11. Dual space

Dual space

Let V be a vector space over K, then the space dual to V is V^{*}=L(V,K), the space of all linear mappings into the field called also linear functionals or linear forms.

Notice that for any subspace W of V and v\in V\setminus W, there is a fuctional \varphi such that \varphi[W]=\{0\} and\varphi(v)=1. To see this, complete a basis of W to a basis V, is such a way that vector v is in the completed basis and define the values of the fuctional in this basis accordingly.

E.g., let V=\mathbb{R}^3 and W=\text{lin}((1,1,0),(0,0,1)) and v=(1,-1,0). Then functional \varphi, to satisfy the above conditions has to be the following \varphi((1,1,0))=0,\varphi((0,0,1))=0 and \varphi((1,-1,0))=1. Thus, \varphi((x,y,z))=x/2-y/2.

Dual basis

Therefore, given a basis of a finite-dimensional space V: v_1,\ldots, v_n, \dim V=n, we can find functionals \varphi_1,\ldots, \varphi_n, such that \varphi_i(v_i)=1, but for i\neq j, \varphi_i(v_j)=0. Notice, that those functionals are a basis of space V^{*} and this basis is called the dual basis to v_1,\ldots, v_n. Ususaly we denote \varphi_i by v_i^*

E.g., for basis v_1=(1,1,0),v_2=(0,0,1),v_3=(1,-1,0), the dual basis is v_1^*((x,y,z))=x/2+y/2, v_2^*((x,y,z))=z and v_3^*((x,y,z))=x/2-y/2.

Notice also that given a functional \varphi \in V^*, its coordinates in the basis (v^*_1,\ldots, v^*_n) are the values of this functional on vectors v_1,\ldots, v_n. Indeed, if \varphi(v_i)=a_i we get that \varphi = a_1 v_1^*+\ldots +a_n v_n^*, because the left and right side of this equation are equal when calculated on argument equal to v_i for any i\in \{1,\ldots, n\}.

The existence of the dual basis shows that if \dim V<\infty, then V\simeq V^*, and for given basis v_1,\ldots, v_n we have an isomorphism \Phi\colon V\to V^* defined as \Phi(v_i)=v_i^* for all i=1,\ldots,n.

This is not true in the case of infinitely dimensional spaces. Notice that, for example, the dual space (K^\infty_c)^* to the space of all sequences which are zeroes from some point on is isomorphic to K^\infty, so it is not isomorphic to K^\infty_c. Indeed, the isomorphism \Phi\colon K^\infty\to (K^\infty_c)^* given as \Phi((a_n))=\phi_{(a_n)}, where (a_n)\in K^\infty, \phi_{(a_n)}\colon K^\infty_c\to K (so \phi_{(a_n)}\in (K^\infty_c)^*) defined as \phi_{(a_n)}((b_n))=\sum_{i=1}^\infty a_ib_i (the series is convergent because sequence (b_n) equals zero from some point on).

Space V^{**}

Notice that every vector v\in V is related to a functional \psi_v\colon V^*\to K (so \psi_v\in V^{**}) given by the formula \psi_v(\varphi)=\varphi(v). Moreover, the mapping \Phi\colon V\to V^{**} defined as \Phi(v)=\psi_v is linear (indeed \Phi(v+w)(\varphi)=\psi_{v+w}(\varphi)=\varphi(v+w)=\varphi(v)+\varphi(w)=\psi_v(\varphi)+\psi_w(\varphi)=(\Phi(v)+\Phi(w))(\varphi) and \Phi(av)(\varphi)=\psi_{av}(\varphi)=\varphi(av)=a\varphi(v)=a\psi_v(\varphi)=(a\Phi(v))(\varphi)), and actually is a monomorphism. Thus, if \dim V<\infty, and so \dim V=\dim V^*=\dim V^{**}, then \Phi is an isomorphism!

Dual mapping

If \varphi\colon V\to W is a linear mapping, then \varphi^*\colon W^*\to V^* given as \varphi^*(f)=f\circ\varphi is called its dual mapping.

It is easy to notice, that (id_V)^*=id_{V^*}, (\psi\circ \phi)^*=\phi^*\circ \psi^*, and that if \varphi is an isomorphism, then \varphi^* is an isomorphism as well.

Moreover, one can check that if \varphi is a monomorphism, then \varphi^* is an epimorphism. On the other hand, if \varphi is an epimorphism, then \varphi^* is a monomorphism.

Some things get easier under the assumption that the dimensions are finite. It becomes clear that then r(\varphi)=r(\varphi^*), and if \mathcal{A} is a basis of V, and \oB is a basis of W, and \mathcal{A}^* and \mathcal{B}^* are respective dual bases then (M(\varphi)_{\mathcal{A}}^{\mathcal{B}})^T=M(\varphi^*)_{\mathcal{B}^*}^{\mathcal{A}^*}, where T is the transposition, meaning that the rows are swapped with columns. Indeed, if a_{ij} is the entry in the i-th row and j-th column of M(\varphi)_{\mathcal{A}}^{\mathcal{B}}, we get a_{ij}=\beta^*_i(\varphi(\alpha_j))=(\varphi^*(\beta^*_i))(\alpha_j), and this is the entry in j-th row and i-th column of matrix M(\varphi^*)_{\mathcal{B}^*}^{\mathcal{A}^*}, where \alpha_i are vectors from \mathcal{A}, and \beta_i are vectors from \mathcal{B}.

It is also worth mentioning that in the finite dimensional case, after identifying V=V^{**} we get \varphi^{**}=\varphi.

Dual space to an unitary space

Notice that if (V,\xi) is an unitary space, we get an easy identification between V and V^*. Consider a mapping \xi'\colon V\to V^* defined as (\xi''(v))(w)=\xi(w,v). One can prove that \xi' is a bijection and \xi'(av+bw)=\bar{a}v+\bar{b}w, so \xi' is an isomorphism V\to V^\star, where V^\star is the space V^* with modified operation of multiplication (a\cdot f)(v)=\bar{a}\cdot f(v). From this point on we will be identifying v\in V with \xi'(v)\in V^*.

It is easy to prove that assuming this identification for any endomorphisms \varphi,\phi\colon V\to V, \phi=\varphi^* if and only if \xi(\varphi(v),w)=\xi(v,\phi(w)) for any v,w\in V. Therefore, \varphi=\varphi^{**}. The mapping \varphi is self-adjoint if and only if \varphi=\varphi^*. Moreover, a mapping \varphi is self-adjoint if and only if its matrix in an orthonormal basis is hermitian.

There is a common property of self-adjoint endomorphisms and unitary automorphism. For both of them we get \varphi\circ \varphi^*=\varphi^*\circ \varphi. The mappings satisfying this condition are called normal endomorphims. One can prove that an endomorphism has an orthonormal basis consisting of eigenvectors if and only if it is normal.