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For a vector subspace $U\le V$,In finite dimension, $\dim U^{\circ }=\dim V-\dim U$.

An annihilator of a vector subspace collects the linear functionals that ignore every input from that vector subspace. For $U=\mathrm{span}\{(1,2)\}\subseteq {\mathbb{R}}^2$, a linear functional $\phi (x,y)=ax+by$ vanishes on $U$ exactly when $a+2b=0$. Hence $U^{\circ }$ is the one-dimensional span of $(x,y)\mapsto 2x-y$. The zero level set in the linear functional diagram is precisely $U$.

For a linear map between finite-dimensional vector spaces, the image of its dual map is the annihilator of a vector subspace given by its kernel. Inclusion follows by composition. Conversely a linear functional vanishing on the kernel descends to the image and extends to the target by basis extension. The identity and the rank-nullity theorem show that a linear map and its dual have equal rank.

For a vector subspace $U$ of a finite-dimensional vector space $V$, its annihilator of a vector subspace $U^0\subseteq V^{\ast }$ determines $U$ exactly: $U={\bigcap }_{f\in U^0}\ker f$. Indeed, a basis of $U$ can be extended to one of $V$. If $v\notin U$, include $v$ among the added basis vectors and define a linear functional taking value one at $v$ and zero on all the other basis vectors. This functional vanishes on $U$ but not at $v$, excluding every outside vector from the intersection.

It suffices to intersect the kernels of any basis $g_1,\dots ,g_k$ of $U^0$: every other annihilating functional is their linear combination. The dual basis for a basis of $V$ extending a basis of $U$ shows $k=\dim V-\dim U$. If $U=V$, the empty intersection is $V$. This translates a subspace description by spanning vectors into an exact description by independent linear equations.

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 $T^{\prime }$. Indeed, for $\phi \in U^0$ and $u\in U$,For a finite-dimensional vector space, the converse also holds. If $T^{\prime }(U^0)\subseteq U^0$, every $\phi \in U^0$ 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 $Tu\in U$. ThusThe dimension identity $\dim U^0=\dim V-\dim U$ exchanges a one-dimensional invariant subspace with an invariant hyperplane in the dual space.

For a bilinear form $B$ on a finite-dimensional vector space $V$, let $\Phi :V\to V^{\ast }$ send $u$ to $B(u,\cdot )$. Under the bidual evaluation map, $U^{\perp }$ is the annihilator of a vector subspace $\mathrm{im}(\Phi {|}_U)$, so $\dim U^{\perp }=\dim V-\mathrm{rank}(\Phi {|}_U)$. The rank-nullity theorem on $U$ now gives the displayed equality. In particular the sum of dimensions is at least $\dim V$. If $U\cap U^{\perp }=0$, the dimension formula for a sum of subspaces forces $V=U\oplus U^{\perp }$. The exact formula records the extra dimensions contributed by the left radical $\ker \Phi$, and does not assume an inner product.

The annihilator of a vector subspace $W\le V^{\ast }$ isIf $\dim W=k$, extend a basis of $W$ to one of $V^{\ast }$. In the resulting dual basis of $V^{\ast \ast }$, the last $n-k$ coordinate functionals span $W^{\circ }$. Hence $\boxed{\dim W+\dim W^{\circ }=n}$.

The bidual evaluation map $j:V\to V^{\ast \ast }$ is $j(x)(\varphi )=\varphi (x)$. It is a linear isomorphism in finite dimension. Its inverse image of the annihilator isThis identification uses evaluation only; no choice of inner product is involved.

The annihilator of a vector subspace is the vector subspace of the dual space consisting of the linear functionals that vanish on $U$:It is a vector subspace because every linear combination of functionals vanishing on $U$ still vanishes there.

The span $U$ is the coordinate plane $\{(x_1,x_2,0,0{)}^T\}$. Consequently its annihilator of a vector subspace is $U^{\circ }=\mathrm{span}\{{\epsilon }_3,{\epsilon }_4\}$. The dual map is composition with $T$, soIt follows thatEquivalently, $T^{\prime }(a{\epsilon }_3+b{\epsilon }_4)$ sends $x$ to $a{x}_3$.

The annihilator of a vector subspace consists of the linear functionals that vanish on every element of that vector subspace:For the dimension identity, choose a basis $w_1,\dots ,w_k$ of $W$ and use basis extension to obtain $w_1,\dots ,w_n$ for $V$. A linear functional is determined by its values on these $n$ basis vectors. Belonging to the annihilator of a vector subspace sets its first $k$ values to zero and leaves its last $n-k$ values arbitrary. Equivalently the last $n-k$ elements of the dual basis form a basis of $W^0$.

For a linear functional $\lambda \in V^{\prime }$, the definition of the dual map givesThus the kernel of a linear map is precisely the annihilator of a vector subspaceWrite $n=\dim U$, $m=\dim V$ and $r=\dim \mathrm{im}T$. The rank-nullity theorem, the dimension formula for an annihilator of a vector subspace, and $\dim V^{\prime }=m$ giveApplying the same dimension formula in $U$ givesFinally, if $\alpha =T^{\prime }\lambda$ and $u\in \ker T$, then $\alpha (u)=\lambda (Tu)=\lambda (0)=0$. Therefore $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^0$. These are finite-dimensional vector subspaces of equal dimension, so inclusion is equality:This is the dual image and kernel annihilator identity.

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The annihilator of a vector subspace isIt is a vector subspace of $V^{\prime }$: linear combinations of functionals vanishing on $U$ also vanish on $U$.

Choose a basis $u_1,\dots ,u_m$ of $U$. Because $v\notin U$, the list $u_1,\dots ,u_m,v$ is linearly independent. Extend it to a basis of $V$. A linear functional is specified uniquely by its values on a basis, so define $f$ to be zero on the $u_j$ and on every added basis vector other than $v$, and set $f(v)=1$.

Then $f$ vanishes on every linear combination of the $u_j$, so $f\in U^0$, whileThis proves that the annihilator of a vector subspace separates every vector outside the subspace from that subspace; no inner product or preferred coordinates are needed.

The dual basis is $E^{\ast }=\{e_1^{\ast },\dots ,e_n^{\ast }\}$, where $e_i^{\ast }(e_j)={\delta }_{ij}$ and each coordinate linear functional extends linearly. Thus $e_i^{\ast }({\sum }_j{a}_j{e}_j)=a_i$. These functionals are a basis of $V^{\ast }$ because a linear functional is uniquely determined by its values on $E$.

For a linear subspace $U$ of dimension $k$, extend a basis $u_1,\dots ,u_k$ to one of $V$. In its dual basis, the annihilator of a vector subspace $U^{\circ }$ is spanned by the last $n-k$ coordinate functionals. Hence $\boxed{\dim U+\dim U^{\circ }=\dim V}$.

The transpose in the source is the algebraic dual map $T^{\ast }:V^{\ast }\to V^{\ast }$ defined by $T^{\ast }f=f\circ T$. Composition of linear maps is linear, so $T^{\ast }f$ is a linear functional; and $T^{\ast }(af+bg)=a{T}^{\ast }f+b{T}^{\ast }g$, proving linearity in its input. If $T(e_j)={\sum }_i{a}_{ij}{e}_i$, thenThe coefficient of $e_j^{\ast }$ in $T^{\ast }{e}_i^{\ast }$ is consequently $a_{ij}$, so $\boxed{[T^{\ast }{]}_{E^{\ast }}=([T{]}_E{)}^{\mathsf{T}}}$. No inner product is involved.

A linear functional $f$ annihilates $\mathrm{Im}T$ exactly when $f(Tv)=0$ for all $v$, equivalently $T^{\ast }f=0$. Thus $(\mathrm{Im}T{)}^{\circ }=\ker T^{\ast }$. If $v\in \ker T$, then $(T^{\ast }f)(v)=f(Tv)=0$, giving $\mathrm{Im}{T}^{\ast }\subseteq (\ker T{)}^{\circ }$. The first equality and the dimension formula give $\dim \ker T^{\ast }=n-\dim \mathrm{Im}T$. By the rank-nullity theorem,The inclusion is therefore an equality:

Write $V^{\ast }=V^{\prime }$ for the dual space. Define $\theta (v)(f)=f(v)$. For fixed $v$, this is a linear functional on $V^{\ast }$; evaluation is also linear in $v$, so $\theta :V\to V^{\ast \ast }$ is a linear map. If $v\ne 0$, extend $v$ to a basis and take its first coordinate linear functional $f$; then $f(v)=1$. Thus $\theta$ is injective. A dual basis has the same number of elements as the original basis, so $\dim V=\dim V^{\ast }=\dim V^{\ast \ast }$. The rank-nullity theorem now makes $\theta$ an linear isomorphism.

The annihilator of a vector subspace of a linear subspace $U\le V$ isExtend a basis $v_1,\dots ,v_k$ of $U$ to a basis $v_1,\dots ,v_n$ of $V$, with dual basis $f_1,\dots ,f_n$. A linear functional belongs to $U^{\circ }$ exactly when its first $k$ coordinates in this dual basis vanish. Hence $U^{\circ }$ has basis $f_{k+1},\dots ,f_n$, provingThe dual map is ${\alpha }^{\ast }(h)=h\circ \alpha$; here this is the algebraic dual map, denoted ${\alpha }^{\prime }$ in the source. If $v\in \ker \alpha$, then $({\alpha }^{\prime }h)(v)=h(\alpha v)=0$. Consequently $\mathrm{im}{\alpha }^{\prime }\subseteq (\ker \alpha {)}^{\circ }$. Moreover,because coordinate linear functionals separate vectors. This proves $(\mathrm{im}{\alpha }^{\prime }{)}^{\circ }=\theta (\ker \alpha )$, including the requested second inclusion. Taking dimensions givesThe first inclusion has equal dimensions on both sides, soFor a matrix $A=(a_{ij})$ representing $\alpha$, the matrix of ${\alpha }^{\prime }$ in the corresponding dual bases is $A^{\mathsf{T}}$: its $ji$ entry is $h_i(\alpha v_j)=a_{ij}$. Thus the column rank of $A^{\mathsf{T}}$ is the row rank of $A$, while the column rank of $A$ is $\dim \mathrm{im}\alpha$. Applying the established equality to these two linear maps proves row rank equals column rank over every field.

The dual space is the vector space of all linear functionals on $V$:Addition and scalar multiplication are defined pointwise. The annihilator of a vector subspace isIt is a vector subspace of $V^{\prime }$: linear combinations of functionals vanishing on $U$ also vanish on $U$.

Choose a basis $u_1,\dots ,u_m$ of $U$. Because $v\notin U$, the list $u_1,\dots ,u_m,v$ is linearly independent. Extend it to a basis of $V$. A linear functional is specified uniquely by its values on a basis, so define $f$ to be zero on the $u_j$ and on every added basis vector other than $v$, and set $f(v)=1$.

Then $f$ vanishes on every linear combination of the $u_j$, so $f\in U^0$, whileThis proves that the annihilator of a vector subspace separates every vector outside the subspace from that subspace; no inner product or preferred coordinates are needed.

If $\varphi =T^{\prime }\psi$ and $v\in \ker T$, then $\varphi (v)=\psi (Tv)=0$. Thus $\mathrm{im}{T}^{\prime }\subseteq (\ker T{)}^{\circ }$, using the annihilator of a vector subspace.

Conversely, let $\varphi$ vanish on the kernel of a linear map $\ker T$. Define a linear functional on $\mathrm{im}T$ by ${\psi }_0(Tv)=\varphi (v)$. This is well-defined: $Tv=Tw$ implies $v-w\in \ker T$, so $\varphi (v)=\varphi (w)$. Extend ${\psi }_0$ to a linear functional $\psi$ on $W$, using the extension argument from part (i). Then $T^{\prime }\psi =\varphi$, provingThe dimension formula from part (i) and the rank-nullity theorem now giveTo deduce the eigenvalue statement, apply this equality of rank of a linear map to $T-\lambda I$. Its dual map is $T^{\prime }-\lambda I$, so their kernel of a linear map dimensions agree. Thus the same $\lambda$ are eigenvalues, with the same geometric multiplicities.

For algebraic multiplicities as well, part (a)(iii) givesso equality of rank applies to every power. HenceAfter extending to a field where the characteristic polynomials split, take $r\ge \dim V$. These kernel of a linear map spaces are the generalized eigenspaces; the primary decomposition theorem for linear maps identifies their dimensions with the algebraic multiplicities. The same proof works over the original field for every eigenvalue in it. We obtainIn fact the argument also includes the zero eigenvalue.

The annihilator of a vector subspace isChoose a basis $e_1,\dots ,e_r$ of $U$ and extend it to a basis $e_1,\dots ,e_d$ of $V$. In the dual basis, precisely $f_{r+1},\dots ,f_d$ span $U^{\circ }$. ThereforeThe canonical restriction linear transformationhas kernel of a linear map $U^{\circ }$. It is surjective: extend any linear functional on $U$ by assigning arbitrary values, for instance zero, to the additional basis vectors. The First isomorphism theorem for vector spaces gives the explicit linear isomorphismIt is natural because the displayed map itself makes no basis choice; a basis was used only to prove surjectivity. In particular, $U^{\prime }$ is canonically a quotient vector space of $V^{\prime }$, rather than canonically a subspace of $V^{\prime }$.

Let $F$ be the common field. The dual map is $S^{\prime }(f)=f\circ S$ for $f\in W^{\prime }$, where $W^{\prime }$ and $V^{\prime }$ are the dual spaces.

First assume $S^{\prime }$ is injective. If $S$ were not surjective, its image of a linear map $\mathrm{im}S$ would be a proper vector subspace of $W$. Choose a basis $w_1,\dots ,w_r$ of $\mathrm{im}S$ and use basis extension to obtain a basis $w_1,\dots ,w_m$ of $W$, with $r<m$. There is a linear functional $f\in W^{\prime }$ withIt is nonzero and vanishes on $\mathrm{im}S$, so $S^{\prime }(f)=f\circ S=0$, contradicting injectivity. Equivalently, the annihilator of a vector subspace $\mathrm{im}S$ would give a nonzero kernel for $S^{\prime }$. Therefore

Now assume $S^{\prime }$ is surjective. If $0\ne v\in \ker S$, basis extension gives a linear functional $h\in V^{\prime }$ with $h(v)=1$: include $v$ in a basis and take its coordinate functional. By surjectivity, $h=S^{\prime }(f)$ for some $f\in W^{\prime }$. Buta contradiction. Hence $\ker S=0$, which proves

For the given nondegenerate bilinear form, define the linear mapBy part (a), $\Phi$ is a linear isomorphism. The paper's bilinear orthogonal complement uses $U$ in the first argument; with this convention,Here $U^0$ is the annihilator of a vector subspace $U$. Since $\Phi$ is a linear isomorphism, its restriction gives a linear isomorphism $U^{\perp }\to U^0$.

For completeness, choose a basis $u_1,\dots ,u_k$ of $U$ and extend it to a basis $u_1,\dots ,u_n$ of $V$. In its dual basis $u_1^{\ast },\dots ,u_n^{\ast }$, a linear functional vanishes on $U$ exactly when its first $k$ coefficients are zero. ThereforeConsequentlyThe argument includes $U=0$ and $U=V$, and requires no symmetry assumption on the bilinear form.