Dual Maps and Covectors, linear algebra II lesson

A bakery turns an order of loaves and cakes into amounts of flour and sugar, and a price list turns amounts of flour and sugar into a cost. You can price an order without ever listing its ingredients, because the two steps combine into a single price list on orders that gives the same total. The map that moves a measurement from the outputs of a linear map back to its inputs is the dual map. This lesson shows why the dual map reverses the order of composition, why its matrix is the transpose, and why the coordinates of a measurement change by the inverse of the matrix that changes vector coordinates. Throughout, $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ , and a tuple such as $(3, 2)$ stands for a column vector.

Key terms

Covector

Definition Let $V$ be a vector space over $\mathbb{F}$ . A covector on $V$ is an element of the dual space $V^{\vee} = \operatorname{Hom}(V, \mathbb{F})$ , that is, a linear functional $\ell \colon V \to \mathbb{F}$ . The word covector is another name for a linear functional, used when its coordinates are compared with those of vectors. If $V$ is finite-dimensional with ordered basis $\mathcal{B} = (b_1, \dots, b_n)$ and dual basis $\mathcal{B}^{\vee} = (\beta_1, \dots, \beta_n)$ , where $\beta_i(b_j) = \delta_{ij}$ , then every covector satisfies

\begin{gathered} \ell = \sum_{j=1}^{n} \ell(b_j)\, \beta_j, \\[0.4em] \ell(v) = r_{\mathcal{B}}\, [v]_{\mathcal{B}} \quad \text{for all } v \in V, \end{gathered}

where $r_{\mathcal{B}} = \begin{bmatrix} \ell(b_1) & \cdots & \ell(b_n) \end{bmatrix}$ is the evaluation row of $\ell$ in the basis $\mathcal{B}$ . The coefficient column of $\ell$ in the dual basis is the transpose, $[\ell]_{\mathcal{B}^{\vee}} = r_{\mathcal{B}}^{\mathsf{T}}$ . Over $\mathbb{C}$ the row multiplies the coordinate column with no conjugation.

In plain words A covector is a linear measurement, a rule that assigns a scalar to every vector. In a basis it is recorded by what it reads on each basis vector, and those readings form a row. The row is a record of a measurement, not an arrow in $V$ . Turning a covector into a vector needs extra structure, such as an inner product.

Example On $\mathcal{P}_2$ with basis $(1, t, t^2)$ , integration over $[0, 1]$ ,

\ell(p) = \int_0^1 p(t)\, dt,

is a covector, because the integral of a sum is the sum of the integrals and constants factor out. It reads $1$ , $\tfrac{1}{2}$ and $\tfrac{1}{3}$ on the basis, so its evaluation row is $\begin{bmatrix} 1 & \tfrac{1}{2} & \tfrac{1}{3} \end{bmatrix}$ and $\ell = \beta_1 + \tfrac{1}{2}\beta_2 + \tfrac{1}{3}\beta_3$ . The polynomial $p = 4 + 6t - 3t^2$ has coordinate column $(4, 6, -3)$ , and the row gives $4 + 3 - 1 = 6$ . Integrating directly gives $\left[4t + 3t^2 - t^3\right]_0^1 = 6$ as well.

Pullback of a functional

Definition Let $V$ and $W$ be vector spaces over $\mathbb{F}$ , let $T \in \operatorname{Hom}(V, W)$ and let $\ell \in W^{\vee}$ . The pullback of $\ell$ through $T$ is the composite $\ell \circ T \colon V \to \mathbb{F}$ , given by $(\ell \circ T)(v) = \ell(Tv)$ . It is linear, since for all $x, y \in V$ and scalars $a, b$

\ell(T(ax + by)) = a\,\ell(Tx) + b\,\ell(Ty),

so $\ell \circ T$ belongs to $V^{\vee}$ . In coordinates, let $V = \mathbb{F}^n$ and $W = \mathbb{F}^m$ , let $Tx = Ax$ with $A \in \mathbb{F}^{m \times n}$ , and let $\ell(w) = rw$ for a row $r \in \mathbb{F}^{1 \times m}$ . Then $(\ell \circ T)(x) = r(Ax) = (rA)x$ , so the pullback has the row $rA$ .

In plain words To measure an output, you first send the input through the map and then read the measurement. The pullback combines both steps into one measurement on the inputs, with exactly the same reading on every input.

Example A loaf uses $3$ units of flour and $1$ of sugar, a cake uses $2$ of flour and $4$ of sugar, and the prices are $2$ per unit of flour and $5$ per unit of sugar. The order map and the price row are

A = \begin{bmatrix} 3 & 2 \\ 1 & 4 \end{bmatrix}, \qquad r = \begin{bmatrix} 2 & 5 \end{bmatrix}.

The pullback row is $rA = \begin{bmatrix} 11 & 24 \end{bmatrix}$ , so a loaf costs $11$ and a cake $24$ . For an order of $2$ loaves and $1$ cake the ingredients are $A(2, 1) = (8, 6)$ , which cost $2 \cdot 8 + 5 \cdot 6 = 46$ . The pullback gives $11 \cdot 2 + 24 \cdot 1 = 46$ without computing the ingredients.

Dual map

Definition Let $V$ and $W$ be vector spaces over $\mathbb{F}$ and $T \in \operatorname{Hom}(V, W)$ . The dual map of $T$ is

T^{\vee} \colon W^{\vee} \to V^{\vee}, \qquad T^{\vee}(\ell) = \ell \circ T.

It is linear. For $\ell_1, \ell_2 \in W^{\vee}$ and scalars $a, b$ , evaluating at any $v \in V$ gives $(a\ell_1 + b\ell_2)(Tv) = a\,\ell_1(Tv) + b\,\ell_2(Tv)$ , so $T^{\vee}(a\ell_1 + b\ell_2) = a\,T^{\vee}(\ell_1) + b\,T^{\vee}(\ell_2)$ . Over $\mathbb{C}$ the dual map is complex-linear, with no conjugation anywhere. It uses no inner product, so it is a different object from the adjoint $T^{*}$ , which depends on one.

In plain words Vectors travel forward through $T$ , from $V$ to $W$ . Measurements travel backward through $T^{\vee}$ , from measurements on $W$ to measurements on $V$ . The dual map handles every output measurement at once and returns its pullback.

Example Let $D \colon \mathcal{P}_2 \to \mathcal{P}_1$ be differentiation, $D(p) = p'$ , and let $\operatorname{ev}_1 \in \mathcal{P}_1^{\vee}$ be evaluation at $1$ , so $\operatorname{ev}_1(q) = q(1)$ . Then $D^{\vee}(\operatorname{ev}_1)$ is the functional $p \mapsto p'(1)$ on $\mathcal{P}_2$ , which reads $b + 2c$ on $p = a + bt + ct^2$ . Evaluation at $0$ , $\operatorname{ev}_0(q) = q(0)$ , pulls back to $p \mapsto p'(0)$ , which reads $b$ . By linearity $D^{\vee}(\operatorname{ev}_1 - \operatorname{ev}_0)$ reads $p'(1) - p'(0) = 2c$ .

Dual map of a composition

Definition Let $V$ , $W$ and $Z$ be vector spaces over $\mathbb{F}$ , with $T \in \operatorname{Hom}(V, W)$ and $S \in \operatorname{Hom}(W, Z)$ . Then

(ST)^{\vee} = T^{\vee} S^{\vee}

as maps from $Z^{\vee}$ to $V^{\vee}$ . To prove it, take $\ell \in Z^{\vee}$ and $v \in V$ . Then

\begin{aligned} \big((ST)^{\vee}\ell\big)(v) &= \ell(S(Tv)) \\[0.4em] &= (S^{\vee}\ell)(Tv) \\[0.4em] &= \big(T^{\vee}(S^{\vee}\ell)\big)(v), \end{aligned}

and two functionals that agree at every vector are equal. The dual of the identity $I_V$ is the identity of $V^{\vee}$ , so if $T$ is an isomorphism, then $T^{\vee}$ is one too, with $(T^{\vee})^{-1} = (T^{-1})^{\vee}$ . In matrices, if $T$ and $S$ act by $A$ and $B$ , an output row $r$ on $Z$ pulls back to $(rB)A = rBA$ .

In plain words A measurement taken after $S$ meets $S$ first on its way back. So it is pulled back through $S$ first and through $T$ second, the reverse of the order in which vectors travel.

Example Let $T \colon \mathbb{R}^3 \to \mathbb{R}^2$ and $S \colon \mathbb{R}^2 \to \mathbb{R}^2$ act by

A = \begin{bmatrix} 1 & 0 & 1 \\ 0 & 2 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 1 & 1 \\ 1 & -1 \end{bmatrix},

and take the output row $r = \begin{bmatrix} 1 & 2 \end{bmatrix}$ . Pulling back through $S$ gives $rB = \begin{bmatrix} 3 & -1 \end{bmatrix}$ , and then through $T$ gives $(rB)A = \begin{bmatrix} 3 & -2 & 2 \end{bmatrix}$ . Check at $v = (1, 1, 1)$ . Forward, $Av = (2, 3)$ and $B(2, 3) = (5, -1)$ , which $r$ reads as $5 - 2 = 3$ . The input row reads $3 - 2 + 2 = 3$ . The other order fails outright, because $rA$ is a $1 \times 3$ row and cannot be multiplied by the $2 \times 2$ matrix $B$ .

Matrix of the dual map

Definition Let $V$ and $W$ be finite-dimensional over $\mathbb{F}$ with ordered bases $\mathcal{B} = (b_1, \dots, b_n)$ and $\mathcal{C} = (c_1, \dots, c_m)$ , and dual bases $\mathcal{B}^{\vee} = (\beta_1, \dots, \beta_n)$ and $\mathcal{C}^{\vee} = (\gamma_1, \dots, \gamma_m)$ . If $T \in \operatorname{Hom}(V, W)$ has matrix $A = [T]_{\mathcal{C} \leftarrow \mathcal{B}}$ , so that $Tb_j = \sum_{i} a_{ij} c_i$ , then

[T^{\vee}]_{\mathcal{B}^{\vee} \leftarrow \mathcal{C}^{\vee}} = A^{\mathsf{T}}.

The reason is one evaluation, $(T^{\vee}\gamma_i)(b_j) = \gamma_i(Tb_j) = a_{ij}$ . So $T^{\vee}\gamma_i = \sum_{j} a_{ij}\beta_j$ , and its coefficient column is row $i$ of $A$ written as a column, which is column $i$ of $A^{\mathsf{T}}$ . In evaluation rows the same fact says that if $\ell$ has row $r$ in $\mathcal{C}$ , then $T^{\vee}\ell$ has row $rA$ in $\mathcal{B}$ . This holds over $\mathbb{C}$ with the plain transpose. The conjugate transpose $A^{*}$ belongs to the adjoint, which comes later, and represents it when both bases are orthonormal.

In plain words Each output coordinate measurement $\gamma_i$ pulls back to the measurement that reads row $i$ of $A$ . Writing those rows as columns produces the transpose. The transpose is what pulling back looks like in coordinates.

Example For $D \colon \mathcal{P}_2 \to \mathcal{P}_1$ with bases $(1, t, t^2)$ and $(1, t)$ , the images $D(1) = 0$ , $D(t) = 1$ and $D(t^2) = 2t$ give

A = \begin{bmatrix} 0 & 1 & 0 \\ 0 & 0 & 2 \end{bmatrix}, \qquad A^{\mathsf{T}} = \begin{bmatrix} 0 & 0 \\ 1 & 0 \\ 0 & 2 \end{bmatrix}.

Evaluation at $1$ on $\mathcal{P}_1$ is $\operatorname{ev}_1 = \gamma_1 + \gamma_2$ , with coefficient column $(1, 1)$ , and $A^{\mathsf{T}}(1, 1) = (0, 1, 2)$ . So $D^{\vee}(\operatorname{ev}_1) = \beta_2 + 2\beta_3$ , which reads $b + 2c$ on $a + bt + ct^2$ , as found directly in the dual map example. For a complex case, let $T(z) = Az$ on $\mathbb{C}^2$ , where $A$ has first row $\begin{bmatrix} i & 1 \end{bmatrix}$ and second row $\begin{bmatrix} 0 & 1 \end{bmatrix}$ , and let $\ell(w) = w_1$ . The pullback reads $\ell(Az) = iz_1 + z_2$ , with coefficient column $(i, 1) = A^{\mathsf{T}}(1, 0)$ . The conjugate transpose would give $A^{*}(1, 0) = (-i, 1)$ , which reads $-i$ at $z = (1, 0)$ , while the true reading there is $i$ .

Covector coordinate change

Definition Let $V$ be a vector space over $\mathbb{F}$ of finite dimension $n$ with ordered bases $\mathcal{B}$ and $\mathcal{C}$ , and let $P = P_{\mathcal{C} \leftarrow \mathcal{B}}$ , so that $[v]_{\mathcal{C}} = P[v]_{\mathcal{B}}$ for every $v \in V$ . If $\ell \in V^{\vee}$ has evaluation rows $r_{\mathcal{B}}$ and $r_{\mathcal{C}}$ , then

\begin{gathered} r_{\mathcal{C}} = r_{\mathcal{B}}\, P^{-1}, \\[0.4em] [\ell]_{\mathcal{C}^{\vee}} = (P^{-1})^{\mathsf{T}}\, [\ell]_{\mathcal{B}^{\vee}}. \end{gathered}

To derive the row law, use that the reading $\ell(v)$ does not depend on the basis. For every $v$ ,

r_{\mathcal{C}}[v]_{\mathcal{C}} = r_{\mathcal{B}}[v]_{\mathcal{B}} = r_{\mathcal{B}} P^{-1} [v]_{\mathcal{C}}.

As $v$ runs over $V$ , the column $[v]_{\mathcal{C}}$ runs over all of $\mathbb{F}^n$ , and a row is determined by its products with all columns, so $r_{\mathcal{C}} = r_{\mathcal{B}}P^{-1}$ . Transposing gives the law for coefficient columns in the dual bases.

In plain words The vector and the covector stay fixed, and only their coordinates change. Vector coordinates change by $P$ , so the evaluation row of a covector must be multiplied on the right by $P^{-1}$ for the reading to stay the same. If the coefficient column $c$ of a covector and the coordinate column $x$ of a vector were both multiplied by $P$ , the reading $c^{\mathsf{T}}x$ would become $c^{\mathsf{T}}P^{\mathsf{T}}Px$ , which matches for every pair only when $P^{\mathsf{T}}P = I$ .

Example In $\mathbb{R}^2$ let $\mathcal{B}$ be the standard basis and $\mathcal{C} = ((1, 0), (1, 1))$ . Writing $(x, y) = x'(1, 0) + y'(1, 1)$ gives $y' = y$ and $x' = x - y$ , so

P = \begin{bmatrix} 1 & -1 \\ 0 & 1 \end{bmatrix}, \qquad P^{-1} = \begin{bmatrix} 1 & 1 \\ 0 & 1 \end{bmatrix}.

For $\ell(x, y) = 2x - y$ , the row $r_{\mathcal{B}} = \begin{bmatrix} 2 & -1 \end{bmatrix}$ becomes $r_{\mathcal{C}} = r_{\mathcal{B}}P^{-1} = \begin{bmatrix} 2 & 1 \end{bmatrix}$ , which matches the direct readings $\ell(1, 0) = 2$ and $\ell(1, 1) = 1$ . The vector $v = (3, 2)$ has $[v]_{\mathcal{C}} = P(3, 2) = (1, 2)$ , and the new row reads $2 \cdot 1 + 1 \cdot 2 = 4 = \ell(3, 2)$ . The wrong law $r_{\mathcal{B}}P = \begin{bmatrix} 2 & -3 \end{bmatrix}$ would read $2 - 6 = -4$ .

Check yourself

Let $T \colon \mathbb{R}^2 \to \mathbb{R}^3$ act by the matrix below, and let $\ell(u) = u_1 - u_2 + 2u_3$ . Find $T^{\vee}(\ell)$ as a formula in $(x, y)$ and check it at $v = (2, 1)$ .

A = \begin{bmatrix} 1 & 1 \\ 2 & 0 \\ -1 & 1 \end{bmatrix}

Show the answer

The row of $\ell$ is $r = \begin{bmatrix} 1 & -1 & 2 \end{bmatrix}$ , and

\begin{aligned} rA &= \begin{bmatrix} 1 - 2 - 2 & 1 + 0 + 2 \end{bmatrix} \\[0.4em] &= \begin{bmatrix} -3 & 3 \end{bmatrix}, \end{aligned}

so $T^{\vee}(\ell)(x, y) = -3x + 3y$ . At $v = (2, 1)$ the forward route gives $Av = (3, 4, -1)$ and $\ell(Av) = 3 - 4 - 2 = -3$ . The pullback gives $-6 + 3 = -3$ , the same reading.

Maps $T$ and $S$ on $\mathbb{R}^2$ act by the matrices $A$ and $B$ below, and vectors pass through $T$ first and then $S$ . An output sensor on the final space has row $r = \begin{bmatrix} 1 & 3 \end{bmatrix}$ . A student computes the input row as $rAB$ . Find the correct row and show with $v = (1, 1)$ that the student’s row gives the wrong reading.

A = \begin{bmatrix} 1 & 2 \\ 0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix}

Show the answer

The sensor is pulled back through $S$ first, so the correct row is $(rB)A$ . Here $rB = \begin{bmatrix} 3 & 1 \end{bmatrix}$ and $(rB)A = \begin{bmatrix} 3 & 7 \end{bmatrix}$ . Forward, $Av = (3, 1)$ and $B(3, 1) = (1, 3)$ , which the sensor reads as $1 + 9 = 10$ . The correct row gives $3 + 7 = 10$ . The student’s row is $(rA)B$ , with $rA = \begin{bmatrix} 1 & 5 \end{bmatrix}$ and $(rA)B = \begin{bmatrix} 5 & 1 \end{bmatrix}$ , which reads $5 + 1 = 6$ . The identity $(ST)^{\vee} = T^{\vee}S^{\vee}$ is exactly the statement that $rBA$ is right.

In $\mathbb{R}^2$ let $\mathcal{B}$ be the standard basis and $\mathcal{C} = ((2, 1), (1, 1))$ , and let $\ell(x, y) = x + 3y$ . Find the row of $\ell$ in $\mathcal{C}$ from the change-of-basis matrix $P = P_{\mathcal{C} \leftarrow \mathcal{B}}$ , check it by direct evaluation, and confirm that the pairing with $v = (1, 2)$ is unchanged. What goes wrong if the row is multiplied by $P$ instead of $P^{-1}$ ?

Show the answer

The basis matrix with columns $(2, 1)$ and $(1, 1)$ has determinant $1$ , and $P = P_{\mathcal{C} \leftarrow \mathcal{B}}$ is its inverse,

P = \begin{bmatrix} 1 & -1 \\ -1 & 2 \end{bmatrix}, \qquad P^{-1} = \begin{bmatrix} 2 & 1 \\ 1 & 1 \end{bmatrix}.

Then $r_{\mathcal{C}} = \begin{bmatrix} 1 & 3 \end{bmatrix}P^{-1} = \begin{bmatrix} 5 & 4 \end{bmatrix}$ , and directly $\ell(2, 1) = 5$ and $\ell(1, 1) = 4$ . The vector has $[v]_{\mathcal{C}} = P(1, 2) = (-1, 3)$ , and indeed $-(2, 1) + 3(1, 1) = (1, 2)$ . The new row reads $-5 + 12 = 7$ , which equals $\ell(1, 2) = 1 + 6 = 7$ . Multiplying by $P$ instead gives $\begin{bmatrix} 1 & 3 \end{bmatrix}P = \begin{bmatrix} -2 & 5 \end{bmatrix}$ , which reads $2 + 15 = 17$ on $(-1, 3)$ , so the measurement would no longer be the same functional.

Dual Maps and Covectors

Key terms

Covector

Pullback of a functional

Dual map

Dual map of a composition

Matrix of the dual map

Covector coordinate change

Check yourself

How this lesson connects

Builds on

Used later in