Dual Spaces and Dual Bases, linear algebra II lesson

Two sensors read a vector $(x, y)$ in the plane. One reports $2x - y$ and the other $x + y$ , so at $(3, 1)$ they report $5$ and $4$ . The readings are scalars, not new vectors. What kind of object is a sensor, and can sensors be added and scaled like vectors? The answer is the dual space of all linear measurements, and when the original space is finite-dimensional, each of its bases has a matching basis of the dual space.

Key terms

Linear functional

Definition Let $V$ be a vector space over $\mathbb{F}$ , where $\mathbb{F}$ is $\mathbb{R}$ or $\mathbb{C}$ . A linear functional on $V$ is a linear map $\ell: V \to \mathbb{F}$ , so

\ell(av + bw) = a\,\ell(v) + b\,\ell(w)

for all $v, w \in V$ and $a, b \in \mathbb{F}$ . Over $\mathbb{C}$ the scalars come out unchanged, with no complex conjugate. Every linear functional sends $0$ to $0$ .

In plain words A functional measures a vector and returns one number, and the measurement respects addition and scaling. In the real plane, the inputs with a fixed reading of a nonzero functional form a line, and the lines for different readings are parallel.

Example The map $\ell(x, y) = 2x - y$ is a linear functional on $\mathbb{R}^2$ , and $\ell(3, 1) = 5$ . Moving along the line $2x - y = 5$ keeps the reading, and moving across it changes the reading. On a nonzero space a length is not a linear functional. For $v \ne 0$ we have $\lVert -v \rVert = \lVert v \rVert > 0$ , while linearity would require the reading of $-v$ to be the negative of the reading of $v$ . The map $(x, y) \mapsto x + 1$ is not one either, because it sends $0$ to $1$ .

Dual space

Definition The dual space of $V$ is the set of all linear functionals on $V$ ,

V^{\vee} = \operatorname{Hom}(V, \mathbb{F}),

with the pointwise operations $(\ell + m)(v) = \ell(v) + m(v)$ and $(a\ell)(v) = a\,\ell(v)$ . It is a vector space over $\mathbb{F}$ , the case $W = \mathbb{F}$ of the space of linear maps $\operatorname{Hom}(V, W)$ . Its zero vector is the functional that reads $0$ on every input. Many books write $V^{*}$ for the dual space. This course keeps $V^{\vee}$ , so that the dual is not confused with the adjoint $T^{*}$ , which needs an inner product and comes later.

For $V = \mathbb{F}^n$ with column vectors, every $\ell \in V^{\vee}$ is multiplication by exactly one row, namely $\ell(x) = rx$ with $r = \begin{bmatrix} \ell(e_1) & \cdots & \ell(e_n) \end{bmatrix}$ . Over $\mathbb{C}$ this row carries no conjugation. A row represents a functional, but a functional is not a vector of $V$ . Turning one into a vector needs an extra choice, such as an inner product.

In plain words Measurements form a vector space of their own. Two sensors are added by adding their readings, and a sensor is scaled by scaling its readings. A functional is a different kind of object from the vectors it measures.

Example On $\mathbb{R}^2$ the functionals $\ell(x, y) = 2x - y$ and $m(x, y) = x + y$ add to $(\ell + m)(x, y) = 3x$ . At $(3, 1)$ the readings are $5$ and $4$ , and the sum reads $9 = 3 \cdot 3$ . In standard coordinates, $\ell$ is represented by the row $\begin{bmatrix} 2 & -1 \end{bmatrix}$ multiplying the column of $(x, y)$ .

Coordinate functional

Definition Let $V$ be finite-dimensional over $\mathbb{F}$ with ordered basis $\mathcal{B} = (b_1, \dots, b_n)$ . Every $v \in V$ has a unique expansion $v = c_1 b_1 + \cdots + c_n b_n$ . The $i$ th coordinate functional is

\beta_i: V \to \mathbb{F}, \qquad \beta_i(v) = c_i,

the $i$ th entry of the coordinate column $[v]_{\mathcal{B}}$ . It is linear because the coordinates of a sum or a scalar multiple are the sum or the multiple of the coordinates. It satisfies

\beta_i(b_j) = \delta_{ij} = \begin{cases} 1, & i = j, \\ 0, & i \ne j. \end{cases}

When $n \ge 2$ , each $\beta_i$ depends on the whole basis, not only on $b_i$ .

In plain words A coordinate functional reads off one coordinate of a vector in a chosen basis. Change any one basis vector and every coordinate functional may change.

Example In $\mathbb{R}^2$ take $\mathcal{B} = ((1, 1), (1, -1))$ . Writing $(x, y) = a(1, 1) + b(1, -1)$ gives $a + b = x$ and $a - b = y$ , so

\begin{gathered} \beta_1(x, y) = \frac{x + y}{2}, \\[0.4em] \beta_2(x, y) = \frac{x - y}{2}. \end{gathered}

Check that $\beta_1$ reads $1$ on $(1, 1)$ and $0$ on $(1, -1)$ , while $\beta_2$ reads $0$ and $1$ . With the basis $((1, 1), (0, 1))$ the first basis vector is the same, but the first coordinate functional becomes $(x, y) \mapsto x$ .

Dual basis

Definition Let $V$ be finite-dimensional over $\mathbb{F}$ with basis $b_1, \dots, b_n$ . There is exactly one list $\beta_1, \dots, \beta_n$ in $V^{\vee}$ with $\beta_i(b_j) = \delta_{ij}$ , namely the coordinate functionals, and it is a basis of $V^{\vee}$ , called the dual basis. Every $\ell \in V^{\vee}$ expands as

\ell = \sum_{i=1}^{n} \ell(b_i)\,\beta_i.

The expansion holds because $\ell(v) = \sum_i c_i\,\ell(b_i) = \sum_i \ell(b_i)\,\beta_i(v)$ for $v = \sum_i c_i b_i$ , so the $\beta_i$ span. They are independent, because applying $\sum_i a_i \beta_i = 0$ to $b_j$ gives $a_j = 0$ . The list is unique, because a linear functional is determined by its values on a basis. Hence $\dim V^{\vee} = \dim V$ . This equality does not single out a preferred isomorphism between $V$ and $V^{\vee}$ , since matching $b_i$ with $\beta_i$ depends on the basis. The finite-dimensional hypothesis matters. On the space of all real polynomials, evaluation at $1$ adds up all the coefficients and is not a finite combination of coefficient functionals.

In plain words The dual basis is the list of coordinate readers for a basis, one for each basis vector. The coefficients of any measurement in the dual basis are just its readings on the basis vectors.

Example With $\mathcal{B} = ((1, 1), (1, -1))$ and $\ell(x, y) = 2x - y$ , the readings on the basis are $\ell(1, 1) = 1$ and $\ell(1, -1) = 3$ , so $\ell = \beta_1 + 3\beta_2$ . Indeed

\frac{x + y}{2} + \frac{3(x - y)}{2} = 2x - y.

The vector $(3, 1)$ has coordinate column $(2, 1)$ in $\mathcal{B}$ , and the coefficient row of $\ell$ times this column gives

\begin{bmatrix} 1 & 3 \end{bmatrix} \begin{bmatrix} 2 \\ 1 \end{bmatrix} = 5 = \ell(3, 1).

The coordinates of the vector and the coefficients of the functional both depend on the basis, but the reading does not.

Evaluation functional

Definition Fix $s \in \mathbb{F}$ . On $\mathcal{P}_d$ , the polynomials over $\mathbb{F}$ of degree at most $d$ together with the zero polynomial, evaluation at $s$ is the linear functional $\operatorname{ev}_s(p) = p(s)$ . It is linear because substituting $s$ respects sums and scalar multiples of polynomials. For the monomial basis $b_k = t^{k-1}$ with $k = 1, \dots, d + 1$ , the dual basis functionals $\beta_k$ read the coefficients. The values on the basis are $\operatorname{ev}_s(b_k) = s^{k-1}$ , where $s^0 = 1$ even for $s = 0$ because $b_1$ is the constant polynomial $1$ . The expansion in the dual basis therefore gives

\operatorname{ev}_s = \sum_{k=1}^{d+1} s^{k-1} \beta_k.

The point $s$ is fixed in advance. A rule that picks the point depending on the polynomial itself is a different map and needs its own linearity check.

In plain words Plugging in a fixed number is a measurement of a polynomial. In coefficient coordinates it weights each coefficient by a power of that number.

Example On $\mathcal{P}_2$ with basis $1, t, t^2$ , write $p = a + bt + ct^2$ , so $\beta_1(p) = a$ , $\beta_2(p) = b$ and $\beta_3(p) = c$ . Evaluation at $2$ gives $a + 2b + 4c$ , so

\operatorname{ev}_2 = \beta_1 + 2\beta_2 + 4\beta_3.

For $p = 3 - t + 2t^2$ this gives $3 + 2(-1) + 4(2) = 9$ , which matches direct substitution, $p(2) = 3 - 2 + 8 = 9$ .

Conjugate-linear map

Definition Let $V$ and $W$ be vector spaces over $\mathbb{C}$ . A map $f: V \to W$ is conjugate-linear if

\begin{gathered} f(v + w) = f(v) + f(w), \\[0.4em] f(av) = \overline{a}\,f(v) \end{gathered}

for all $v, w \in V$ and $a \in \mathbb{C}$ . A map that is both linear and conjugate-linear is zero, since $i f(v) = f(iv) = -i f(v)$ forces $f(v) = 0$ . The elements of $V^{\vee}$ are linear, so a nonzero conjugate-linear map from $V$ to $\mathbb{C}$ never belongs to the dual space. The same two conditions written over $\mathbb{R}$ would only repeat linearity, because $\overline{a} = a$ for real $a$ .

In plain words A conjugate-linear map lets a scalar out only after conjugating it. Over the complex numbers that small change is enough to break linearity, unless the map is zero.

Example Complex conjugation $z \mapsto \overline{z}$ on $\mathbb{C}$ is conjugate-linear but not linear. It sends $i \cdot 1$ to $-i$ , while $i$ times its value at $1$ is $i$ . By contrast, $\varphi(z_1, z_2) = z_1 + i z_2$ is a linear functional on $\mathbb{C}^2$ , with the unconjugated row $\begin{bmatrix} 1 & i \end{bmatrix}$ . Later the inner product $\langle x, z \rangle = x^{*} z$ , which is conjugate-linear in its first argument, will represent this functional as $\varphi(z) = \langle x, z \rangle$ with $x = (1, -i)$ . The conjugation belongs to that representation, not to the functional.

Check yourself

Which of these maps on $\mathbb{R}^2$ are linear functionals? The maps are $f(x, y) = 3x - 2y$ , $g(x, y) = x + 1$ , $h(x, y) = \lvert x \rvert$ and $k(x, y) = xy$ .

Show the answer

Only $f$ is a linear functional. The map $g$ sends $(0, 0)$ to $1$ rather than $0$ . The map $h$ gives $h(-1, 0) = 1 = h(1, 0)$ , but linearity would require $h(-1, 0) = -h(1, 0)$ . The map $k$ gives $k(2, 2) = 4$ , but linearity would require $k(2, 2) = 2k(1, 1) = 2$ .

In $\mathbb{R}^2$ take the basis $b_1 = (1, 0)$ , $b_2 = (1, 1)$ . Find the dual basis, and explain why $\beta_1$ is not the functional $(x, y) \mapsto x$ even though $b_1 = e_1$ . Then write $\ell(x, y) = 2x - y$ in the dual basis and check the result at $(3, 2)$ .

Show the answer

Writing $(x, y) = a(1, 0) + b(1, 1)$ gives $b = y$ and $a = x - y$ , so

\beta_1(x, y) = x - y, \qquad \beta_2(x, y) = y.

The functional $(x, y) \mapsto x$ reads $1$ on $b_2 = (1, 1)$ , but the dual basis requires $\beta_1(b_2) = 0$ , so $\beta_1$ depends on $b_2$ as well as on $b_1$ . The readings $\ell(b_1) = 2$ and $\ell(b_2) = 1$ give $\ell = 2\beta_1 + \beta_2$ , and indeed $2(x - y) + y = 2x - y$ . At $(3, 2)$ the coordinates are $1$ and $2$ , and $2 \cdot 1 + 1 \cdot 2 = 4$ , which equals $\ell(3, 2) = 6 - 2$ .

On $\mathbb{C}^2$ let $\varphi(z_1, z_2) = z_1 + i z_2$ and $\psi(z_1, z_2) = \overline{z_1} + i\,\overline{z_2}$ . Which of them belongs to $(\mathbb{C}^2)^{\vee}$ ?

Show the answer

Only $\varphi$ belongs to the dual space. It is additive and satisfies $\varphi(az) = a\,\varphi(z)$ for every complex $a$ . The map $\psi$ is additive, but $\psi(az) = \overline{a}\,\psi(z)$ , so it is conjugate-linear. With $a = i$ and $z = e_1$ we get $\psi(ie_1) = \overline{i} = -i$ , while $i\,\psi(e_1) = i$ . A nonzero conjugate-linear map is not linear, so $\psi$ is not in $(\mathbb{C}^2)^{\vee}$ .

Dual Spaces and Dual Bases

Key terms

Linear functional

Dual space

Coordinate functional

Dual basis

Evaluation functional

Conjugate-linear map

Check yourself

How this lesson connects

Builds on

Used later in