Definition. Let $V$ and $W$ be two vector spaces over the same field $\mathbb{F}$. A transformation $A: V \to W$ is called a linear operator if
$$A ( \alpha x + \beta y) = \alpha A x + \beta A y, \quad \forall x, y \in V, \quad \forall \alpha, \beta \in \mathbb{F}.$$
The definition allows the possibility $V = W$; then the condition that $V$ and $W$ are vector spaces over the same field is automatically filled. In this case it is said to be that $A$ is a linear operator on $V$.
If $V \neq W$, the definition of a linear operator requires that the spaces must be over the same field. Namely, the scalars $\alpha$ and $\beta$ on the left side of equality $A (\alpha x + \beta y) = \alpha A x + \beta A y$ multiply vectors in $V$, while those same scalars multiply the images of these vectors ($Ax$ and $Ay$) in $W$, on the right side of equality.
Note. A function $A: V \to W$ is called a linear transformation if
$$A( \alpha x + \beta y) = \alpha A x + \beta A y, \quad \forall x, y \in V, \quad \forall \alpha, \beta \in \mathbb{F}.$$
From here it follows immediately
(i) $A (x +y) = Ax + By, \quad \forall x, y \in V$, if we take $\alpha = \beta = 1$;
(ii) $A ( \alpha x) = \alpha A x , \quad \forall x \in V, \forall \alpha \in \mathbb{F}$, if we take $\beta = 0$.
The property (i) is called additivity, and the property (ii) is called homogeneity. Therefore, each linear operator is additive and homogeneous. The reverse is also true.
Every linear operator maps the zero vector into the zero vector: $A0 = 0$.
If $A: V \to W$ is a linear operator, then
$$A ( \alpha_1 x_1 + \alpha_2 x_2 + \ldots + \alpha_n x_n) = \alpha_1 A x_1 + \alpha_2 A x_2 + \ldots + \alpha_n A x_n, \quad \forall x_i \in V, \forall \alpha_i \in \mathbb{F},$$
that is, linear operators respect the linear combinations.
Suppose now that $A: V \to W$ is a linear operator and $\{b_1, b_2, \ldots, b_n \}, n \in \mathbb{N}$, a basis for $V$. We choose the vector $x \in V$ and write it in the form
$$x = \alpha_1 b_1 + \alpha_2 b_2 + \ldots + \alpha_n b_n.$$
Then $Ax$ can be expressed as
$$Ax = \alpha_1 A b_1+ \alpha_2 A b_2 + \ldots + \alpha_n A b_n.$$
From here we can conclude that if we know the vectors $Ab_1, \ldots, Ab_n$, then implicitly we know the vector $x$, for each vector $x \in V$.
Looking from another angle, this observation leads us to the following proposition.
Proposition 1. Let $V$ and $W$ be two vector spaces over the same field $\mathbb{F}$ and $\{b_1, \ldots, b_n \}$ be any basis for $V$ and $(w_1, \ldots, w_n)$ any ordered $n$- tuple of vectors in $W$. There there is a unique linear operator $A: V \to W$ such that $Ab_i = w_i, \quad \forall i = 1, \ldots, n$.
Since $(w_1, \ldots, w_n)$ is the ordered $n$- tuple, it is possible that some of the vectors $w_i$ are repeated.
We can now explore some basic properties of linear operators.
Firstly, linear operators kept the properties of subspaces.
Proposition 2. Let $A: V \to W$ be a linear operator.
(1) If $K$ is a subspace of $V$, then $A(K)$ is a subspace of $W$.
(2) If $L$ is a subspace of $W$, then $A^{-1}(L)$ is a subspace of $V$.
Consider two special cases: $K = V \le V$ and $L = \{0 \} \le W$. This leads us to the definition of the rank and nullity – the central concepts in the study of linear operators.
Definition. Let $V$ and $W$ be a finite dimensional vector spaces and let $A: V \to W$ be a linear operator. The range ( or image) of the linear operator $A$ is the set of all vectors $w \in W$ such that $w = Av$, for some $v \in V$, that is,
$$Im (A) = A(V) = \{ Av : v \in V\}.$$
The kernel of the linear operator $A$ is the set of all vectors $v \in V$ such that $Av = 0$, that is,
$$\ker (A) = A^{-1}\{0 \} = \{ v \in V: Av = 0 \}.$$
The number $\dim (\ker (A))$ is called the nullity of the operator $A$ and it is denoted with $null (A)$, and the number $\dim (Im(A))$ is called the rank of the operator $A$ denoted with $r(A)$.
It is routine to show that $\ker (A)$ is a subspace of $V$ and $Im (A)$ is a subspace of $W$. Moreover, we have the following proposition.
Proposition 3. Let $A: V \to W$ be a linear operator.
(i) $A$ is surjective if and only if $Im (A) = W$.
(ii) $A$ is injective if and only if $\ker (A) = \{0\}$.
Proposition 4. Let $V$ and $W$ be a finite dimensional vector spaces and let $A: V \to W$ be a linear operator. Then $A$ is injective if and only if for each linearly independent set $S$ in $V$ the set $A(S) = \{Av: v \in S \}$ is linearly independent in $W$.
Now we can write the the rank – nullity theorem.
Theorem 1. If $V$ and $W$ are finite dimensional vector spaces and $A: V \to W$ is a linear operator, then
$$r(A) + null (A) = \dim V.$$
Definition. A linear operator $A: V \to W$ is called
(i) a monomorphism if $A$ is injective;
(ii) an epimorphism if $A$ is surjective;
(iii) an isomorphism if $A$ is bijective.
A consequence of the rank – nullity theorem is the following corollary.
Corollary 1. Let $A: V \to W$ be a linear operator and let $\dim V = \dim W < \infty$. The following conditions are equivalent to each other:
(1) $A$ is a monomorphism.
(2) $A$ is an epimorphism.
(3) $A$ is an isomorphism.
Two vector spaces $V$ and $W$ are called isomorphic if there is an isomorphism $A: V \to W$.
Proposition 5. Let $A: V \to W$ and $B: W \to Z$ be linear operators. Then the composition $B \circ A: V \to Z$ is also a linear operator. Specially, the composition of two isomorphisms is again an isomorphism.
The following proposition characterizes the isomorphisms of the finite dimensional vector spaces.
Proposition 6. Let $A: V \to W$ be a linear operator and $\dim V = n$. Then the following claims are equivalent to each other:
(1) $A$ is an isomorphism.
(2) For each basis $\{b_1, b_2, \ldots, b_n \}$ of $V$, the set $\{Ab_1, Ab_2, \ldots, Ab_n \}$ is a basis for $W$.
(3) There is a basis $\{e_1, e_2, \ldots, e_n \}$ of $V$ such that $\{Ae_1, Ae_2, \ldots, Ae_n \}$ is a basis for $W$.
Proposition 7. Let $V$ and $W$ be a finite dimensional vector spaces over the same field. Then $V$ and $W$ are isomorphic if and only if $\dim V = \dim W$.
The space of linear operators
When $V$ and $W$ are the vector spaces over the same field, we can observe the set of all linear operators from $V$ into $W$, denoted by $L (V, W)$. This set is always non empty.
We want now to introduce the structure of a vector space in the $L (V, W)$.
Definition. Let $V$ and $W$ be two vector spaces over the same field $\mathbb{F}$.
- For $A, B \in L(V, W)$ we define $A + B: V \to W$ as $(A + B) x = Ax + Bx$.
- For $A \in L (V, W)$ and $\alpha \in \mathbb{F}$ we define $\alpha A: V \to W$ as $(\alpha A) x = \alpha A x$.
This means that the linear operators may be added using point wise addition, and they can be multiplied by scalars in a similar way.
Theorem 2. If $V$ and $W$ are the vector spaces over the same field $\mathbb{F}$, then $L (V, W)$ is also a vector space over the field $\mathbb{F}$.
Theorem 3. Suppose $\dim V = n$ and $\dim W = m$. Then $ L (V, W)$ has finite dimension $n \cdot m$, that is,
$$\dim L (V, W) = \dim V \cdot \dim W.$$
Consider yet another special case when $V = W$. This we will denote by $L(V)$, instead $L(V, V)$.
Of course, $\dim L(V,V)=n^2$
Proposition 8. Let $V$ be vector space. For $L(V)$ we have:
(1) $L(V)$ is a vector space
(2) $\forall A,B \in L(V) AB \in L(V)$
(3) $A(BC)=(AB)C, \forall A, B, C \in L(V)$
(4) $A(B+C)=AB+AC, (A+B)C=AC+BC, \forall A, B, C \in L(V)$
(5) $(\alpha A)B= \alpha(AB) = A(\alpha B), \forall A, B \in L(V)$
(6) $\exists! I \in L(V)$ such that $ AI=IA=A, \forall A \in L(V)$
