3.7 Dual spaces and quotient spaces

Chapter 2.3 studied linear maps from one space to another. This chapter takes the simplest possible target, the real numbers themselves. A linear map from $V$ to $\mathbb R$ is called a linear functional, and the set of all of them is again a vector space, the dual space of $V$. It explains something you have been writing since 2.5 without a reason: why the transpose appears whenever a map is asked what it does to functionals, and why in calculus a point is a column while a derivative is a row. The second half builds the quotient space $V/U$, the exact way to say “treat two vectors as the same when they differ by something in $U$”. It reuses the machinery of equivalence relations from 0.4, and it turns rank–nullity into a statement about isomorphism.

The picture

Take the function $\varphi(x, y) = x + 2y$ on the plane. It is linear: it sends $(0, 0)$ to $0$, and it respects sums and multiples. It reads one number off each point. Look at where it is constant. The points with $\varphi = 0$ lie on the line $x + 2y = 0$. The points with $\varphi = 1$ lie on a second line $x + 2y = 1$, the first line moved along. Every value $c$ gives one more parallel line. So a functional on the plane is a family of equally spaced parallel lines, and its value at a point says which line the point is on.

The visualizer below shows the same thing for the map $T(x, y) = (x + 2y,\ 0.5x + y)$. Its second output is half of the first, so $T$ is the functional $x + 2y$ followed by a stretch onto a line.

The red line is the null space: the vectors along $(-2, 1)$ that $T$ sends to $O$. Now take any point off the red line and look at the whole line through it that is parallel to the red line. Every point of that parallel line lands on the same point of the green line, because two points on it differ by a multiple of $(-2, 1)$, and $T$ kills that. So $T$ cannot tell the points of a parallel line apart. Each of these parallel lines is a coset of the null space, and the set of all of them, with a way to add and scale them, is the quotient space. The picture says that the plane, with the red line collapsed to a point, is the green line. That sentence is the first isomorphism theorem, and the rest of the chapter makes it precise.

The idea, in words

Linear functionals and the dual space

How many functionals are there, and what do they look like? On $\mathbb R^n$ the answer is a row of numbers.

This is the reason rows and columns are different things. A column $x$ is a vector. A row $a$ is a functional, a machine that eats a column and returns a number. The two are never added to each other, and a row eats a column and not another row.

By the proposition, $(\mathbb R^n)^*$ is the space of $1 \times n$ rows. It has dimension $n$, like $\mathbb R^n$. Is that a coincidence of $\mathbb R^n$, or true for every finite-dimensional $V$? To prove it for every $V$ we need a basis of $V^*$, built from a basis of $V$.

The dual basis

For the standard basis of $\mathbb R^n$ the dual basis is the coordinate functionals $x \mapsto x_i$, which are the rows $e_1^{\mathsf T}, \dots, e_n^{\mathsf T}$. For a different basis it is different, and computing it is the first thing to learn.

The dual map and the transpose

A linear map $T: V \to W$ turns vectors of $V$ into vectors of $W$. What does it do to functionals? If $\psi$ is a functional on $W$ (it reads a number off vectors of $W$), then reading $\psi$ at $T(v)$ is a way of reading a number off $v \in V$. Doing so produces a functional on $V$.

Notice the direction. $T$ goes from $V$ to $W$; $T^{\mathsf T}$ goes from $W^*$ back to $V^*$. This reversal is why transposing a product reverses the order. The next theorem says that in coordinates, $T^{\mathsf T}$ is the transpose matrix of 2.5.

Annihilators, and a second proof that row rank equals column rank

2.6 proved this fact by elimination. The proof here needs no elimination and no pivots, and it shows what is going on: the rows of $A$ are functionals on the input space, and the number of independent ones equals the number of independent columns, because both count the dimension of the range.

import numpy as np

B = np.array([[1, 1],
              [1, 2]])               # columns are b1 = (1,1) and b2 = (1,2)
Binv = np.linalg.inv(B)              # its rows are the dual basis
print("dual basis, one functional per row:")
print(np.round(Binv) + 0)
print("B^-1 B, entry (i,j) = phi_i(b_j):")
print(np.round(Binv @ B) + 0)
x = np.array([5, 3])
print("coordinates of (5,3):", np.round(Binv @ x) + 0)

A = np.array([[1, 2], [0, 1], [3, 0]])
x, y = np.array([2, -1]), np.array([1, -1, 2])
print("(Ax).y =", (A @ x) @ y, "   x.(A^T y) =", x @ (A.T @ y))
print("rank A =", np.linalg.matrix_rank(A), "  rank A^T =", np.linalg.matrix_rank(A.T))

Cosets and the quotient space

Now the second construction. The idea is the one from 0.5: in $\mathbb Z_n$ two integers count as the same when they differ by a multiple of $n$. In a vector space $V$, pick a subspace $U$, and call two vectors the same when they differ by a vector of $U$.

In $\mathbb R^2$ take the line $U = \{t(1, 2) : t \in \mathbb R\}$, the line $y = 2x$ through $O$. The coset $(0, 3) + U$ is the set of points $(t, 3 + 2t)$, the line $y = 2x + 3$: the line $U$ moved up by $3$. The coset $(5, 13) + U$ is the points $(5 + t, 13 + 2t)$, which satisfy $y = 2x + 3$ as well, since $13 + 2t = 2(5 + t) + 3$. So $(5, 13) + U = (0, 3) + U$: the same line has many representatives. The cosets of a line through the origin are the lines parallel to it, and the coset of $O$ is the line itself, $O + U = U$. In $\mathbb R^3$ the cosets of a plane through the origin are the parallel planes.