Theory

Linear maps

The functions between vector spaces that preserve sums and scalar multiples, the matrix that represents them, kernel and image, the rank-nullity theorem, isomorphisms, composition and change of basis.

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Definition and examples
Maps defined on the vectors of a basis

Definition and examples

1. Linear map

Let $V$ and $W$ be real vector spaces. A function $f : V \to W$ is a linear map if for every $u, v \in V$ and every $λ \in R$ the two conditions are satisfied together:

{f (u + v) = f (u) + f (v) f (λ v) = λ f (v)

A linear map from $V$ to itself is called an endomorphism of $V$ . The two conditions are equivalent to the single condition $f (α u + β v) = α f (u) + β f (v)$ for every $u, v \in V$ and $α, β \in R$ ; by induction, $f$ sends every linear combination to the linear combination of the images with the same coefficients. The associated matrix and the construction of $f$ from its values on a basis are based on this property.

2. First properties

Let $f : V \to W$ be linear. Then $f (0_{V}) = 0_{W}$ and $f (- v) = - f (v)$ for every $v \in V$ . Indeed, setting $λ = 0$ in the second condition gives $f (0_{V}) = f (0 \cdot 0_{V}) = 0 \cdot f (0_{V}) = 0_{W}$ , and setting $λ = - 1$ gives the second equality. The first property provides an immediate check, because a map with $f (0_{V}) \neq = 0_{W}$ is not linear.

The linear maps of the exam

For every $A \in M_{m, n} (R)$ the function $f_{A} : R^{n} \to R^{m}$ , $f_{A} (x) = A x$ , is linear, since the row-by-column product satisfies $A (x + y) = A x + A y$ and $A (λ x) = λ A x$ . All functions such as $f (x, y, z) = (x + 2 y, y - z)$ , in which each component is a polynomial of degree one with no constant term, are of this type.
The derivative $D : R_{n} [x] \to R_{n - 1} [x]$ , $D (p) = p^{'}$ , is linear, since the derivative of a sum is the sum of the derivatives and multiplicative constants can be taken outside.
Transposition $M_{n} (R) \to M_{n} (R)$ , $A \mapsto A^{T}$ , and the trace $M_{n} (R) \to R$ , $A \mapsto tr A$ , are linear.
The rotation by a right angle about the origin, $R (x, y) = (- y, x)$ , is linear. Rotating the parallelogram built on $u$ and $v$ gives the parallelogram built on $R (u)$ and $R (v)$ , so the diagonal $u + v$ goes to the diagonal $R (u) + R (v)$ .

The rotation

R (x, y) = (- y, x)

sends the parallelogram built on

u = (2, 1)

and

v = (1, 2)

to the one built on

R (u)

and

R (v)

, therefore

R (u + v) = R (u) + R (v)

Warning

The condition $f (0_{V}) = 0_{W}$ is necessary and not sufficient for linearity.

$f (x, y) = (x + 1, y)$ is not linear, since $f (0, 0) = (1, 0)$ .
$g (x, y) = (x y, x)$ satisfies $g (0, 0) = (0, 0)$ and is not linear, since $g (2, 2) = (4, 2)$ is different from $2 g (1, 1) = (2, 2)$ .

To rule out linearity a numerical counterexample suffices; to prove it, the two conditions with generic vectors and scalars are needed.

Maps defined on the vectors of a basis

Existence and uniqueness on the vectors of a basis

Let $B = (v_{1}, \dots, v_{n})$ be a basis of $V$ and let $w_{1}, \dots, w_{n}$ be arbitrary vectors of $W$ . Then there exists one and only one linear map $f : V \to W$ such that $f (v_{i}) = w_{i}$ for $i = 1, \dots, n$ , and it is given by

f (a_{1} v_{1} + \dots + a_{n} v_{n}) = a_{1} w_{1} + \dots + a_{n} w_{n} .

A linear map is therefore determined by its values on a basis, and these values can be chosen freely. To compute $f (v)$ one writes $v$ as a combination of the basis vectors and replaces each $v_{i}$ with $w_{i}$ .

Proof

By the uniqueness of coordinates every $v \in V$ can be written in only one way as $v = a_{1} v_{1} + \dots + a_{n} v_{n}$ , so the formula defines a function $f : V \to W$ ; moreover $f (v_{i}) = w_{i}$ , because the coordinates of $v_{i}$ are $1$ in position $i$ and $0$ elsewhere. If $u = b_{1} v_{1} + \dots + b_{n} v_{n}$ , then $α u + β v = \sum_{i} (α b_{i} + β a_{i}) v_{i}$ , therefore

f (α u + β v) = i = 1 \sum n (α b_{i} + β a_{i}) w_{i} = α f (u) + β f (v),

that is, $f$ is linear. If $g$ is another linear map with $g (v_{i}) = w_{i}$ , by linearity $g (v) = \sum_{i} a_{i} g (v_{i}) = \sum_{i} a_{i} w_{i} = f (v)$ for every $v$ , so $g = f$ .

A map assigned on a basis of the plane

We look for a linear $f : R^{2} \to R^{3}$ with $f (1, 1) = (2, 0, 1)$ and $f (1, 2) = (1, 1, 0)$ . The vectors $(1, 1)$ and $(1, 2)$ form a basis of $R^{2}$ , since they are not proportional, so $f$ exists and is unique. We write $(x, y) = a (1, 1) + b (1, 2)$ , that is

{a + b = x a + 2 b = y ⟹ a = 2 x - y, b = y - x,

and by linearity

f (x, y) = (2 x - y) (2, 0, 1) + (y - x) (1, 1, 0) = (3 x - y, - x + y, 2 x - y) .

Values assigned on dependent vectors

If the values are assigned on linearly dependent vectors $u_{1}, \dots, u_{k}$ , every relation among the vectors must also hold among the values: for example, from $u_{3} = u_{1} + u_{2}$ it follows that a linear $f$ satisfies $f (u_{3}) = f (u_{1}) + f (u_{2})$ . If all the relations are respected, $f$ exists, and it is unique if and only if $u_{1}, \dots, u_{k}$ span $V$ ; if a relation is not respected, $f$ does not exist. There is, for example, no linear map $f : R^{2} \to R^{2}$ with $f (1, 0) = (1, 1)$ , $f (0, 1) = (2, 0)$ and $f (1, 1) = (3, 2)$ , since from $(1, 1) = (1, 0) + (0, 1)$ it would follow that $f (1, 1) = (3, 1)$ .

The theory continues with 8 more sections (the associated matrix, kernel and image, injectivity and surjectivity…), plus all worked examples, the solutions and the PDF.

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