Dual Space

This post was machine-translated from the Korean original by Marvin (via Kimi). It may contain errors or awkward phrasing — the Korean original is the source of truth.

Dual Basis

Let \(V\) be a finite-dimensional \(\mathbb{K}\)-vector space. From §Space of Linear Maps, ⁋Proposition 5, taking \(W=\mathbb{K}\), we know that \(V^\ast=\Hom(V,\mathbb{K})\) has the same dimension as \(V\). In particular, if \(\mathcal{B}=\{x_1,\ldots, x_n\}\) is a basis of \(V\), then the collection of linear maps \(\xi^i\) sending \(x_i\) to 1 and every other \(x_j\) to 0

\[\mathcal{B}^\ast=\{\xi^1,\ldots, \xi^n\}\]

forms a basis of \(V^\ast\). We call this the dual basis of \(\mathcal{B}\).

Even when \(V\) is infinite-dimensional, the linear independence of the set \(\mathcal{B}^\ast\) defined as above for any basis \(\mathcal{B}\) follows from exactly the same proof as in §Space of Linear Maps, ⁋Proposition 5, without any modification. Hence \(\dim V\leq\dim V^\ast\) always holds, and in fact, if \(V\) is infinite-dimensional then necessarily \(\dim V<\dim V^\ast\). To see this, it suffices to verify that the function obtained by extending the map sending every element of an arbitrary \(\mathcal{B}\) to \(1\) cannot be expressed as a linear combination of elements of \(\mathcal{B}^\ast\).

Double Dual Space

When \(V\) is finite-dimensional, \(V\) and \(V^\ast\) have the same dimension, and therefore the dual space \(V^{\ast\ast}\) of \(V^\ast\) is also a \(\mathbb{K}\)-vector space with the same dimension as \(V^\ast\). We call this the double dual of \(V\).

To show that \(V\) and \(V^\ast\) are isomorphic, we had to choose a specific basis. By contrast, we can construct an injective linear map from \(V\) to \(V^{\ast\ast}\) that is independent of the choice of basis. Since \(V\) and \(V^{\ast\ast}\) have the same dimension, this injective linear map must be an isomorphism.

Definition 1 Given three \(\mathbb{K}\)-vector spaces \(U,V,W\), a function \(f:U\times V\rightarrow W\) is called bilinear if for any \(u,u_1,u_2\in U\), \(v,v_1,v_2\in V\), and scalar \(\alpha\),

\[f(u_1+u_2,v)=f(u_1,v)+f(u_2,v),\qquad f(u,v_1+v_2)=f(u,v_1)+f(u,v_2),\qquad f(\alpha u,v)=\alpha f(u,v)=f(u,\alpha v)\]

hold.

In other words, for any \(u\in U\) the function \(f(u,-):V\rightarrow W\) is linear, and for any \(v\in V\) the function \(f(-,v):U\rightarrow W\) is also linear.

Definition 2 Given two \(\mathbb{K}\)-vector spaces \(V,W\), a pairing of \(V\) and \(W\) is a bilinear map \((-,-):V\times W\rightarrow \mathbb{K}\). If for every nonzero \(v\in V\) the linear map

\[(v,-): W\rightarrow \mathbb{K}\]

is never the zero function, then this pairing is called non-degenerate on the left; similarly, if for every nonzero \(w\in W\) the map

\[(-,w):U\rightarrow \mathbb{K}\]

is never the zero function, then this pairing is called non-degenerate on the right. A pairing that is non-degenerate on both the left and the right is simply called a non-degenerate pairing.

The notation \((-,-)\) coincides with that for ordered pairs, which may cause some confusion, but the two are easily distinguished from context, and the notation is convenient; we therefore adopt it.

For example, the dot product of vectors \((-,-):\mathbb{R}^n\times\mathbb{R}^n\rightarrow\mathbb{R}\) on \(V=W=\mathbb{R}^n\) is a non-degenerate pairing. First,

\[(v,w_1+w_2)=(v,w_1)+(v,w_2),\qquad (v_1+v_2,w)=(v_1,w)+(v_2,w),\qquad (\alpha v,w)=\alpha(v,w)=(v,\alpha w)\]

hold trivially, so the dot product is indeed a pairing. Moreover, for any nonzero vector \(v\) we have \((v,v)=\lVert v\rVert^2\neq 0\), so the non-degeneracy condition is also satisfied.

When \(V=W\) as above, a pairing is sometimes called a bilinear form. The example we use in this post is the following.

Example 3 For any \(\mathbb{K}\)-vector space \(V\) and its dual space \(V^\ast\), define \((-,-):V\times V^\ast\rightarrow \mathbb{K}\) by

\[(v,f)=f(v)\]

Then \((-,-)\) is a non-degenerate pairing. First, for fixed \(v\in V\) and arbitrary \(f,g\in V^\ast\),

\[(v,f+g)=(f+g)(v)=f(v)+g(v)=(v,f)+(v,g)\]

and for fixed \(f\in V^\ast\) and arbitrary \(v_1,v_2\in V\),

\[(v_1+v_2,f)=f(v_1+v_2)=f(v_1)+f(v_2)=(v_1,f)+(v_2,f)\]

hold. Similarly, for any scalar \(\alpha\),

\[(\alpha v,f)=f(\alpha v)=\alpha f(v)=(\alpha f)(v)=(v,\alpha f)\]

holds, so \((-,-)\) is a pairing. Furthermore, \((-,-)\) is non-degenerate. If any \(f\in V^\ast\) is nonzero, then there exists \(v\) with \(f(v)\neq 0\), so \((-,-)\) is non-degenerate on the right. Also, for any nonzero vector \(v\), choose a basis \(\mathcal{B}\) containing \(v\) and construct its dual basis as above; for the element \(\xi\) of the dual basis corresponding to \(v\), we have \((v,\xi)=1\), so \((-,-)\) is also non-degenerate on the left.

Thus \((-,-)\) is a non-degenerate pairing. We call this the canonical pairing.

Proposition 4 Given two \(\mathbb{K}\)-vector spaces \(V,W\) with a non-degenerate pairing \((-,-):V\times W\rightarrow \mathbb{K}\), the function \(V\rightarrow W^\ast\) defined by

\[v\mapsto (v,-)\]

is an injective linear map. Similarly, the function \(W\rightarrow V^\ast\) defined by

\[w\mapsto (-,w)\]

is also an injective linear map.

Proof

That these functions are linear maps is obvious since \((-,-)\) is linear in each argument. That they are injective is exactly the statement that \((-,-)\) is non-degenerate on the left and on the right.

Corollary 5 Two finite-dimensional \(\mathbb{K}\)-vector spaces \(V,W\) with a non-degenerate pairing \((-,-):V\times W\rightarrow \mathbb{K}\) are isomorphic.

Proof

This is immediate from the two inequalities

\[\dim V\leq\dim W^\ast=\dim W,\qquad \dim W\leq\dim V^\ast=\dim V.\]

In this sense, a non-degenerate pairing between two finite-dimensional vector spaces is sometimes called a perfect pairing. Applying this corollary to the canonical pairing of Example 3 with \(W=V^\ast\), we obtain an isomorphism from \(V\) to \(V^{\ast\ast}\). Explicitly, this function is the evaluation map

\[\ev_v:f\mapsto f(v)\]

for any \(f\in V^\ast\). The discussion above shows that the map \(V\rightarrow V^{\ast\ast}\) defined by \(v\mapsto \ev_v\) is an isomorphism.

Dual Map

Let \(V,W\) be two \(\mathbb{K}\)-vector spaces and let \(L:V\rightarrow W\) be a linear map. Then we can define a function \(L^\ast:W^\ast\rightarrow V^\ast\) by

\[L^\ast(f)=f\circ L\]

In diagram form, this is as follows.

Equivalently, by the canonical pairing defined above, \(L^\ast\) can be characterized by

\[(Lv, f)=(v,L^\ast f)\qquad\text{for all $v\in V$ and $f\in W^\ast$}\tag{1}\]

Here the pairing \((-,-)\) on the left-hand side is the canonical pairing of \(W\), and the pairing \((-,-)\) on the right-hand side is the canonical pairing of \(V\).

Now suppose in particular that \(V\) and \(W\) are both finite-dimensional \(\mathbb{K}\)-vector spaces. Let \(\mathcal{B}=\{x_1,\ldots, x_n\}\) be a basis of \(V\) and \(\mathcal{C}=\{y_1,\ldots, y_m\}\) a basis of \(W\), with dual bases

\[\mathcal{B}^\ast=\{\xi^1,\ldots,\xi^n\},\qquad\mathcal{C}^\ast=\{\upsilon^1,\ldots,\upsilon^m\}\]

respectively. By §Fundamental Theorem of Linear Algebra, every linear map can be represented by a matrix once bases are chosen; hence we can represent \(L^\ast\) as a matrix with respect to the bases \(\mathcal{C}^\ast\) and \(\mathcal{B}^\ast\).

First, suppose \(L\) is represented by the following matrix with respect to \(\mathcal{B}\) and \(\mathcal{C}\):

\[[L]_\mathcal{C}^\mathcal{B}=\begin{pmatrix}\alpha_{11}&\alpha_{12}&\cdots&\alpha_{1n}\\\alpha_{21}&\alpha_{22}&\cdots&\alpha_{2n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m1}&\alpha_{m2}&\cdots&\alpha_{mn}\end{pmatrix}\]

That is,

\[\begin{aligned}L(x_1)&=\alpha_{11}y_1+\alpha_{21}y_2+\cdots+\alpha_{m1}y_m\\L(x_2)&=\alpha_{12}y_1+\alpha_{22}y_2+\cdots+\alpha_{m2}y_m\\&\phantom{a}\vdots\\L(x_n)&=\alpha_{1n}y_1+\alpha_{2n}y_2+\cdots+\alpha_{mn}y_m\end{aligned}\]

holds. To represent \(L^\ast\) as a matrix, we need to know the image of each element of \(\mathcal{C}^\ast\). For any \(\upsilon^k\in\mathcal{C}^\ast\), we have \(L^\ast(\upsilon^k)=\upsilon^k\circ L\), and as an element of \(V^\ast\) this function is expressed as a linear combination of elements of \(\mathcal{B}^\ast\):

\[\begin{aligned}L^\ast(\upsilon^1)&=\beta_{11}\xi^1+\beta_{21}\xi^2+\cdots+\beta_{n1}\xi^n\\L^\ast(\upsilon^2)&=\beta_{12}\xi^1+\beta_{22}\xi^2+\cdots+\beta_{n2}\xi^n\\&\phantom{a}\vdots\\L^\ast(\upsilon^m)&=\beta_{1m}\xi^1+\beta_{2m}\xi^2+\cdots+\beta_{nm}\xi^n\end{aligned}\]

To find the coefficients \(\beta_{lk}\), substitute \(x_l\) into both sides of

\[L^\ast(\upsilon^k)=\beta_{1k}\xi^1+\cdots+\beta_{lk}\xi^l+\cdots+\beta_{nk}\xi^n\]

The right-hand side yields \(\beta_{lk}\), while the left-hand side becomes

\[L^\ast(\upsilon^k)(x_l)=\upsilon^k(L(x_l))=\upsilon^k(\alpha_{1l}y_1+\cdots+\alpha_{ml}y_m)=\alpha_{kl}\]

Thus \(\beta_{lk}=\alpha_{kl}\), and the matrix representation of \(L^\ast\) is

\[[L^\ast]_{\mathcal{B}^\ast}^{\mathcal{C}^\ast}=\begin{pmatrix}\alpha_{11}&\alpha_{21}&\cdots&\alpha_{m1}\\\alpha_{12}&\alpha_{22}&\cdots&\alpha_{m2}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{1n}&\alpha_{2n}&\cdots&\alpha_{mn}\end{pmatrix}=\begin{pmatrix}\alpha_{11}&\alpha_{12}&\cdots&\alpha_{1n}\\\alpha_{21}&\alpha_{22}&\cdots&\alpha_{2n}\\\vdots&\vdots&\ddots&\vdots\\\alpha_{m1}&\alpha_{m2}&\cdots&\alpha_{mn}\end{pmatrix}^t=\bigl([L]_\mathcal{C}^\mathcal{B}\bigr)^t\]

In other words, the transpose of a matrix is precisely the matrix representing the dual map.

Orthogonal Complement

First, the following proposition holds.

Proposition 6 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces, and let \(L:V\rightarrow W\) be a linear map and \(L^\ast:W^\ast\rightarrow V^\ast\) its dual.

1. If \(L\) is injective, then \(L^\ast\) is surjective.
2. If \(L\) is surjective, then \(L^\ast\) is injective.

Proof

Both claims are immediate from §Space of Linear Maps, ⁋Corollary 2.

1. If \(L\) is injective, there exists \(R:W\rightarrow V\) satisfying \(R\circ L=\id_V\). Then for any \(f\in V^\ast\), the composition \(f\circ R\) is a function from \(W\) to \(\mathbb{K}\), i.e., an element of \(W^\ast\), and

   \[L^\ast(f\circ R)=(f\circ R)\circ L=f\circ(R\circ L)=f\circ\id_V=f\]

   so \(L^\ast\) is surjective.

2. If \(L\) is surjective, there exists \(S:W\rightarrow V\) satisfying \(L\circ S=\id_W\). Thus, if some \(f\in W^\ast\) satisfies \(L^\ast(f)=0\),

   \[L^\ast(f)=f\circ L=0\implies 0=(f\circ L)\circ S=f\circ(L\circ S)=f\circ\id_W=f\]

   so \(L^\ast\) is injective.

This proposition suggests a specific relationship between \(\ker L\) and \(\im L^\ast\), and between \(\im L\) and \(\ker L^\ast\). We begin with the following definition.

Definition 7 Let \(V\) be a \(\mathbb{K}\)-vector space with canonical pairing \((-,-)\), and consider an arbitrary subset \(S\subseteq V\). The collection of \(f\in V^\ast\) satisfying \((v,f)=0\) for all \(v\in S\) is called the orthogonal complement or annihilator of \(S\), and is denoted by \(S^\perp\).

Similarly, given an arbitrary subset \(T\subseteq V^\ast\), the collection of \(v\in V\) satisfying \((v,f)=0\) for all \(f\in T\) is called the orthogonal complement of \(T\) and is denoted by \(T^\perp\).

In particular, when \(S\) or \(T\) is a singleton we sometimes write \(v^\perp\) or \(f^\perp\).

That \(v^\perp\) is a subspace of \(V^\ast\) for any \(v\in V\) is obvious from the bilinearity of \((-,-)\). Now, from the equality

\[S^\perp=\bigcap_{v\in S}v^\perp\]

and §Basis of a Vector Space, ⁋Lemma 3, we see that \(S^\perp\) is a subspace of \(V^\ast\). Similarly, for any \(T\subseteq V^\ast\), the set \(T^\perp\) is a subspace of \(V\).

Proposition 8 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces, and let \(L:V\rightarrow W\) be a linear map and \(L^\ast:W^\ast\rightarrow V^\ast\) its dual. For any subspace \(U\subseteq V\) and its orthogonal complement \(U^\perp\),

\[L(U)^\perp=(L^\ast)^{-1}(U^\perp)\]

holds.

Proof

Using equation (1), which defines the dual map \(L^\ast\) via the canonical pairing, for any \(\upsilon\in W^\ast\) we have

\[\upsilon\in L(U)^\perp\iff (L(u),\upsilon)=0\text{ for all $u\in U$}\iff (u, L^\ast(\upsilon))=0\text{ for all $u\in U$}\iff L^\ast(\upsilon)\in U^\perp\]

as desired.

Corollary 9 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces, and let \(L:V\rightarrow W\) be a linear map and \(L^\ast:W^\ast\rightarrow V^\ast\) its dual. Then \((\im L)^\perp=\ker(L^\ast)\).

Proof

It suffices to take \(U=V\) in Proposition 8. From the non-degeneracy of the canonical pairing \((-,-)\), we have \(U^\perp=\{0\}\), yielding the desired result.

In Proposition 8, we could instead start with \(U\subseteq W^\ast\) rather than \(U\subseteq V\). In that case we obtain the following proposition.

Proposition 10 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces, and let \(L:V\rightarrow W\) be a linear map and \(L^\ast:W^\ast\rightarrow V^\ast\) its dual. For any subspace \(U\subseteq W^\ast\) and its orthogonal complement \(U^\perp\),

\[\bigl(L^\ast(U)\bigr)^\perp=L^{-1}(U^\perp)\]

holds.

Proof

This too follows from equation (1): for any \(x\in V\),

\[x\in \bigl(L^\ast(U)\bigr)^\perp\iff (x, L^\ast(\upsilon))=0\text{ for all $\upsilon\in U$}\iff (L(x),\upsilon)=0\text{ for all $\upsilon\in U$}\iff L(x)\in U^\perp\]

which gives the claim.

Corollary 11 Let \(V,W\) be two \(\mathbb{K}\)-vector spaces, and let \(L:V\rightarrow W\) be a linear map and \(L^\ast:W^\ast\rightarrow V^\ast\) its dual. Then \(\bigl(\im L^\ast\bigr)^\perp=\ker L\).

Proof

It suffices to take \(U=W^\ast\) in Proposition 10.

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References

[Lee] 이인석, 선형대수와 군, 서울대학교 출판문화원, 2005.
[Bou] Bourbaki, N. Algebra I, Elements of Mathematics. Springer-Verlag Berlin, 1998.

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