Chapter 13A Quotient Spaces and Dual Spaces¶
Prerequisites: Vector spaces and subspaces (Chapter 4) · Linear maps (Chapter 5) · Dimension formula (Chapter 5)
Chapter arc: Equivalence classes and quotient spaces → Dimension formula and isomorphism theorems → Linear functionals and dual spaces → Dual basis → Annihilators → Transpose maps → Double dual and canonical isomorphism → Applications in finite dimensions
Further connections：Quotient spaces are widely used in algebraic topology (construction of homology groups) and functional analysis (quotient Banach spaces); dual spaces underlie distribution theory (Schwartz distributions = dual of test function spaces) and weak topologies; in algebraic geometry, duality manifests as Serre duality
Quotient spaces and dual spaces are two deep and elegant constructions in linear algebra. The quotient space simplifies the structure of a vector space by "collapsing a subspace to a point," embodying the algebraic ideas of equivalence relations and congruence. The dual space examines a vector space from the perspective of "measurement," organizing linear functionals themselves into a new vector space. Their interplay — annihilators, transpose maps, the double dual — reveals the deep symmetry underlying linear algebra.
13A.1 Quotient Space Construction¶
Core question: Given a subspace \(W\), can we treat vectors in \(V\) that are "equivalent modulo \(W\)" as a single element? → Cosets → The quotient space \(V/W\)
Cosets and equivalence relations¶
Definition 13A.1 (Coset)
Let \(V\) be a vector space over \(\mathbb{F}\) and \(W\) a subspace of \(V\). For \(v \in V\), the set
is called the coset (or affine subset) of \(v\) with respect to \(W\). The element \(v\) is called a representative of the coset.
Theorem 13A.1 (Equivalent characterizations of cosets)
Let \(u, v \in V\) and \(W\) be a subspace of \(V\). The following are equivalent:
- \(u + W = v + W\);
- \(u - v \in W\);
- \(u \in v + W\).
Proof
\((1) \Rightarrow (3)\): \(u = u + 0 \in u + W = v + W\).
\((3) \Rightarrow (2)\): \(u \in v + W\) means \(u = v + w\) for some \(w \in W\), so \(u - v = w \in W\).
\((2) \Rightarrow (1)\): Let \(u - v = w_0 \in W\). For any \(u + w \in u + W\), we have \(u + w = v + w_0 + w \in v + W\) (since \(w_0 + w \in W\)). Similarly \(v + W \subseteq u + W\). \(\blacksquare\)
Note
The essence of cosets is an equivalence relation: define \(u \sim v\) if and only if \(u - v \in W\). Then \(\sim\) is an equivalence relation on \(V\), and \(v + W\) is precisely the equivalence class of \(v\).
Definition 13A.2 (Quotient space)
Let \(W\) be a subspace of \(V\). The set of all cosets
equipped with the operations:
- Addition: \((u + W) + (v + W) = (u + v) + W\);
- Scalar multiplication: \(\lambda(v + W) = \lambda v + W\), \(\lambda \in \mathbb{F}\),
forms a vector space over \(\mathbb{F}\), called the quotient space of \(V\) by \(W\). The zero element of \(V/W\) is \(0 + W = W\).
Theorem 13A.2 (Well-definedness of quotient space operations)
The addition and scalar multiplication on \(V/W\) are independent of the choice of representatives.
Proof
Suppose \(u + W = u' + W\) and \(v + W = v' + W\). Then \(u - u' \in W\) and \(v - v' \in W\).
Addition: \((u + v) - (u' + v') = (u - u') + (v - v') \in W\), so \((u + v) + W = (u' + v') + W\).
Scalar multiplication: \(\lambda u - \lambda u' = \lambda(u - u') \in W\), so \(\lambda u + W = \lambda u' + W\). \(\blacksquare\)
Example 13A.1
Let \(V = \mathbb{R}^3\) and \(W = \{(x, y, 0) : x, y \in \mathbb{R}\}\) (the \(xy\)-plane).
The coset \((a, b, c) + W = \{(x, y, c) : x, y \in \mathbb{R}\}\), i.e., the horizontal plane at height \(c\). Two cosets are equal if and only if their third coordinates agree. Thus \(V/W \cong \mathbb{R}\), with the isomorphism \((a, b, c) + W \mapsto c\).
Example 13A.2
Let \(V = \mathbb{R}^2\) and \(W = \{(t, t) : t \in \mathbb{R}\}\) (the line through the origin with slope \(1\)).
The coset \((a, b) + W = \{(a + t, b + t) : t \in \mathbb{R}\}\), a family of lines with slope \(1\). Two cosets coincide if and only if \(b - a\) is the same. Hence \(V/W \cong \mathbb{R}\).
The quotient map¶
Definition 13A.3 (Quotient map)
The quotient map \(\pi: V \to V/W\) is defined by
It is a surjective linear map with \(\ker \pi = W\).
Theorem 13A.3 (Properties of the quotient map)
The quotient map \(\pi: V \to V/W\) satisfies:
- \(\pi\) is a linear map;
- \(\pi\) is surjective;
- \(\ker \pi = W\).
Proof
-
\(\pi(u + v) = (u + v) + W = (u + W) + (v + W) = \pi(u) + \pi(v)\); \(\pi(\lambda v) = \lambda v + W = \lambda(v + W) = \lambda \pi(v)\).
-
For any \(v + W \in V/W\), \(\pi(v) = v + W\).
-
\(\pi(v) = 0_{V/W} = W\) if and only if \(v + W = W\) if and only if \(v \in W\). \(\blacksquare\)
13A.2 Dimension and Isomorphism Theorems¶
Core question: What is the dimension of \(V/W\)? → Dimension formula → The first isomorphism theorem — one of the most important structural results in linear algebra
Theorem 13A.4 (Dimension of the quotient space)
Let \(V\) be a finite-dimensional vector space and \(W\) a subspace of \(V\). Then
Proof
The quotient map \(\pi: V \to V/W\) is surjective with \(\ker \pi = W\). By the rank-nullity theorem:
Therefore \(\dim(V/W) = \dim V - \dim W\). \(\blacksquare\)
Note
\(\dim(V/W)\) is also called the codimension of \(W\) in \(V\), denoted \(\operatorname{codim} W\).
Theorem 13A.5 (First isomorphism theorem)
Let \(T: V \to U\) be a linear map. Then there exists a unique isomorphism
such that \(\widetilde{T}(v + \ker T) = T(v)\). That is, \(V/\ker T \cong \operatorname{im} T\).
Proof
Well-definedness: If \(v + \ker T = v' + \ker T\), then \(v - v' \in \ker T\), so \(T(v) = T(v')\).
Linearity: \(\widetilde{T}((u + \ker T) + (v + \ker T)) = \widetilde{T}((u+v) + \ker T) = T(u+v) = T(u) + T(v) = \widetilde{T}(u + \ker T) + \widetilde{T}(v + \ker T)\). Scalar multiplication is similar.
Injectivity: \(\widetilde{T}(v + \ker T) = 0\) means \(T(v) = 0\), i.e., \(v \in \ker T\), so \(v + \ker T = 0 + \ker T\).
Surjectivity: For any \(T(v) \in \operatorname{im} T\), \(\widetilde{T}(v + \ker T) = T(v)\).
Uniqueness: Determined uniquely by \(\widetilde{T} \circ \pi = T\). \(\blacksquare\)
Example 13A.3
Let \(T: \mathbb{R}^3 \to \mathbb{R}^2\) be defined by \(T(x, y, z) = (x + y, y + z)\).
\(\ker T = \{(x, y, z) : x + y = 0,\; y + z = 0\} = \{(t, -t, t) : t \in \mathbb{R}\}\), of dimension \(1\).
\(\operatorname{im} T = \mathbb{R}^2\) (one can verify \(T\) is surjective).
The first isomorphism theorem gives \(\mathbb{R}^3/\ker T \cong \mathbb{R}^2\), and \(\dim(\mathbb{R}^3/\ker T) = 3 - 1 = 2 = \dim \mathbb{R}^2\).
Theorem 13A.6 (Second isomorphism theorem)
Let \(U, W\) be subspaces of \(V\). Then
In particular, \(\dim(U + W) - \dim W = \dim U - \dim(U \cap W)\).
Proof
Define \(\varphi: U \to (U + W)/W\) by \(\varphi(u) = u + W\). This is a linear map.
Surjectivity: For \((u + w) + W \in (U+W)/W\), we have \((u + w) + W = u + W = \varphi(u)\).
Kernel: \(\varphi(u) = W\) if and only if \(u \in W\), i.e., \(u \in U \cap W\). So \(\ker \varphi = U \cap W\).
By the first isomorphism theorem, \(U/(U \cap W) \cong (U + W)/W\). \(\blacksquare\)
Theorem 13A.7 (Third isomorphism theorem)
Let \(W \subseteq U\) be subspaces of \(V\). Then
Proof
Define \(\varphi: V/W \to V/U\) by \(\varphi(v + W) = v + U\).
Well-definedness: If \(v + W = v' + W\), then \(v - v' \in W \subseteq U\), so \(v + U = v' + U\).
Linearity and surjectivity are immediate.
Kernel: \(\varphi(v + W) = U\) if and only if \(v \in U\), i.e., \(v + W \in U/W\). So \(\ker \varphi = U/W\).
By the first isomorphism theorem, \((V/W)/(U/W) \cong V/U\). \(\blacksquare\)
13A.3 Linear Functionals and the Dual Space¶
Core question: What is special about linear maps \(V \to \mathbb{F}\)? → Linear functionals → All linear functionals form the dual space \(V^*\)
Definition 13A.4 (Linear functional)
Let \(V\) be a vector space over \(\mathbb{F}\). A linear map \(\varphi: V \to \mathbb{F}\) is called a linear functional on \(V\).
Example 13A.4
The following are all linear functionals:
- \(\varphi: \mathbb{R}^n \to \mathbb{R}\), \(\varphi(x_1, \ldots, x_n) = a_1 x_1 + \cdots + a_n x_n\) (a linear combination of coordinates).
- \(\varphi: C[0,1] \to \mathbb{R}\), \(\varphi(f) = \int_0^1 f(t)\, dt\) (definite integration).
- \(\varphi: \mathbb{F}[x]_{\leq n} \to \mathbb{F}\), \(\varphi(p) = p(c)\) (evaluation at the point \(c\)).
Definition 13A.5 (Dual space)
The set of all linear functionals on \(V\), equipped with pointwise addition and scalar multiplication, forms a vector space
called the dual space (or algebraic dual) of \(V\).
Theorem 13A.8 (Dimension of the dual space)
If \(V\) is finite-dimensional, then
Proof
\(V^* = \mathcal{L}(V, \mathbb{F})\). By the dimension formula for spaces of linear maps, \(\dim \mathcal{L}(V, \mathbb{F}) = \dim V \cdot \dim \mathbb{F} = \dim V \cdot 1 = \dim V\). \(\blacksquare\)
Theorem 13A.9 (Kernel of a linear functional)
Let \(\varphi \in V^*\) be a nonzero linear functional. Then \(\ker \varphi\) is a subspace of \(V\) of codimension \(1\) (a hyperplane):
Conversely, every codimension-\(1\) subspace of \(V\) is the kernel of some nonzero linear functional.
Proof
\(\varphi \neq 0\) implies \(\operatorname{im} \varphi = \mathbb{F}\) (since \(\operatorname{im} \varphi\) is a subspace of \(\mathbb{F}\), and a nonzero subspace of \(\mathbb{F}\) must be \(\mathbb{F}\) itself). By the rank-nullity theorem, \(\dim \ker \varphi = \dim V - \dim \operatorname{im} \varphi = \dim V - 1\).
Conversely, let \(\dim W = \dim V - 1\). Choose \(v_0 \in V \setminus W\), so that \(V = W \oplus \operatorname{span}(v_0)\). Define \(\varphi(w + cv_0) = c\). Then \(\varphi\) is a nonzero linear functional with \(\ker \varphi = W\). \(\blacksquare\)
Example 13A.5
In \(\mathbb{R}^3\), the linear functional \(\varphi(x, y, z) = 2x - y + 3z\) has kernel
a plane through the origin (a \(2\)-dimensional subspace, codimension \(1\)).
13A.4 The Dual Basis¶
Core question: How does a basis of \(V\) determine a basis of \(V^*\)? → Definition and construction of the dual basis → Dual basis vectors as coordinate extraction maps
Definition 13A.6 (Dual basis)
Let \(\mathcal{B} = \{e_1, e_2, \ldots, e_n\}\) be a basis for the finite-dimensional vector space \(V\). The dual basis \(\mathcal{B}^* = \{e_1^*, e_2^*, \ldots, e_n^*\}\) of \(V^*\) is defined by the condition
Theorem 13A.10 (Existence and uniqueness of the dual basis)
The dual basis \(\mathcal{B}^*\) exists, is unique, and forms a basis for \(V^*\).
Proof
Existence and uniqueness: For each \(i\), \(e_i^*\) is uniquely determined by its values on the basis \(\{e_1, \ldots, e_n\}\). Defining \(e_i^*(e_j) = \delta_{ij}\) and extending linearly gives a well-defined linear functional.
It is a basis: Suppose \(\sum c_i e_i^* = 0\) (the zero functional). Evaluating at \(e_j\): \(\sum c_i e_i^*(e_j) = c_j = 0\). So \(e_1^*, \ldots, e_n^*\) are linearly independent. Since \(\dim V^* = n\), they form a basis. \(\blacksquare\)
Theorem 13A.11 (Coordinate interpretation of the dual basis)
Let \(v = \sum_{i=1}^n a_i e_i \in V\). Then
That is, \(e_i^*\) extracts the \(i\)-th coordinate of \(v\) with respect to the basis \(\mathcal{B}\).
Furthermore, any \(\varphi \in V^*\) can be written as \(\varphi = \sum_{i=1}^n \varphi(e_i)\, e_i^*\).
Proof
\(e_i^*(v) = e_i^*\!\left(\sum_j a_j e_j\right) = \sum_j a_j\, e_i^*(e_j) = \sum_j a_j\, \delta_{ij} = a_i\).
For the second claim, let \(\psi = \sum_i \varphi(e_i)\, e_i^*\). On basis vectors: \(\psi(e_j) = \sum_i \varphi(e_i)\, \delta_{ij} = \varphi(e_j)\). Hence \(\psi = \varphi\). \(\blacksquare\)
Example 13A.6
\(V = \mathbb{R}^3\) with the standard basis \(\{e_1, e_2, e_3\}\). The dual basis is
These are the three coordinate functions.
Example 13A.7
\(V = \mathbb{R}^2\) with basis \(\{v_1, v_2\} = \{(1, 1), (1, -1)\}\). Find the dual basis.
Let \(v_1^*(x, y) = ax + by\). From \(v_1^*(1, 1) = 1\) and \(v_1^*(1, -1) = 0\): \(a + b = 1\), \(a - b = 0\), giving \(a = b = \frac{1}{2}\).
Let \(v_2^*(x, y) = cx + dy\). From \(v_2^*(1, 1) = 0\) and \(v_2^*(1, -1) = 1\): \(c + d = 0\), \(c - d = 1\), giving \(c = \frac{1}{2}\), \(d = -\frac{1}{2}\).
Therefore \(v_1^*(x, y) = \frac{x + y}{2}\) and \(v_2^*(x, y) = \frac{x - y}{2}\).
13A.5 Annihilators¶
Core question: Which functionals in \(V^*\) vanish on a given subspace \(U\)? → The annihilator \(U^0\) → Dimension formula for annihilators → Annihilators establish a duality between subspaces of \(V\) and subspaces of \(V^*\)
Definition 13A.7 (Annihilator)
Let \(U\) be a subspace of \(V\). The annihilator of \(U\) is
\(U^0\) is a subspace of \(V^*\).
Theorem 13A.12 (Dimension of the annihilator)
Let \(V\) be finite-dimensional and \(U\) a subspace of \(V\). Then
Proof
Choose a basis \(\{e_1, \ldots, e_k\}\) for \(U\) and extend it to a basis \(\{e_1, \ldots, e_k, e_{k+1}, \ldots, e_n\}\) for \(V\). Let \(\{e_1^*, \ldots, e_n^*\}\) be the dual basis.
Claim: \(U^0 = \operatorname{span}\{e_{k+1}^*, \ldots, e_n^*\}\).
On one hand, for \(i > k\) and \(j \leq k\): \(e_i^*(e_j) = 0\), so \(e_i^* \in U^0\).
On the other hand, let \(\varphi = \sum_{i=1}^n c_i e_i^* \in U^0\). For \(j \leq k\): \(\varphi(e_j) = c_j = 0\). So \(\varphi = \sum_{i=k+1}^n c_i e_i^*\).
Hence \(U^0 = \operatorname{span}\{e_{k+1}^*, \ldots, e_n^*\}\) and \(\dim U^0 = n - k = \dim V - \dim U\). \(\blacksquare\)
Theorem 13A.13 (Duality properties of annihilators)
Let \(V\) be finite-dimensional and \(U, W\) subspaces of \(V\). Then:
- \(U \subseteq W \Leftrightarrow W^0 \subseteq U^0\).
- \((U + W)^0 = U^0 \cap W^0\).
- \((U \cap W)^0 = U^0 + W^0\).
- \((U^0)^0 = U\) (under the canonical isomorphism \(V \cong V^{**}\)).
Proof
-
If \(U \subseteq W\) and \(\varphi \in W^0\), then \(\varphi\) vanishes on \(W\), hence on \(U\), so \(\varphi \in U^0\). The converse follows by dimension counting.
-
\(\varphi \in (U + W)^0\) iff \(\varphi(u + w) = 0\) for all \(u \in U, w \in W\), iff \(\varphi(u) = 0\) for all \(u \in U\) and \(\varphi(w) = 0\) for all \(w \in W\), iff \(\varphi \in U^0 \cap W^0\).
-
By (2), \((U^0 \cap W^0)^0 = ((U+W)^0)^0 = U + W\) (using (4)). Taking annihilators again and using (4), or by a direct dimension count:
By (2), \(\dim(U^0 \cap W^0) = \dim(U + W)^0 = n - \dim(U+W)\). Substituting and simplifying gives \(\dim(U^0 + W^0) = n - \dim(U \cap W) = \dim(U \cap W)^0\). The inclusion relation then yields equality.
- See the discussion of the double dual in Section 13A.7. \(\blacksquare\)
Example 13A.8
\(V = \mathbb{R}^3\), \(U = \operatorname{span}\{(1, 0, 1), (0, 1, 1)\}\). Find \(U^0\).
Let \(\varphi(x, y, z) = ax + by + cz \in U^0\). From \(\varphi(1, 0, 1) = a + c = 0\) and \(\varphi(0, 1, 1) = b + c = 0\): \(a = -c\), \(b = -c\). Taking \(c = 1\): \(\varphi(x, y, z) = -x - y + z\).
Check: \(\dim U^0 = 1 = 3 - 2 = \dim V - \dim U\).
Example 13A.9
Quotient spaces and annihilators: the dual space \((V/U)^*\) is naturally isomorphic to \(U^0\).
Define \(\Phi: U^0 \to (V/U)^*\) by \(\Phi(\varphi)(v + U) = \varphi(v)\). This is well-defined (\(\varphi \in U^0\) ensures independence of the representative) and is an isomorphism. This follows from the equality of dimensions (\(\dim U^0 = \dim V - \dim U = \dim(V/U) = \dim(V/U)^*\)) together with injectivity.
13A.6 The Transpose (Dual) Map¶
Core question: How does a linear map \(T: V \to W\) induce a map from \(W^*\) to \(V^*\)? → The transpose map \(T^*\) → Transpose maps reverse direction → Relationship to matrix transpose
Definition 13A.8 (Transpose map)
Let \(T: V \to W\) be a linear map. The transpose (or dual map) \(T^*: W^* \to V^*\) is defined by
That is, for \(v \in V\), \((T^*\psi)(v) = \psi(T(v))\).
Note
Notice the reversal of direction: \(T: V \to W\) induces \(T^*: W^* \to V^*\) (a contravariant functor).
Theorem 13A.14 (Properties of the transpose map)
Let \(S, T: V \to W\) and \(R: W \to U\) be linear maps, and \(\lambda \in \mathbb{F}\). Then:
- \(T^*\) is a linear map.
- \((S + T)^* = S^* + T^*\).
- \((\lambda T)^* = \lambda T^*\).
- \((R \circ T)^* = T^* \circ R^*\) (note the reversal of order).
- \((\operatorname{id}_V)^* = \operatorname{id}_{V^*}\).
Proof
-
\(T^*(\psi_1 + \psi_2) = (\psi_1 + \psi_2) \circ T = \psi_1 \circ T + \psi_2 \circ T = T^*\psi_1 + T^*\psi_2\). Scalar multiplication is similar.
-
For \(\varphi \in U^*\): \((R \circ T)^*(\varphi) = \varphi \circ (R \circ T) = (\varphi \circ R) \circ T = T^*(R^*(\varphi)) = (T^* \circ R^*)(\varphi)\).
The remaining parts are similar. \(\blacksquare\)
Theorem 13A.15 (Kernel and image of the transpose)
Let \(T: V \to W\) be a linear map between finite-dimensional spaces. Then:
- \(\ker T^* = (\operatorname{im} T)^0\).
- \(\operatorname{im} T^* = (\ker T)^0\).
- \(\operatorname{rank} T^* = \operatorname{rank} T\).
Proof
-
\(\psi \in \ker T^*\) iff \(T^*\psi = 0\), i.e., \(\psi \circ T = 0\), i.e., \(\psi(T(v)) = 0\) for all \(v \in V\), i.e., \(\psi\) vanishes on \(\operatorname{im} T\), i.e., \(\psi \in (\operatorname{im} T)^0\).
-
First, \(\operatorname{im} T^* \subseteq (\ker T)^0\): if \(\varphi = T^*\psi\), then for \(v \in \ker T\), \(\varphi(v) = \psi(T(v)) = \psi(0) = 0\), so \(\varphi \in (\ker T)^0\).
Dimension count: \(\dim \operatorname{im} T^* = \dim W^* - \dim \ker T^* = \dim W - \dim(\operatorname{im} T)^0 = \dim W - (\dim W - \dim \operatorname{im} T) = \operatorname{rank} T\). And \(\dim(\ker T)^0 = \dim V - \dim \ker T = \operatorname{rank} T\). Since the dimensions are equal and we have the inclusion, equality holds.
- From (2), \(\operatorname{rank} T^* = \dim \operatorname{im} T^* = \operatorname{rank} T\). \(\blacksquare\)
Theorem 13A.16 (Transpose map and matrix transpose)
Let \(\mathcal{B}\) be a basis for \(V\), \(\mathcal{C}\) a basis for \(W\), and \(\mathcal{B}^*, \mathcal{C}^*\) the corresponding dual bases. If the matrix of \(T: V \to W\) with respect to \(\mathcal{B}, \mathcal{C}\) is \(A\), then the matrix of \(T^*: W^* \to V^*\) with respect to \(\mathcal{C}^*, \mathcal{B}^*\) is \(A^T\) (the transpose of \(A\)).
Proof
Let \(A = (a_{ij})\), so that \(T(e_j) = \sum_i a_{ij} f_i\). Then
On the other hand, if \(T^* f_i^* = \sum_j b_{ji}\, e_j^*\), then \((T^* f_i^*)(e_j) = b_{ji}\).
Therefore \(b_{ji} = a_{ij}\), so the matrix of \(T^*\) is \(A^T\). \(\blacksquare\)
Example 13A.10
Let \(T: \mathbb{R}^2 \to \mathbb{R}^3\), \(T(x, y) = (x + y, 2x, y)\). With respect to the standard bases, the matrix of \(T\) is
The matrix of \(T^*: (\mathbb{R}^3)^* \to (\mathbb{R}^2)^*\) with respect to the dual standard bases is
Verification: \(T^*(\varphi)(x, y) = \varphi(x + y, 2x, y)\). If \(\varphi(a, b, c) = \alpha a + \beta b + \gamma c\), then \(T^*\varphi(x, y) = \alpha(x+y) + 2\beta x + \gamma y = (\alpha + 2\beta)x + (\alpha + \gamma)y\), consistent with the matrix multiplication by \(A^T\).
13A.7 The Double Dual and Canonical Isomorphism¶
Core question: What is the relationship between \(V^{**} = (V^*)^*\) and \(V\)? → Canonical isomorphism → "Canonical" means independent of the choice of basis
Definition 13A.9 (Evaluation map)
For \(v \in V\), define \(\hat{v}: V^* \to \mathbb{F}\) by
Then \(\hat{v}\) is a linear functional on \(V^*\), i.e., \(\hat{v} \in V^{**}\).
Theorem 13A.17 (Canonical isomorphism)
The map \(\iota: V \to V^{**}\) defined by \(\iota(v) = \hat{v}\) is an injective linear map. When \(V\) is finite-dimensional, \(\iota\) is an isomorphism, called the canonical isomorphism.
Proof
Linearity: \(\widehat{u + v}(\varphi) = \varphi(u + v) = \varphi(u) + \varphi(v) = \hat{u}(\varphi) + \hat{v}(\varphi)\), so \(\widehat{u+v} = \hat{u} + \hat{v}\). Scalar multiplication is similar.
Injectivity: Suppose \(\iota(v) = 0\), i.e., \(\hat{v} = 0\). Then \(\varphi(v) = 0\) for all \(\varphi \in V^*\). If \(v \neq 0\), extend \(v\) to a basis \(\{v, e_2, \ldots, e_n\}\) of \(V\) and let \(v^*\) be the first dual basis vector. Then \(v^*(v) = 1 \neq 0\), a contradiction. Hence \(v = 0\).
Isomorphism in finite dimensions: \(\dim V^{**} = \dim V^* = \dim V\). An injective linear map between spaces of equal dimension is automatically an isomorphism. \(\blacksquare\)
Note
The word "canonical" has a precise mathematical meaning (natural transformation in category theory): the definition of \(\iota\) does not depend on a choice of basis and is compatible with linear maps. Concretely, if \(T: V \to W\), then \(T^{**} \circ \iota_V = \iota_W \circ T\) (a commutative diagram).
By contrast, although \(V\) and \(V^*\) are isomorphic in finite dimensions (having equal dimension), there is no canonical isomorphism between them — any isomorphism requires a choice of basis or the introduction of an inner product.
Theorem 13A.18 (Dual of the dual basis)
Let \(\{e_1, \ldots, e_n\}\) be a basis for \(V\) and \(\{e_1^*, \ldots, e_n^*\}\) the dual basis. Under the canonical isomorphism \(\iota: V \to V^{**}\),
where \(\{e_1^{**}, \ldots, e_n^{**}\}\) is the dual basis of \(\{e_1^*, \ldots, e_n^*\}\). That is, the basis of \(V\) maps under the canonical isomorphism to the dual of the dual basis.
Proof
We need to verify \(\hat{e}_i(e_j^*) = \delta_{ij}\). By definition, \(\hat{e}_i(e_j^*) = e_j^*(e_i) = \delta_{ji} = \delta_{ij}\). Therefore \(\hat{e}_i\) is precisely the \(i\)-th element of the dual basis of \(\{e_1^*, \ldots, e_n^*\}\). \(\blacksquare\)
Example 13A.11
Application of the canonical isomorphism: proving annihilator property (4) — \((U^0)^0 = U\).
Under the canonical isomorphism \(\iota: V \to V^{**}\), \((U^0)^0 = \{\Phi \in V^{**} : \Phi(\varphi) = 0 \text{ for all } \varphi \in U^0\}\).
If \(v \in U\), then for any \(\varphi \in U^0\), \(\hat{v}(\varphi) = \varphi(v) = 0\), so \(\iota(U) \subseteq (U^0)^0\).
Dimension count: \(\dim(U^0)^0 = \dim V^* - \dim U^0 = \dim V - (\dim V - \dim U) = \dim U = \dim \iota(U)\).
Hence \(\iota(U) = (U^0)^0\), i.e., under the canonical isomorphism, \((U^0)^0\) corresponds to \(U\).
13A.8 Applications in Finite Dimensions¶
Core question: How does dual space theory apply to concrete linear algebra problems? → Dual description of linear systems → Dual perspective on change of basis → Dual characterization of rank
Dual description of linear systems¶
Theorem 13A.19 (Solution space and annihilators)
The solution space \(S\) of the homogeneous system \(A\mathbf{x} = \mathbf{0}\) (\(A\) an \(m \times n\) matrix) satisfies
where \(\varphi_i \in (\mathbb{F}^n)^*\) is the linear functional defined by the \(i\)-th row of \(A\). Setting \(W = \operatorname{span}\{\varphi_1, \ldots, \varphi_m\} \subseteq (\mathbb{F}^n)^*\), we have \(S = W^0\) (viewing \(W\) as a subspace of \((\mathbb{F}^n)^*\) and \(W^0\) as the corresponding subspace of \(\mathbb{F}^n\) under the canonical isomorphism).
Therefore \(\dim S = n - \dim W = n - \operatorname{rank} A\).
Proof
\(v \in S\) iff \(Av = 0\), i.e., the inner product of each row of \(A\) with \(v\) is zero, i.e., \(\varphi_i(v) = 0\) for all \(i\). Vanishing on all \(\varphi_i\) is equivalent to vanishing on \(\operatorname{span}\{\varphi_i\}\).
Through the canonical isomorphism \(\iota: \mathbb{F}^n \to (\mathbb{F}^n)^{**}\), \(S\) is precisely the annihilator of \(W\) in \((\mathbb{F}^n)^{**}\) pulled back to \(\mathbb{F}^n\) via \(\iota\), i.e., \(S = W^0\) (in the generalized sense). Hence \(\dim S = n - \dim W = n - \operatorname{rank} A\). \(\blacksquare\)
Dual perspective on change of basis¶
Theorem 13A.20 (Change-of-basis formula for dual bases)
Let \(\mathcal{B} = \{e_1, \ldots, e_n\}\) and \(\mathcal{B}' = \{e_1', \ldots, e_n'\}\) be two bases for \(V\) with transition matrix \(P\) (i.e., \(e_j' = \sum_i p_{ij} e_i\)). Then the transition matrix for the dual bases is \((P^{-1})^T\):
Proof
Let \(e_j'^* = \sum_i q_{ij}\, e_i^*\). From \(e_j'^*(e_k') = \delta_{jk}\):
Hence \(Q^T P = I\), i.e., \(Q^T = P^{-1}\), so \(Q = (P^{-1})^T\). \(\blacksquare\)
Dual characterization of rank¶
Theorem 13A.21 (Dual proof that row rank equals column rank)
The row rank of an \(m \times n\) matrix \(A\) equals its column rank.
Proof
View \(A\) as a linear map \(T: \mathbb{F}^n \to \mathbb{F}^m\). The column rank is \(\dim \operatorname{im} T = \operatorname{rank} T\).
The row rank is \(\dim \operatorname{span}\{\text{rows of } A\}\). The rows of \(A\) define linear functionals \(\varphi_1, \ldots, \varphi_m\) in \((\mathbb{F}^n)^*\). The image of \(A^T: (\mathbb{F}^m)^* \to (\mathbb{F}^n)^*\) is precisely \(\operatorname{span}\{\varphi_1, \ldots, \varphi_m\}\) (since \(T^*(f_i^*) = f_i^* \circ T = \varphi_i\), where \(\{f_1^*, \ldots, f_m^*\}\) is the standard basis of \((\mathbb{F}^m)^*\)). Hence the row rank equals \(\operatorname{rank} T^*\).
By Theorem 13A.15 (3), \(\operatorname{rank} T^* = \operatorname{rank} T\). \(\blacksquare\)
Example 13A.12
Dual theory in Lagrange interpolation.
Let \(V = \mathbb{F}[x]_{\leq n}\) (the space of polynomials of degree \(\leq n\)), and choose \(n + 1\) distinct points \(c_0, c_1, \ldots, c_n \in \mathbb{F}\). Define the evaluation functionals \(\varepsilon_i: V \to \mathbb{F}\) by \(\varepsilon_i(p) = p(c_i)\).
\(\{\varepsilon_0, \varepsilon_1, \ldots, \varepsilon_n\}\) is a basis for \(V^*\) (since if \(\sum a_i \varepsilon_i = 0\), then the polynomial \(\sum a_i \ell_i(x) = 0\), where \(\ell_i\) are the Lagrange basis polynomials, so \(a_i = 0\)).
The dual basis of this set (under \(V^{**} \cong V\)) is precisely the Lagrange basis polynomials \(\ell_0, \ell_1, \ldots, \ell_n\), since \(\varepsilon_i(\ell_j) = \ell_j(c_i) = \delta_{ij}\).
This is the dual-space interpretation of the polynomial interpolation from Chapter 0.
Example 13A.13
Let \(T: V \to V\) be a linear operator (\(V\) finite-dimensional) with \(T^2 = T\) (\(T\) is idempotent). Prove \((T^*)^2 = T^*\).
By Theorem 13A.14 (4): \((T^*)^2 = T^* \circ T^* = (T \circ T)^* = (T^2)^* = T^*\).
Furthermore, \(V = \ker T \oplus \operatorname{im} T\), and in the dual space \(V^* = \ker T^* \oplus \operatorname{im} T^*\). By Theorem 13A.15:
- \(\ker T^* = (\operatorname{im} T)^0\);
- \(\operatorname{im} T^* = (\ker T)^0\).
This demonstrates the perfect symmetry of the idempotent decomposition in the dual space.