5.5 Duality for Subspaces and Quotients

A subspace $U$ of $V$ gives two new vector spaces: $U$ itself, and the quotient $V/U$ of Chapter 4. Each has a dual space, and so far $U^{*}$ and $(V/U)^{*}$ have nothing to do with $V^{*}$ . This section connects them. The answer is a swap: the dual of the quotient is a subspace of $V^{*}$ , namely the annihilator $U^{0}$ , and the dual of the subspace is a quotient of $V^{*}$ , namely $V^{*}/U^{0}$ . Both identifications come from the First Isomorphism Theorem, and neither needs a basis. At the end we use them to describe hyperplanes, the subspaces cut out by a single equation, in any dimension.

5.5.1. Restricting functionals to a subspace

Let $U$ be a subspace of $V$ . A functional $\varphi \in V^{*}$ measures every vector of $V$ , so in particular it measures the vectors of $U$ . Forgetting the rest gives the restriction $\varphi|_U \colon U \to F$ , $u \mapsto \varphi(u)$ , which is a functional on $U$ . This defines the restriction map $\rho \colon V^{*} \to U^{*}, \qquad \rho(\varphi) = \varphi|_U .$ It is not a new kind of map. Let $\iota \colon U \to V$ , $\iota(u) = u$ , be the inclusion. Its dual map (Definition 5.4.1) sends $\varphi$ to $\varphi \circ \iota$ , and $(\varphi \circ \iota)(u) = \varphi(u)$ . So $\rho = \iota'$ , and in particular $\rho$ is linear.

Two questions decide what $\rho$ does. Which functionals restrict to zero? By Definition 5.2.1, exactly those in $U^{0}$ . And does every functional on $U$ come from one on $V$ ? That is an extension problem, and it has a positive answer in every dimension.

Theorem 5.5.1 (Dual of a Subspace)

Let $V$ be a vector space over $F$ , of any dimension, and let $U$ be a subspace of $V$ . The restriction map $\rho \colon V^{*} \to U^{*}$ , $\rho(\varphi) = \varphi|_U$ , is linear and surjective, and $\operatorname{ker} \rho = U^{0}$ . Hence $V^{*}/U^{0} \to U^{*}, \qquad \varphi + U^{0} \mapsto \varphi|_U$ is a well-defined isomorphism, and $U^{*} \cong V^{*}/U^{0}$ .

Idea. The kernel is the definition of $U^{0}$ read aloud. The real content is surjectivity: given $\psi$ on $U$ , we must extend it to all of $V$ . A basis tells us how. Extend a basis of $U$ to a basis of $V$ ; a linear map may be prescribed freely on a basis, so declare the extension to agree with $\psi$ on the basis of $U$ and to be $0$ on the new basis vectors. Then the First Isomorphism Theorem turns “surjective with kernel $U^{0}$ ” into the isomorphism.

Proof. We saw above that $\rho = \iota'$ is linear. For $\varphi \in V^{*}$ , $\rho(\varphi) = 0$ means $\varphi(u) = 0$ for every $u \in U$ , which is $\varphi \in U^{0}$ by Definition 5.2.1. So $\operatorname{ker} \rho = U^{0}$ .

Surjective. Let $\psi \in U^{*}$ . By Theorem 1.8.3, $U$ has a basis $B_U$ . It is a linearly independent subset of $V$ , so by Theorem 1.8.2 there is a basis $B$ of $V$ with $B_U \subseteq B$ . By Theorem 2.1.6 there is a linear map $\varphi \colon V \to F$ with $\varphi(b) = \psi(b) \quad \text{for } b \in B_U, \qquad \varphi(b) = 0 \quad \text{for } b \in B \setminus B_U .$ Then $\varphi|_U$ and $\psi$ are linear maps $U \to F$ that agree on the basis $B_U$ of $U$ , so $\varphi|_U = \psi$ by the uniqueness part of Theorem 2.1.6. Thus $\rho(\varphi) = \psi$ , and $\rho$ is surjective. (If $V$ is finite-dimensional, the finite versions Corollary 1.5.5, Theorem 1.6.5 and Theorem 2.1.5 suffice, and no choice principle is needed.)

By the First Isomorphism Theorem (Theorem 4.4.7), $\varphi + \operatorname{ker}\rho \mapsto \rho(\varphi)$ is a well-defined isomorphism $V^{*}/\operatorname{ker} \rho \to \operatorname{im} \rho$ . Since $\operatorname{ker} \rho = U^{0}$ and $\operatorname{im} \rho = U^{*}$ , this is the claimed isomorphism.

In words: a functional on $U$ is a functional on $V$ , up to adding something that vanishes on $U$ . Two extensions of the same $\psi$ differ by an element of $U^{0}$ , and every element of $U^{0}$ can be added.

When $V$ is finite-dimensional, taking dimensions recovers the count of this chapter. By Corollary 5.1.4, $\dim U^{*} = \dim U$ and $\dim V^{*} = \dim V$ , and by Theorem 4.4.5, $\dim(V^{*}/U^{0}) = \dim V - \dim U^{0}$ . So the isomorphism says $\dim U = \dim V - \dim U^{0}$ , which is Theorem 5.2.2 again.

5.5.2. Functionals on a quotient

Now the other space. A functional $\psi$ on $V/U$ measures cosets. Composing with the quotient map $\pi \colon V \to V/U$ , $\pi(v) = v + U$ , turns it into a measurement of vectors, $v \mapsto \psi(v + U)$ . This functional on $V$ cannot tell apart two vectors in the same coset, so in particular it is zero on $U$ . The theorem says that nothing is lost and nothing else appears: the functionals on $V/U$ are exactly the functionals on $V$ that vanish on $U$ .

Theorem 5.5.2 (Dual of a Quotient)

Let $V$ be a vector space over $F$ , of any dimension, let $U$ be a subspace, and let $\pi \colon V \to V/U$ be the quotient map. Then the dual map $\pi' \colon (V/U)^{*} \to V^{*}, \qquad \pi'(\psi) = \psi \circ \pi ,$ is injective, and its image is $U^{0}$ . Hence $\pi'$ is an isomorphism from $(V/U)^{*}$ onto $U^{0}$ , and $(V/U)^{*} \cong U^{0}$ .

Idea. Injectivity uses only that $\pi$ is surjective: if $\psi \circ \pi$ is zero, then $\psi$ is zero on every coset. For the image, one inclusion is “ $\pi$ kills $U$ ”. The other inclusion asks us to build a functional on $V/U$ from a functional on $V$ that vanishes on $U$ , and that is exactly what the universal property of the quotient does.

Proof. The map $\pi'$ is linear by Definition 5.4.1.

Injective. Suppose $\pi'(\psi) = \psi \circ \pi = 0$ . Every element of $V/U$ is $\pi(v)$ for some $v \in V$ , because $\pi$ is surjective (Theorem 4.4.4 (c)). Then $\psi(\pi(v)) = 0$ for every $v$ , so $\psi = 0$ . By Theorem 2.2.3, $\pi'$ is injective.

$\operatorname{im} \pi' \subseteq U^{0}$ . For $u \in U$ , $\pi(u) = U$ is the zero vector of $V/U$ , so $(\psi \circ \pi)(u) = \psi(0) = 0$ . Hence $\psi \circ \pi \in U^{0}$ .

$U^{0} \subseteq \operatorname{im} \pi'$ . Let $\varphi \in U^{0}$ . Then $U \subseteq \operatorname{ker} \varphi$ , so by Theorem 4.4.10 (a) there is a linear map $\psi \colon V/U \to F$ with $\psi \circ \pi = \varphi$ . This $\psi$ lies in $(V/U)^{*}$ , and $\pi'(\psi) = \varphi$ .

So $\pi'$ is an injective linear map with image $U^{0}$ ; regarded as a map onto $U^{0}$ it is a linear bijection, hence an isomorphism.

Concretely, the inverse isomorphism $U^{0} \to (V/U)^{*}$ is $\varphi \mapsto \bar\varphi$ , where $\bar\varphi(v + U) = \varphi(v)$ . The formula is well defined precisely because $\varphi$ vanishes on $U$ : if $v - v' \in U$ , then $\varphi(v) - \varphi(v') = \varphi(v - v') = 0$ .

The two theorems fit together as a mirror:

	subspace $U \subseteq V$	quotient $V/U$
its dual is	a quotient of $V^{}$ : $U^{} \cong V^{*}/U^{0}$	a subspace of $V^{}$ : $(V/U)^{} \cong U^{0}$
the map	$\iota'$ : restrict	$\pi'$ : compose with $\pi$
the map is	surjective	injective

The last row is no accident: $\iota$ is injective and $\pi$ is surjective, and dualizing swaps the two properties.

Alternatively, in finite dimension both theorems follow from Theorem 5.4.3, which says $\operatorname{ker} T' = (\operatorname{im} T)^{0}$ and $\operatorname{im} T' = (\operatorname{ker} T)^{0}$ . For $T = \pi$ : $\operatorname{im} \pi = V/U$ , whose annihilator is $\{0\}$ , and $\operatorname{ker} \pi = U$ , so $\pi'$ is injective with image $U^{0}$ . For $T = \iota$ : $\operatorname{im} \iota = U$ , so $\operatorname{ker} \iota' = U^{0}$ , and $\operatorname{ker} \iota = \{0\}$ , whose annihilator is all of $U^{*}$ , so $\iota'$ is surjective. The proofs above avoid dimension counts, which is why they work in every dimension.

Quick check. Let $V$ be a vector space of dimension $6$ and $U$ a subspace of dimension $2$ . Find $\dim U^{0}$ , $\dim (V/U)^{*}$ , $\dim U^{*}$ , and $\dim(V^{*}/U^{0})$ , and check that the two theorems are consistent with these numbers.

Solution. By Theorem 5.2.2, $\dim U^{0} = 6 - 2 = 4$ . By Theorem 4.4.5 and Corollary 5.1.4, $\dim(V/U)^{*} = \dim(V/U) = 4$ , matching $(V/U)^{*} \cong U^{0}$ . Next, $\dim U^{*} = \dim U = 2$ , and $\dim(V^{*}/U^{0}) = 6 - 4 = 2$ , matching $U^{*} \cong V^{*}/U^{0}$ .

5.5.3. A line in $\mathbb{R}^3$ , in coordinates

The theorems name their isomorphisms without choosing anything. To see them act, we now choose bases and compute. Recall that a functional on $\mathbb{R}^3$ is determined by its values on $\mathbf{e}_1, \mathbf{e}_2, \mathbf{e}_3$ (Theorem 5.1.1). So every $\varphi \in (\mathbb{R}^3)^{*}$ has the form $\varphi(x, y, z) = ax + by + cz$ , with $(a, b, c) = (\varphi(\mathbf{e}_1), \varphi(\mathbf{e}_2), \varphi(\mathbf{e}_3))$ . We write such a functional simply as $ax + by + cz$ .

Example (Both isomorphisms for a line)

Let $\mathbf{u} = (1, 2, -1)$ and $U = \operatorname{span}(\mathbf{u}) \subseteq \mathbb{R}^3$ .

(a) Find a basis of $U^{0}$ .
(b) Find a basis of $\mathbb{R}^3/U$ , its dual basis $(\psi_1, \psi_2)$ of $(\mathbb{R}^3/U)^{*}$ , and the images $\pi'(\psi_1), \pi'(\psi_2)$ .
(c) Describe the restriction map $\rho \colon (\mathbb{R}^3)^{*} \to U^{*}$ and the isomorphism $(\mathbb{R}^3)^{*}/U^{0} \to U^{*}$ .

Solution.

(a) By Theorem 5.2.3 (b), $U^{0} = \{\mathbf{u}\}^{0}$ , so $ax + by + cz \in U^{0}$ exactly when $a + 2b - c = 0$ , that is, $c = a + 2b$ . Taking $(a, b) = (1, 0)$ and $(0, 1)$ gives $\varphi_1 = x + z, \qquad \varphi_2 = y + 2z .$ They are independent (look at the coefficients of $x$ and $y$ ), and $\dim U^{0} = 3 - 1 = 2$ by Theorem 5.2.2, so $(\varphi_1, \varphi_2)$ is a basis of $U^{0}$ (Theorem 1.6.4).
(b) The list $(\mathbf{u}, \mathbf{e}_1, \mathbf{e}_2)$ is independent: in $a\mathbf{u} + b\mathbf{e}_1 + c\mathbf{e}_2 = (a + b, 2a + c, -a) = \mathbf{0}$ , the third entry gives $a = 0$ , and then $b = c = 0$ . It has length 3, so it is a basis of $\mathbb{R}^3$ , and by Theorem 4.4.5 $(\mathbf{e}_1 + U, \mathbf{e}_2 + U)$ is a basis of $\mathbb{R}^3/U$ . Its dual basis consists of the $\psi_i$ with $\psi_i(\mathbf{e}_j + U) = \delta_{ij}$ (Theorem 5.1.3).

The functional $\pi'(\psi_1) = \psi_1 \circ \pi$ on $\mathbb{R}^3$ takes the values $\psi_1(\mathbf{e}_1 + U) = 1$ at $\mathbf{e}_1$ , $\psi_1(\mathbf{e}_2 + U) = 0$ at $\mathbf{e}_2$ , and $\psi_1(U) = 0$ at $\mathbf{u}$ . Writing it as $ax + by + cz$ , this says $a = 1$ , $b = 0$ , and $a + 2b - c = 0$ , so $c = 1$ . Similarly $\pi'(\psi_2)$ has $a = 0$ , $b = 1$ , $c = 2$ . Hence $\pi'(\psi_1) = x + z = \varphi_1, \qquad \pi'(\psi_2) = y + 2z = \varphi_2 .$ So $\pi'$ carries the dual basis of $(\mathbb{R}^3/U)^{*}$ onto the basis of $U^{0}$ found in (a), as Theorem 5.5.2 predicts. Read backwards: $\psi_1((x, y, z) + U) = x + z$ and $\psi_2((x, y, z) + U) = y + 2z$ . These are well defined, since adding $t\mathbf{u} = (t, 2t, -t)$ changes $x + z$ by $t - t = 0$ and $y + 2z$ by $2t - 2t = 0$ .
(c) The space $U^{*}$ is one-dimensional, with basis the functional $\xi$ determined by $\xi(\mathbf{u}) = 1$ (the dual basis of $(\mathbf{u})$ ). A functional $\varphi = ax + by + cz$ restricts to $\varphi|_U = \varphi(\mathbf{u})\,\xi = (a + 2b - c)\,\xi$ , since both sides take the value $a + 2b - c$ at $\mathbf{u}$ . So $\rho$ is “evaluate at $\mathbf{u}$ ”. Its kernel is $\{ a + 2b - c = 0 \} = U^{0}$ , and it is surjective, since $\rho(x) = \xi$ . The isomorphism of Theorem 5.5.1 is $(ax + by + cz) + U^{0} \ \mapsto\ (a + 2b - c)\,\xi .$ For instance $x + U^{0} = -z + U^{0}$ , because $x - (-z) = \varphi_1 \in U^{0}$ ; both restrict to $\xi$ .

5.5.4. Why “natural” matters here

In the example, bases entered only when we wanted numbers. The maps $\rho$ and $\pi'$ were defined before any basis was chosen: restrict a functional, or compose it with $\pi$ . That is what makes the two isomorphisms natural, in the sense of the discussion of the double dual. They are part of the structure of $V$ and $U$ , not of a coordinate system.

There is a tempting shortcut that is not natural. Since $U^{*}$ is a quotient of $V^{*}$ , one might try to find a copy of $U^{*}$ inside $V^{*}$ , as a subspace. This can be done, but only by making a choice. Pick a complement $W$ of $U$ , so $V = U \oplus W$ . A functional that vanishes on $W$ is determined by its values on $U$ , so the restriction $\rho$ maps $W^{0}$ isomorphically onto $U^{*}$ (Exercise (C2: Complements and a choice of copy of $U^{*}$ )). But different complements give different subspaces $W^{0}$ .

Warning. $U^{*}$ is not a subspace of $V^{*}$ . Its elements are functionals on $U$ , which cannot be evaluated at vectors outside $U$ . To place $U^{*}$ inside $V^{*}$ one must extend each functional, and extensions are not unique. In the example above, the complement $W_1 = \operatorname{span}(\mathbf{e}_1, \mathbf{e}_2)$ of $U$ gives $W_1^{0} = \operatorname{span}(z)$ , while $W_2 = \operatorname{span}(\mathbf{e}_1, \mathbf{e}_3)$ gives $W_2^{0} = \operatorname{span}(y)$ . These are different lines in $(\mathbb{R}^3)^{*}$ , and both restrict isomorphically onto $U^{*}$ : $z|_U = -\xi$ and $y|_U = 2\xi$ . The choice-free statement is $U^{*} \cong V^{*}/U^{0}$ , a quotient.

The same caution applies on the other side, in reverse. $(V/U)^{*}$ is naturally a subspace of $V^{*}$ , namely $U^{0}$ , but $V/U$ itself is not a subspace of $V$ . Duality turns the quotient, which has no natural home inside $V$ , into a subspace, which does.

5.5.5. Codimension and hyperplanes

The quotient measures how far a subspace is from filling the whole space. In $\mathbb{R}^3$ a plane through the origin misses “one direction” and a line misses “two directions”. For finite-dimensional $V$ the number of missing directions is $\dim V - \dim U$ . But this formula is useless in $F[x]$ , where $\dim V$ and $\dim U$ may both be infinite, although $U = \{ p : p(0) = 0 \}$ misses only one direction: every polynomial is a constant plus an element of $U$ . The quotient gives the right number in every case.

The codimension of $U$ counts the directions in $V$ that $U$ is missing.

Definition 5.5.1 (Codimension)

Let $U$ be a subspace of a vector space $V$ over $F$ . If $V/U$ is finite-dimensional, the codimension of $U$ in $V$ is $\operatorname{codim} U \mathrel{:=} \dim(V/U) .$ If $V/U$ is infinite-dimensional, we say that $U$ has infinite codimension.

In words: we do not compare the sizes of $U$ and $V$ . We collapse $U$ to zero and measure what is left. The codimension depends on the ambient space, so when that is unclear we write “codimension of $U$ in $V$ ”.

Definition 5.5.2 (Hyperplane)

A hyperplane in $V$ is a subspace $H$ of $V$ of codimension exactly $1$ , that is, with $\dim(V/H) = 1$ .

Examples.

Finite dimension. If $V$ is finite-dimensional, then $\operatorname{codim} U = \dim V - \dim U$ by Theorem 4.4.5. The hyperplanes of $F^n$ are its subspaces of dimension $n - 1$ : the lines through $\mathbf{0}$ in $F^2$ and the planes through $\mathbf{0}$ in $F^3$ .
Trace. In $M_n(F)$ with $n \ge 1$ , the trace-zero matrices are the kernel of $\operatorname{tr} \colon M_n(F) \to F$ . This map is linear and surjective, since $\operatorname{tr}(c\mathbf{E}_{11}) = c$ , so $M_n(F)/\operatorname{ker}\operatorname{tr} \cong F$ by Theorem 4.4.7, and the trace-zero matrices form a hyperplane.
Polynomials. In $F[x]$ , let $U = \{ p : p(0) = 0 \}$ . Evaluation at $0$ is a linear map $F[x] \to F$ with kernel $U$ , and it is surjective because the constant $c$ maps to $c$ . By Theorem 4.4.7, $F[x]/U \cong F$ , so $U$ is a hyperplane in the infinite-dimensional space $F[x]$ . In the same way $\{ p : p(0) = p(1) = 0 \}$ has codimension 2 (Exercise (B3: Dimension computations)).
Degenerate cases. $\operatorname{codim} V = 0$ , since $V/V$ is the zero space, and $\operatorname{codim}\{0\} = \dim V$ when $V$ is finite-dimensional. In particular $V$ itself is not a hyperplane, and the only hyperplane of a one-dimensional space is $\{0\}$ . The zero space $\{0\}$ has no hyperplanes at all.

Nonexample by minimal change. In $\mathbb{R}^3$ , the plane $\{ x + y + z = 0 \}$ is a hyperplane. Replace it by the line $\{ x + y + z = 0,\ x - y = 0 \} = \operatorname{span}((1, 1, -2))$ . It is still a subspace, but its codimension is $3 - 1 = 2$ , so the clause “codimension exactly 1” fails. Replace it instead by $\{ x + y + z = 1 \}$ : now it is not a subspace at all, so it has no codimension. It is a coset of a hyperplane, an object we return to in the next section.

Why this definition. Codimension is the dual notion of dimension, and the duality theorems make this precise. If $U$ has finite codimension, then by Theorem 5.5.2, Corollary 5.1.4, and Theorem 2.5.4, $\operatorname{codim} U = \dim(V/U) = \dim (V/U)^{*} = \dim U^{0},$ and this holds even when $V$ is infinite-dimensional. So dimension counts vectors, and codimension counts independent equations. For finite-dimensional $V$ this is Theorem 5.2.2 again: $\dim U^{0} = \dim V - \dim U$ .

Warning. In infinite dimension a hyperplane need not be “smaller” than the whole space. In $F[x]$ , the hyperplane $U = \{ p : p(0) = 0 \} = \{ xq : q \in F[x] \}$ is isomorphic to $F[x]$ itself, via $q \mapsto xq$ : this map is linear, injective (if $xq = 0$ then $q = 0$ ), and onto $U$ . So you cannot detect codimension by comparing $U$ with $V$ up to isomorphism, and the formula $\operatorname{codim} U = \dim V - \dim U$ must never be used outside finite dimension. Always compute $\dim(V/U)$ , or count equations.

Hyperplanes are exactly the subspaces cut out by one nontrivial linear equation. In $F^n$ this is familiar: $a_1x_1 + \dots + a_nx_n = 0$ with the $a_i$ not all zero defines a subspace of dimension $n - 1$ . The quotient makes the statement true in every dimension.

Theorem 5.5.3 (Hyperplanes Are Kernels of Functionals)

Let $V$ be a vector space over $F$ , of any dimension, and let $H$ be a subspace of $V$ . Then $H$ is a hyperplane if and only if $H = \operatorname{ker} \varphi$ for some nonzero $\varphi \in V^{*}$ .

Idea. A nonzero functional maps onto the one-dimensional space $F$ , so the First Isomorphism Theorem makes $V/\operatorname{ker}\varphi$ one-dimensional. Conversely, if $V/H$ is one-dimensional, it “is” $F$ after choosing an isomorphism, and composing that isomorphism with $\pi$ gives a functional whose kernel is $H$ .

Proof. $(\Leftarrow)$ Suppose $\varphi \in V^{*}$ is nonzero. Then $\operatorname{im} \varphi$ is a subspace of $F$ (Theorem 2.2.2) containing a nonzero scalar, so $\dim \operatorname{im}\varphi = 1 = \dim F$ , and $\operatorname{im} \varphi = F$ by Theorem 1.6.8. By Theorem 4.4.7, $V/\operatorname{ker}\varphi \cong F$ , so $\dim(V/\operatorname{ker}\varphi) = 1$ , and $\operatorname{ker}\varphi$ is a hyperplane.

$(\Rightarrow)$ Suppose $\dim(V/H) = 1$ . By Theorem 2.5.4 there is an isomorphism $\theta \colon V/H \to F$ . Let $\varphi = \theta \circ \pi$ , where $\pi \colon V \to V/H$ is the quotient map. It is linear as a composition of linear maps (Theorem 2.4.2), so $\varphi \in V^{*}$ . For $v \in V$ , $\varphi(v) = 0$ means $\theta(v + H) = 0$ , which holds exactly when $v + H$ is the zero coset, since $\theta$ is injective; and $v + H = H$ means $v \in H$ by Lemma 4.4.3 (b). So $\operatorname{ker}\varphi = H$ . Finally $\varphi \ne 0$ : since $V/H \ne \{0\}$ , we have $H \ne V$ , so some $v \notin H = \operatorname{ker}\varphi$ .

The functional is not unique, but almost: two functionals with the same kernel are scalar multiples of each other (Exercise (C1: Functionals with the same kernel); Exercise (B2: Functionals with the same kernel) gives a second proof). So hyperplanes of $V$ correspond to nonzero functionals up to a nonzero scalar, just as the plane $x + y + z = 0$ is also $2x + 2y + 2z = 0$ .

A hyperplane together with any one vector outside it fills the space.

Remark. Let $H = \operatorname{ker}\varphi$ be a hyperplane and $v_0 \notin H$ . Then $V = H \oplus \operatorname{span}(v_0)$ . Indeed $\varphi(v_0) \ne 0$ , and every $v \in V$ splits as $v = \Big( v - \tfrac{\varphi(v)}{\varphi(v_0)}\,v_0 \Big) + \tfrac{\varphi(v)}{\varphi(v_0)}\,v_0 ,$ where the first bracket lies in $\operatorname{ker}\varphi$ because $\varphi$ of it is $\varphi(v) - \varphi(v) = 0$ . And $H \cap \operatorname{span}(v_0) = \{0\}$ , since $\varphi(cv_0) = c\varphi(v_0) = 0$ forces $c = 0$ .

Quick check. Decide which of the following are hyperplanes, and for each hyperplane give a nonzero functional with that kernel.

(a) The symmetric matrices in $M_2(F)$ .
(b) The symmetric matrices in $M_3(F)$ .
(c) $\{ p \in F[x] : p(1) = 0 \}$ in $F[x]$ .

Solution.

(a) Yes. $\mathbf{A}$ is symmetric exactly when $a_{12} - a_{21} = 0$ , so the symmetric matrices are the kernel of $\mathbf{A} \mapsto a_{12} - a_{21}$ , which is linear and nonzero (it takes the value $1$ at $\mathbf{E}_{12}$ ). By Theorem 5.5.3 this is a hyperplane.
(b) No. The symmetric $3 \times 3$ matrices form a subspace of dimension $6$ (choose the entries on and above the diagonal freely), so the codimension is $9 - 6 = 3 \ne 1$ .
(c) Yes. It is the kernel of $p \mapsto p(1)$ , which is linear (Theorem 0.8.5) and nonzero (it sends the constant $1$ to $1$ ).

5.5.6. Exercises

A. Check your understanding

Exercise (A1)

(a) State the theorem identifying $(V/U)^{*}$ with a subspace of $V^{*}$ , including the map.
(b) Define the codimension of a subspace, and define a hyperplane.
(c) True or false: $U^{*}$ is a subspace of $V^{*}$ . Justify your answer.
(d) Let $U$ be a subspace of dimension $2$ in $F^5$ . Find $\dim (F^5/U)^{*}$ and $\operatorname{codim} U$ .
(e) Is $\{ (x_1, x_2, x_3, x_4) \in \mathbb{R}^4 : x_1 = x_2 = 0 \}$ a hyperplane in $\mathbb{R}^4$ ? Justify your answer.
(f) True or false: every nonzero linear functional $\varphi \colon V \to F$ is surjective. Justify your answer.

Solution.

(a) For a subspace $U$ of $V$ with quotient map $\pi$ , the dual map $\pi' \colon (V/U)^{*} \to V^{*}$ , $\psi \mapsto \psi \circ \pi$ , is injective with image $U^{0}$ ; so $(V/U)^{*} \cong U^{0}$ (Theorem 5.5.2).
(b) If $V/U$ is finite-dimensional, $\operatorname{codim} U = \dim(V/U)$ ; otherwise $U$ has infinite codimension. A hyperplane is a subspace of codimension exactly $1$ .
(c) False. Elements of $U^{*}$ are defined only on $U$ , so they are not elements of $V^{*}$ . What is true is $U^{*} \cong V^{*}/U^{0}$ , a quotient (Theorem 5.5.1).
(d) $\dim(F^5/U)^{*} = \dim(F^5/U) = 5 - 2 = 3$ , by Corollary 5.1.4 and Theorem 4.4.5. So $\operatorname{codim} U = 3$ .
(e) No. The subspace is $\operatorname{span}(\mathbf{e}_3, \mathbf{e}_4)$ , of dimension 2, so its codimension is $4 - 2 = 2$ .
(f) True, in any dimension. This is Proposition 5.1.2: $\operatorname{im}\varphi$ is a subspace of the one-dimensional space $F$ containing a nonzero element, so it is $F$ .

B. Practice

Exercise (B1: The dual of a quotient of

\mathbb{R}^4

)

Let $U = \operatorname{span}((1, -1, 0, 2), (0, 1, 1, -1)) \subseteq \mathbb{R}^4$ , and write functionals on $\mathbb{R}^4$ as $a_1x_1 + a_2x_2 + a_3x_3 + a_4x_4$ .

(a) Find a basis $(\varphi_1, \varphi_2)$ of $U^{0}$ .
(b) Let $\psi_1, \psi_2 \in (\mathbb{R}^4/U)^{*}$ be the functionals with $\pi'(\psi_i) = \varphi_i$ . Compute $\psi_1$ and $\psi_2$ at the coset $(2, 0, 1, 3) + U$ .
(c) Find a basis of $\mathbb{R}^4/U$ whose dual basis is $(\psi_1, \psi_2)$ .

Solution.

(a) A functional lies in $U^{0}$ exactly when it vanishes at both spanning vectors (Theorem 5.2.3 (b)), that is, $a_1 - a_2 + 2a_4 = 0, \qquad a_2 + a_3 - a_4 = 0 .$ The second gives $a_3 = a_4 - a_2$ , and the first gives $a_1 = a_2 - 2a_4$ . Taking $(a_2, a_4) = (1, 0)$ and $(-1, -1)$ gives $(1, 1, -1, 0)$ and $(1, -1, 0, -1)$ , that is, $\varphi_1 = x_1 + x_2 - x_3, \qquad \varphi_2 = x_1 - x_2 - x_4 .$ Check: $\varphi_1(1, -1, 0, 2) = 0$ , $\varphi_1(0, 1, 1, -1) = 0$ , $\varphi_2(1, -1, 0, 2) = 1 + 1 - 2 = 0$ , $\varphi_2(0, 1, 1, -1) = -1 + 1 = 0$ . They are independent (compare the coefficients of $x_3$ and $x_4$ ). The spanning vectors of $U$ are independent (compare first entries, then second), so $\dim U^{0} = 4 - 2 = 2$ by Theorem 5.2.2, and $(\varphi_1, \varphi_2)$ is a basis by Theorem 1.6.4.
(b) By Theorem 5.5.2, $\psi_i(\mathbf{v} + U) = \varphi_i(\mathbf{v})$ . So $\psi_1((2, 0, 1, 3) + U) = 2 + 0 - 1 = 1$ and $\psi_2((2, 0, 1, 3) + U) = 2 - 0 - 3 = -1$ .
(c) We need cosets $c_1, c_2$ with $\psi_i(c_j) = \delta_{ij}$ . The vector $-\mathbf{e}_3$ has $\varphi_1(-\mathbf{e}_3) = 1$ and $\varphi_2(-\mathbf{e}_3) = 0$ , and $-\mathbf{e}_4$ has $\varphi_1(-\mathbf{e}_4) = 0$ , $\varphi_2(-\mathbf{e}_4) = 1$ . Put $c_1 = -\mathbf{e}_3 + U$ and $c_2 = -\mathbf{e}_4 + U$ . If $ac_1 + bc_2 = 0$ , applying $\psi_1$ and $\psi_2$ gives $a = 0$ and $b = 0$ ; so $(c_1, c_2)$ is independent, and since $\dim(\mathbb{R}^4/U) = 2$ it is a basis (Theorem 1.6.4). By construction $\psi_i(c_j) = \delta_{ij}$ , so $(\psi_1, \psi_2)$ is its dual basis (Theorem 5.1.3).

Exercise (B2: Functionals with the same kernel)

Let $V$ be a vector space over $F$ , of any dimension, and let $\varphi, \psi \in V^{*}$ be nonzero. Prove that $\operatorname{ker}\varphi = \operatorname{ker}\psi$ if and only if $\psi = c\varphi$ for some nonzero $c \in F$ . (Exercise (C1: Functionals with the same kernel) proved $(\Rightarrow)$ by a different route; here use the hyperplane decomposition.)

Hint: the remark after Theorem 5.5.3.

Solution. $(\Leftarrow)$ If $\psi = c\varphi$ with $c \ne 0$ , then $\psi(v) = 0$ if and only if $c\varphi(v) = 0$ , if and only if $\varphi(v) = 0$ , since $c$ is invertible. So $\operatorname{ker}\psi = \operatorname{ker}\varphi$ .

$(\Rightarrow)$ Suppose $\operatorname{ker}\varphi = \operatorname{ker}\psi = H$ . Since $\varphi \ne 0$ , choose $v_0$ with $\varphi(v_0) \ne 0$ ; then $v_0 \notin H$ , so also $\psi(v_0) \ne 0$ . Put $c = \psi(v_0)/\varphi(v_0)$ , which is nonzero, and $\chi = \psi - c\varphi \in V^{*}$ . Then $\chi$ vanishes on $H$ , because both $\psi$ and $\varphi$ do, and $\chi(v_0) = \psi(v_0) - c\varphi(v_0) = 0$ . By Theorem 5.5.3 $H$ is a hyperplane, and by the remark after it $V = H \oplus \operatorname{span}(v_0)$ . A linear map vanishing on $H$ and on $v_0$ vanishes on every $h + av_0$ , hence on $V$ . So $\chi = 0$ , that is, $\psi = c\varphi$ .

Exercise (B3: Dimension computations)

(a) Let $\dim V = 7$ and $\dim U = 3$ . Find $\dim U^{*}$ , $\dim(V^{*}/U^{0})$ , $\dim(V/U)^{*}$ , and $\operatorname{codim} U$ .
(b) In $V = F[x]_{\le 3}$ , let $U = \{ p : p(0) = p(1) = 0 \}$ . Using the map $T \colon V \to F^2$ , $T(p) = (p(0), p(1))$ , find $\operatorname{codim} U$ and $\dim U$ , and write $U$ as an intersection of two hyperplanes.
(c) Hence find a basis of $U^{0}$ consisting of evaluation functionals.

Solution.

(a) $\dim U^{*} = 3$ by Corollary 5.1.4. $\dim U^{0} = 7 - 3 = 4$ by Theorem 5.2.2, so $\dim(V^{*}/U^{0}) = 7 - 4 = 3$ (Theorem 4.4.5), consistent with Theorem 5.5.1. $\dim(V/U)^{*} = \dim(V/U) = 4$ , consistent with Theorem 5.5.2, and $\operatorname{codim} U = 4$ .
(b) $T$ is linear because evaluation is (Theorem 0.8.5), and $\operatorname{ker} T = U$ . It is surjective: $T(1) = (1, 1)$ and $T(x) = (0, 1)$ span $F^2$ . By Theorem 4.4.7, $V/U \cong F^2$ , so $\operatorname{codim} U = 2$ , and $\dim U = 4 - 2 = 2$ by Theorem 4.4.5. Writing $\varepsilon_0(p) = p(0)$ and $\varepsilon_1(p) = p(1)$ , we have $U = \operatorname{ker}\varepsilon_0 \cap \operatorname{ker}\varepsilon_1$ , and both kernels are hyperplanes by Theorem 5.5.3, since both functionals send $1$ to $1$ .
(c) Both $\varepsilon_0$ and $\varepsilon_1$ lie in $U^{0}$ . They are independent: if $a\varepsilon_0 + b\varepsilon_1 = 0$ , evaluating at $x$ gives $b = 0$ , and then evaluating at $1$ gives $a = 0$ . Since $\dim U^{0} = 4 - 2 = 2$ , $(\varepsilon_0, \varepsilon_1)$ is a basis of $U^{0}$ (Theorem 1.6.4).

C. Going deeper

Exercise (C1: Subspaces of finite codimension are intersections of hyperplanes)

Let $V$ be a vector space over $F$ , of any dimension, and let $U$ be a subspace of finite codimension $k \ge 1$ .

(a) Prove that there are nonzero $\varphi_1, \dots, \varphi_k \in U^{0}$ with $U = \operatorname{ker}\varphi_1 \cap \dots \cap \operatorname{ker}\varphi_k$ . Deduce that $U$ is an intersection of $k$ hyperplanes.
(b) Prove that $U$ is not an intersection of fewer than $k$ hyperplanes.

Hint: for (a), take a dual basis of $(V/U)^{*}$ ; for (b), consider the map $V \to F^m$ , $v \mapsto (\varphi_1(v), \dots, \varphi_m(v))$ .

Solution.

(a) Let $\pi \colon V \to V/U$ be the quotient map. Since $\dim(V/U) = k$ , choose a basis $(c_1, \dots, c_k)$ of $V/U$ and let $(\psi_1, \dots, \psi_k)$ be its dual basis (Theorem 5.1.3). Put $\varphi_i = \psi_i \circ \pi$ . By Theorem 5.5.2, $\varphi_i \in U^{0}$ , and $\varphi_i \ne 0$ because $\pi'$ is injective and $\psi_i \ne 0$ .
Since each $\varphi_i$ lies in $U^{0}$ , $U \subseteq \operatorname{ker}\varphi_i$ for each $i$ . Conversely, let $v \in \operatorname{ker}\varphi_1 \cap \dots \cap \operatorname{ker}\varphi_k$ , and write $v + U = \sum_j a_jc_j$ . Applying $\psi_i$ gives $a_i = \psi_i(v + U) = \varphi_i(v) = 0$ for every $i$ . So $v + U$ is the zero coset, and $v \in U$ by Lemma 4.4.3 (b). This proves $U = \bigcap_i \operatorname{ker}\varphi_i$ . Each $\operatorname{ker}\varphi_i$ is a hyperplane by Theorem 5.5.3.
(b) Suppose $U = H_1 \cap \dots \cap H_m$ with hyperplanes $H_i$ . By Theorem 5.5.3, $H_i = \operatorname{ker}\varphi_i$ for nonzero $\varphi_i \in V^{*}$ . The map $S \colon V \to F^m$ , $S(v) = (\varphi_1(v), \dots, \varphi_m(v))$ , is linear, since each entry is, and $\operatorname{ker} S = \bigcap_i \operatorname{ker}\varphi_i = U$ . By Theorem 4.4.7, $V/U \cong \operatorname{im} S$ , a subspace of $F^m$ . Hence $k = \dim(V/U) = \dim \operatorname{im} S \le m$ by Theorem 1.6.7. So at least $k$ hyperplanes are needed.

Exercise (C2: Complements and a choice of copy of

U^{*}

)

Let $V$ be finite-dimensional, and let $U, W$ be subspaces with $V = U \oplus W$ .

(a) Prove that $V^{*} = U^{0} \oplus W^{0}$ without using Theorem 5.2.3 (d). (This is Exercise (C3: Dual of a direct sum decomposition) (a), which used it.)
(b) Deduce that the restriction map $\rho \colon V^{*} \to U^{*}$ restricts to an isomorphism $W^{0} \to U^{*}$ .
(c) In $\mathbb{R}^2$ , let $U = \operatorname{span}(\mathbf{e}_1)$ , $W_1 = \operatorname{span}(\mathbf{e}_2)$ , and $W_2 = \operatorname{span}((1, 1))$ . Find $W_1^{0}$ and $W_2^{0}$ , and find the element of each that restricts to the functional $\xi \in U^{*}$ with $\xi(\mathbf{e}_1) = 1$ . Conclude that the copy of $U^{*}$ inside $V^{*}$ depends on the complement.

Solution.

(a) Let $n = \dim V$ . If $\varphi \in U^{0} \cap W^{0}$ , then $\varphi$ vanishes on $U$ and on $W$ , hence on every $u + w$ , that is, on $U + W = V$ ; so $\varphi = 0$ . By Theorem 5.2.2, $\dim U^{0} + \dim W^{0} = (n - \dim U) + (n - \dim W) = 2n - n = n$ , using $\dim U + \dim W = n$ for a direct sum (Theorem 1.7.4). By Theorem 1.7.3, $\dim(U^{0} + W^{0}) = n - 0 = n = \dim V^{*}$ , so $U^{0} + W^{0} = V^{*}$ (Theorem 1.6.8), and the sum is direct because the intersection is $\{0\}$ .
(b) The restriction of $\rho$ to $W^{0}$ is linear. Its kernel is $W^{0} \cap \operatorname{ker}\rho = W^{0} \cap U^{0} = \{0\}$ by Theorem 5.5.1 and (a), so it is injective. Its domain has dimension $n - \dim W = \dim U = \dim U^{*}$ . An injective linear map between spaces of the same finite dimension is an isomorphism (Corollary 2.3.2 (e)).
(c) Write functionals on $\mathbb{R}^2$ as $ax + by$ . $W_1^{0}$ consists of those with $b = 0$ , so $W_1^{0} = \operatorname{span}(x)$ . $W_2^{0}$ consists of those with $a + b = 0$ , so $W_2^{0} = \operatorname{span}(x - y)$ . Restricting to $U$ : $x|_U$ and $(x - y)|_U$ both take the value $1$ at $\mathbf{e}_1$ , so both equal $\xi$ . The elements $x$ and $x - y$ are different, so the subspaces $W_1^{0} \ne W_2^{0}$ of $V^{*}$ are two different copies of $U^{*}$ , one for each complement.

5.5.1. Restricting functionals to a subspace

5.5.2. Functionals on a quotient

5.5.3. A line in R3, in coordinates

5.5.4. Why “natural” matters here

5.5.5. Codimension and hyperplanes

5.5.6. Exercises

A. Check your understanding

B. Practice

C. Going deeper

5.5.3. A line in $\mathbb{R}^3$ , in coordinates