible row and column operations reduce a rank-$r$ matrix to a block identity $\mathrm{diag}(I_r,0)$, showing that $\dim R=r$. The equations require that every functional in $R$ vanish on $x$. The canonical identification ${\mathbb{R}}^n\cong (({\mathbb{R}}^n{)}^{\ast }{)}^{\ast }$ sends $x$ to evaluation $f\mapsto f(x)$: it is injective because coordinate functionals separate points, and is bijective by equal dimensions. Consequently the solution vector space is the annihilator of $R$ under this identification. The dimension formula just proved givesA basis of this kernel consists of exactly $n-r$ linearly independent solutions, and every solution is their linear combination.

The algebraic dual space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all real linear functionals. For a finite basis $e_1,\dots ,e_n$, define the dual basis by $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$, so $e_i^{\ast }(e_j)={\delta }_{ij}$. Every linear functional satisfiesThis equality follows by evaluating both sides on each vector of the basis. It proves spanning directly. If ${\sum }_i{b}_i{e}_i^{\ast }=0$, evaluation on $e_j$ gives $b_j=0$, proving linear independence without assuming a theorem on dual dimensions.

For a vector subspace $U$, its annihilator of a vector subspace isExtend a basis $e_1,\dots ,e_r$ of $U$ to a basis of $V$. Then $U^{\circ }$ has basis $e_{r+1}^{\ast },\dots ,e_n^{\ast }$, so $\boxed{\dim U^{\circ }=n-\dim U}$.

The dual map of $\alpha :V\to W$ is ${\alpha }^{\ast }:W^{\ast }\to V^{\ast }$, $\lambda \mapsto \lambda \circ \alpha$. In the corresponding dual bases its matrix is the transpose of the matrix for $\alpha$. A matrix and its transpose have the same rank, giving $\mathrm{rank}{\alpha }^{\ast }=\mathrm{rank}\alpha$. Directly,Thus $\boxed{\ker {\alpha }^{\ast }=(\mathrm{im}\alpha {)}^{\circ }}$. Also every $\lambda \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}{\alpha }^{\ast }\subseteq (\ker \alpha {)}^{\circ }$. Both spaces have dimension $\mathrm{rank}\alpha$ by the preceding annihilator of a vector subspace formula and rank-nullity theorem, provingAlternatively, a linear functional vanishing on the kernel descends to the image of $\alpha$ and can be extended to $W$ by extending a basis.

For the infinite-dimensional polynomial ring, the coefficient functionals $L_i$ are linearly independent: evaluate a finite relation on each monomial. However, they do not span the algebraic dual space. The linear functional $p\mapsto p(1)$ takes value one on every monomial, while any finite combination of the $L_i$ vanishes on every sufficiently high-degree monomial. Thus the $L_i$ do not form a basis of $V^{\ast }$. An infinite formal sum of coefficient functionals is not a finite linear combination.

The algebraic dual map is $\delta ={\alpha }^{\vee }:W^{\ast }\to V^{\ast }$, defined by $\delta (\phi )=\phi \circ \alpha$. The annihilator of a vector subspace $U\subseteq V$ isEvery functional $\phi \circ \alpha$ vanishes on $\ker \alpha$, so $\mathrm{im}\delta \subseteq (\ker \alpha {)}^{\circ }$. Conversely, let $\lambda \in (\ker \alpha {)}^{\circ }$. Define a functional on $\mathrm{im}\alpha$ by $\tilde{\lambda }(\alpha v)=\lambda (v)$. It is well defined because two preimages differ by an element of $\ker \alpha$. Extend a basis of $\mathrm{im}\alpha$ to one of $W$ and extend the functional arbitrarily, for instance by zero on the extra basis vectors. The resulting $\phi \in W^{\ast }$ satisfies $\delta \phi =\lambda$.

This is the dual image and kernel annihilator identity. ThereforeThe basis extension theorem also gives $\dim U^{\circ }=\dim V-\dim U$. By the rank-nullity theorem,Apply rank-nullity theorem to both maps, using $\dim W^{\ast }=\dim W$, to obtainOver $\mathbb{C}$, this is the algebraic dual, without complex conjugation; the adjoint operator in part (b) uses the inner products instead.

The dual vector space is $V^{\ast }={\mathrm{Hom}}_{\mathbb{R}}(V,\mathbb{R})$, the vector space of all linear functionals on $V$. For a vector subspace $U\le V$, its annihilator of a vector subspace is

Given a basis $x_1,\dots ,x_n$, define $x_i^{\ast }$ by $x_i^{\ast }(x_j)={\delta }_{ij}$ and extend linearly. If ${\sum }_i{a}_i{x}_i^{\ast }=0$, evaluation at $x_j$ gives $a_j=0$, so the $x_i^{\ast }$ are linearly independent. Every $\varphi \in V^{\ast }$ satisfiesso they also span $V^{\ast }$ and therefore form its dual basis.

Write $\varphi =a{x}_1^{\ast }+b{x}_2^{\ast }+c{x}_3^{\ast }+d{x}_4^{\ast }$. The condition $\varphi \in U^0$ isGaussian elimination gives $b=-2c-3d$ and $a=c+2d$. Hence

The dual space isthe vector space of linear functionals on $V$. If $e_1,\dots ,e_n$ is a basis of a finite-dimensional $V$, define $e_i^{\ast }(e_j)={\delta }_{ij}$. Every $f\in V^{\ast }$ has the unique expansionso the dual basis $e_1^{\ast },\dots ,e_n^{\ast }$ proves

For $U\le V$, its annihilator of a vector subspace isIf $u_1,\dots ,u_k$ is a basis of $U$ and is extended to a basis $u_1,\dots ,u_n$ of $V$, thenConsequentlyIf $U\ne V$, this dimension is positive, giving a nonzero functional that vanishes on $U$.

For a linear map $\alpha :V\to W$, the dual map isNowsoEvery $g\circ \alpha$ vanishes on $\ker \alpha$, henceThe two spaces have the same dimension, sinceThus

Let $Q:V\to V/U$ be the quotient map. Since $Q$ is onto, $Q^{\ast }$ is injective, and the preceding identity givesThereforeFor the inclusion $J:U\hookrightarrow V$, the map $J^{\ast }:V^{\ast }\to U^{\ast }$ is restriction to $U$. It is onto and has kernelThe first isomorphism theorem now gives the other duals of a subspace and its quotient:

A linear map on a finite-dimensional vector space over an algebraically closed field has both an upper triangular matrix and a lower triangular matrix in suitable bases. One proof uses invariant subspaces and annihilators to make the induction step entirely in the dual space.

Choose an eigenvector $v$ and let $U=\mathrm{span}\{v\}$. The dual map preserves the annihilator of a vector subspace $U^0$, of dimension one less. By induction, choose a basis of $U^0$ making the restriction upper triangular. Extend this to a basis of the whole dual space; its matrix $C$ remains upper triangular because $U^0$ is an invariant subspace. In the corresponding primal basis, the matrix is $C^T$, which is lower triangular. Reversing the primal basis makes it upper triangular. Dimension zero and one start the induction. This also allows the original eigenvector to be the first vector of the resulting upper triangular basis.