es

Functions are used throughout mathematics, physics, and engineering to study the structures of sets and relationships between sets. You are familiar with the notation y = f(x), where f is a function that acts on numbers, signified by the input variable x, and produces numbers signified by the output variable y. It is a custom to write an input variable to the right of function following all European languages that perform writing in left to right.

In general, a function f : X ↦ Y is a rule that associates with each x in the set X a unique element y = f(x) in Y. We say that f maps the set X into the set Y and maps the element x to the element y. The set X is the domain of f and the set Y is called range or codomain. The set of all outputs of a particular function is its image. In linear algebra, we are interested in functions that map vectors into vectors preserving linear combinations.

Linear Transformations

Linear transformations are functions between vector spaces that preserve their structure, i.e., they are compatible with the operations of sum of vectors and multiplication of a vector by a scalar. Linear transformations are characterized by important properties that the sum of two linear transformations is linear and the composition of two linear transformations is linear.

If U is a vector space and A is any set, we know that the space UA of all mappings f : A → U is a vector space of functions (now vector valued). If A is itself a vector space V, we naturally single out for special study the subset of UV consisting of all linear mappings. We designate this subset as ℒ(V, U) or Hom(V, U).

Let V and U be vector spaces over a scalar field 𝔽 (which is either ℂ or ℝ or ℚ or ℤ). A function T : V ⇾ U is called a linear transformation (also called linear mapping or vector space homomorphism) if T preserves vector addition and scalar multiplication. In other words, for any scalar k and any two vectors v, u ∈ V, the following identities hold:
  1. T(v + u) = T(v) + T(u)      Preservation of addition;
  2. T(kv) = kT(v)            Preservation of scalar multiplication.
Actually, the two required identities in the definition above can be gathered into a single one
\begin{equation} \label{EqTransform.1} T \left( \lambda {\bf x} + {\bf y} \right) = \lambda T \left( {\bf x} \right) + T \left( {\bf y} \right) , \qquad \forall \lambda \in \mathbb{F} , \qquad \forall {\bf x}. {\bf y} \in V. \end{equation}
It is important to note that if T is a linear transformation from V into U, then T maps zero vector into zero vector, i.e., T(0) = 0; it can be seen from the definition because
\[ T(0) = T(0+0) = T(0) + T(0) . \]
Example 1: Our first two examples are related to affine transformations that include linear maps.
  1. Here is a familiar example: A bank delivers an interest based on fixed rate on deposits. After a certain time t lapse, an initial deposit P (known as principle) yields an interest I = P r t. So proportional deposits will produce proportional interests.
  2. Ohm's law states that the current intensity i through a conductor between two points is directly proportional to the voltage V across the two points. Introducing the constant of proportionality, the resistance R, one arrives at the usual mathematical equation that describes this relationship: \[ V = R\, i . \]
  3. For one mole of a perfect gas in a fixed volume V, the pressure p is proportional to the absolute temperature T, which is expressed by the ideal gas law: \[ p V = n\,RT , \] where n is the amount of substance and R is the ideal gas constant.
  4. In physics, Hooke's law is an empirical law which states that the force (F) needed to extend or compress a spring by some distance (x) scales linearly with respect to that distance—that is, \[ F = k\, x \] where k is a constant factor characteristic of the spring (i.e., its stiffness), and x is small compared to the total possible deformation of the spring. The law is named after 17th-century British physicist Robert Hooke. He first stated the law in 1676 as a Latin anagram.
  5. In control theory, certain physical systems called plants, are studied. A plant is a device having an output depending on an input. Typically, the plant could be a thermostat controlling the temperature in a car, where the input would consist of both outside and inside temperature, and output is directed to the ventilation and climate control system of the car. Another example of plant is a linear amplifier, namely an electronic device which amplifies a signal in a faithful way, at least in a certain domain of frequencies. Such a plant can be considered as a black box, having a certain number of input ports, where the signals si enter, and a (possibly different) number of exit ports producing outputs e j . The plant has linear characteristics when
    • An amplified signal produces an amplified output by the same factor.
    • Superposed input signals produce superposed outputs
    The action of the black box is represented by a linear map f
   ■
End of Example 1
The following terms are also employed:
  • homomorphism (from Greek homoios morphe, “similar form”) for a transformation preserving linear operations;
  • endomorphism for linear operator acting in a vector space;
  • automorphism for bijective linear operator acting in a vector space;
  • isomorphism for bijective linear transformation;
  • monomorphism (or embedding) for injective linear transformation;
  • epimorphism for surjective linear transformation.
Theorem 1: Let V be a finite-dimensional vector space of dimension n≥1, and let β = { v1, v2, … , vn } be a basis for V. Let U be any vector space, and let { u1, u2, … , un } be a list of vectors from U. The function T : V ↦ U defined by
\[ T \left( a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_n {\bf v}_n \right) = a_1 {\bf u}_1 + a_2 {\bf u}_2 + \cdots + a_n {\bf u}_n \]
is a linear transformation for any n scalars 𝑎₁, 𝑎₂, … , 𝑎n.
Because β is a basis in V, there are unique scalars 𝑎1, 𝑎2, … , 𝑎n such that arbitrary vector v ∈ V is represented as a linear combination of basis vectors: v = 𝑎1v1 + 𝑎2v2 + ⋯ + 𝑎nvn. So there is a unique corresponding element
\[ T \left( {\bf v} \right) = T \left( a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_n {\bf v}_n \right) = a_1 {\bf u}_1 + a_2 {\bf u}_2 + \cdots + a_n {\bf u}_n \]
in U. Hence T really is a function.

To show that T is a linear transformation, take any vectors v = 𝑎1v1 + 𝑎2v2 + ⋯ + 𝑎nvn and w = b1v1 + b2v2 + ⋯ + bnv2 in V and any real number k. We have

\begin{align*} T \left( {\bf v} + {\bf w} \right) &= T \left( a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_n {\bf v}_n + b_1 {\bf v}_1 + b_2 {\bf v}_2 + \cdots + b_n {\bf v}_n \right) \\ &= T \left( \left[ a_1 + b_1 \right] {\bf v}_1 + \left[ a_2 + b_2 \right] {\bf v}_2 + \cdots + \left[ a_n + b_n \right] {\bf v}_n \right) \\ &= \left[ a_1 + b_1 \right] {\bf u}_1 + \left[ a_2 + b_2 \right] {\bf u}_2 + \cdots + \left[ a_n + b_n \right] {\bf u}_n \\ &= a_1 {\bf u}_1 + a_2 {\bf u}_2 + \cdots + a_n {\bf u}_n + b_1 {\bf u}_1 + b_2 {\bf u}_2 + \cdots + b_n {\bf u}_n \\ &= a_1 T \left( {\bf v}_1 \right) + a_2 T \left( {\bf v}_2 \right) + \cdots + a_n T \left( {\bf v}_n \right) + b_1 T \left( {\bf v}_1 \right) + b_2 T \left( {\bf v}_2 \right) + \cdots + b_n T \left( {\bf v}_n \right) \\ &= T \left( {\bf v} \right) + T \left( {\bf w} \right) \end{align*}
and
\begin{align*} T \left( k\,{\bf v} \right) &= T \left( k\left[ a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_n {\bf v}_n \right] \right) \\ &= T \left( k\,a_1 {\bf v}_1 + k\,a_2 {\bf v}_2 + \cdots + k\,a_n {\bf v}_n \right) \\ &= k\,a_1 {\bf u}_1 + k\,a_2 {\bf u}_2 + \cdots + k\,a_n {\bf u}_n \\ &= k\, T \left( a_1 {\bf v}_1 + a_2 {\bf v}_2 + \cdots + a_n {\bf v}_n \right) = k\, T \left( {\bf v} \right) . \end{align*}
The two required conditions are satisfied, and T is a linear transformation.
   
Example 2:
  1. The differential operator acts on an arbitrary polynomial as \[ \texttt{D}\,\left( a_0 + a_1 x + a_2 x^2 + \cdots + a_n x^n \right) = a_1 + 2\,a_2 x + 3\,a_3 x^2 + \cdots + a\,a_n x^{n-1} . \] Hence, this operator maps the set of all polynomials of degree up to n into a similar set but of degree one less: \[ \texttt{D} \, : \ \mathbb{R}_{\le n} \left[ x \right] \mapsto \mathbb{R}_{\le n-1} \left[ x \right] . \] The derivative operator is a linear transformation, since \[ \texttt{D} \left( \alpha\, f(x) + \beta\, g(x) \right) = \alpha \,f' (x) + \beta\, g' (x) \] for any real numbers α, β ∈ ℝ and any functions f, g.
   ■
End of Example 2
Linear maps transform linear combinations into linear combinations having the same coefficients.
Theorem 2: Let V be a finite-dimensional vector space over the field 𝔽 and let [e₁, e₂, … , en] be an ordered basis for V. Let U be a vector space over the same field 𝔽 and let b₁, b₂, … , bn be any vectors in U. Then there is precisely one linear transformation T from V into U such that \[ T\left( {\bf e}_i \right) = {\bf b}_i , \qquad i = 1, 2, \ldots , n . \]
To prove that there is some linear transformation T with Tei = bi, we proceed as follows. Given a vector b in E, there is a unique n-tuple (x₁, x₂, … , xn) such that \[ {\f v} = x_1 {\bf e}_1 + x_2 {\bf e}_2 + \cdots + x_n {\bf e}_n . \] For this vector v, we define \[ T{\bf v} = x_1 {\bf b}_1 + x_2 {\bf b}_2 + \cdoys + x_n {\bf b}_n . \] Then T is a well-defined by the rule associating with each vector v in E so Tv is a vector in F. From the definition it is clear that Tei = bi for each i. To see that T is linear, let \[ {\bf u} = y_1 {\bf e}_1 + y_2 {\bf e}_2 + \cdots + y_n {\bf e}_n \] be in E and let λ be any scalar. Now \[ \lambda\,{\bf v} + {\bf u} = \left( \lambda x_1 + y_1 , \lambda x_2 + y_2 , \ldots , \lambda x_n + y_n \right) \] and so by definition \[ T \left( \lambda {\bf v} + {\bf u} \right) = \left( \lambda x_1 + y_1 \right) {\bf b}_1 + \cdots + \left( \lambda x_n + y_n \right) {\bf b}_n . \] On the other hand, \begin{align*} \lambda\, T({\bf v}) + T({\bf u}) &= \lambda \sum_{i=1}^n x_i {\bf b}_i + \sum_{i=1}^n y_i {\bf b}_i \\ &= \sum_{i=1}^n \left( \lambda\,x_i + y_i \right) {\bf b}_i \end{align*} and hence \[ T \left( \lambda\,{\bf v} + {\bf u} \right) = \lambda\, T({\bf v}) + T({\bf u}) . \]

If S is a linear transformation from E into F with Sei = bi, i = 1, 2, … , n, then for the vector \( \displaystyle \quad {\bf v} = \sum_{1 \le i \le n} x_i {\bf e}_i \quad \) we have \begin{align*} S\,{\bf v} &= S \left( \sum_{1 \le i \le n} x_i {\bf e}_i \right) \\ &= \sum_{1 \le i \le n} x_i S \left( {\bf e}_i \right) \\ &= \sum_{1 \le i \le n} x_i {\bf b}_i , \end{align*} so that S is exactly the rule T that we defined above. This shows that the linear transformation T with Tei = bi is unique.

   
Example 3:
  1. Let ℭ[𝑎, b] be a set of all real-values continuous functions on the closed interval. Then the integral operator
    \[ \int_a^b f(x)\,{\text d} x \]
    provides a linear transformation from ℭ[𝑎, b] into ℝ.
  2. Let ℭ(ℝ) be the set of all continuous functions on the real axis (−∞, ∞). Then the shift operator
    \[ T(f)(x) = f(x - x_0 ) , \qquad x_0 \in \mathbb{R}, \]
    is a linear transformation in the space ℭ(ℝ).
  3. Let ℘≤n be the linear space of polynomials of degree n or less. Then the derivative operator
    \[ \texttt{D}\left( \sum_{i=0}^n a_i x^i \right) = \sum_{i=1}^n a_i x^{i-1} \]
    provides a transformation from ℘≤n into ℘≤n-1
  4. Let ℳm×n is a set of all m × n matrices with entries from the field 𝔽. Then transformation gives a linear transformation from ℳm×n into ℳn×m.
  5. Let ℭ∞[𝕋] be set of infinitely differentiable periodic functions on the unit circle 𝕋 (one-dimensional torus). Then expansion of a function from ℭ∞[𝕋] into the Fourier series
    \[ f(x) \sim \sum_{n=-\infty}^{\infty} \hat{f}(n) \, e^{-{\bf j} nx} , \qquad \hat{f}(n) = \frac{1}{2\pi} \int_{-\pi}^{\pi} f(x)\, e^{{\bf j} nx} {\text d}x , \]
    provides a linear transformation from ℭ∞[𝕋] into the set of infinite sequences.
   ■
End of Example 3
This theorem provides a really nice way to create linear transformations.

How to Construct Linear Maps?

Here is a general method for the construction of linear transformations.
Theorem 3: Let S = {v₁, v₂, … , vn} be a linearly independent subset of the vector space V. For any family of vectors u₁, u₂, … , un of the same size as S from vector space U, there is a linear transformation T : V ⇾ U such that T(vk) = uk for k = 1, 2, … , n.
It is possible to enlarge the linearly independent family S into a basis {uj}j∈J of U even if U is infinite dimensional space (in this case J is unbounded). Now each element x ∈ U has a unique decomposition \( \displaystyle {\bf x} = \sum_{j\in J} c_j {\bf u}_j \) as a linear combination of the elements of this basis. The components cj are well-defined for all indices j ∈ J.

We define \( \displaystyle T({\bf x}) = \sum_{j\in J} c_j {\bf v}_j \) with the same components. This map is linear. For example, one can check that it is homogeneous by simply observing that the components of \( \displaystyle \lambda\,{\bf x} = \lambda \sum_{j\in J} c_j {\bf v}_j = \sum_{j\in J} c_j \lambda\,{\bf v}_j \) are the scalars λcj, whence \[ T \left( \lambda {\bf x} \right) = \sum_{j\in J} c_j \lambda\,{\bf v}_j = \lambda \sum_{j\in J} c_j {\bf v}_j = \lambda\, T \left( {\bf x} \right) . \] A similar verification shows the additivity of T.

   
Example 4:    ■
End of Example 4
Theorem 4: Let f,g : V ⇾ U be two linear maps. If f and g agree on a subset S c V, then they agree on the linear span of S. If f and g agree on a set of generators of V, then f = g.
The coincidence set of f and g is equal to the subspace ker(f - g) so that both statements follow.
   
Example 5:    ■
End of Example 5

 

Algebra of Linear Maps

In the study of linear transformations from X into Y, it is of fundamental importance that the set of these transformations inherits a natural vector space structure. We employ the following notation ℒ(X, Y) for the set of all linear transformations from X into Y.
For any two linear transformations T, S ∈ ℒ(X, Y), their sum, denoted as T + S, is defined by \[ \left( S + T \right) ({\bf x} ) = S({\bf x} + T({\bf x}, \qquad \forall {\bf x} \in X . \] For a scalar λ ∈ 𝔽 and a linear transformation T ∈ ℒ(X, Y) = Hom(X, Y), their product is defined by \[ \left( \lambda\,T \right) ({\bf x}) = \lambda\,T({\bf x}) , \qquad \forall {\bf x} \in X . \]
Theorem 5: The set of all linear transformations ℒ(X, Y) = Hom(X, Y) between two vector spaces X and Y over the same field is a vector subspace of YX.
The theorem is an easy formality. If S and T are in ℒ(X, Y), then for any elements x, y from vector space X and any scalar α, we have \begin{align*} \left( S + T \right) \left( \alpha\mathbf{x} + \mathbf{y} \right) &= S \left( \alpha\mathbf{x} + \mathbf{y} \right) + T \left( \alpha\mathbf{x} + \mathbf{y} \right) \\ &= \alpha\, S(\mathbf{x}) + S(\mathbf{y}) + \alpha\, T(\mathbf{x}) + T(\mathbf{y}) \\ &= \alpha \left( S + T \right) \left( \mathbf{x} \right) + \left( S + T \right) \left( \mathbf{y} \right) . \end{align*} So S + T is linear and ℒ(X, Y) is closed under addition.

The closure of Hom(X, Y) under multiplication by scalars follows similarly because \begin{align*} \left( \lambda\,T \right) \left( \alpha\mathbf{x} + \mathbf{y} \right) &= \lambda\,T \left( \alpha\mathbf{x} + \mathbf{y} \right) \\ &= \lambda\,T \left( \alpha\mathbf{x} \right) + \lambda\,T \left( \mathbf{y} \right) = \lambda\,\alpha\,T \,\mathbf{x} + \lambda\,T \, \mathbf{y} \\ &= \alpha \left( \lambda\, T \rught) \mathbf{x} + \left( \lambda\, T \rught) \mathbf{y} \end{align*} for any scalars λ and α and any vectors x, y from X.

Since Hom(X, Y) contains the zero transformation, and so is nonempty, it is a subspace.

   
Example 6:    ■
End of Example 6
Theorem 6: The composition of linear maps is linear: if T ∈ ℒ(X, Y) = Hom(X, Y) and S ∈ ℒ(Y, Z) = Hom(Y, Z), then S ∘ T ∈ Hom(X, Z). Moreover, composition is distributive over addition, under the obvious hypotheses on domains and codomains: \[ \left( S_1 + S_2 \right) \circ T = S_1 \circ T + S_2 \circ T \] and \[ S \circ \left( T_1 + T_2 \right) = S \circ T_1 + S \circ T_2 . \] Also composition commutes with scalar multiplication: \[ \lambda \left( S \circ T \right) = \left( \lambda\,S \right) \circ T = S \circ \left( \lambda\,T \right) , \qquad \lambda \in \mathbb{F} . \]
We have \begin{align*} S\circ T \left( \alpha\,\mathbf{x} + \mathbf{y} \right) &= S \left( T \left( \alpha\,\mathbf{x} + \mathbf{y} \right) \right) \\ &= S \left( \alpha\,T(\mathbf{x}) + T(\mathbf{y}) \right) \\ &= \alpha\,S(T(\mathbf{x})) + S(T(\mathbf{y})) \\ &= \alpha\left( S \circ T \right) (\mathbf{x}) + \left( S \circ T \right) (\mathbf{y}) . \end{align*} So S ∘ T is linear.

Now we check two distributive laws.

   
Example 7:    ■
End of Example 7
Corollary 1: It a linear transformation T ∈ Hom(X, Y) is fixed, then composition on the right by T is a linear transformation from the vector space Hom(Y, Z) to the vector space Hom(X, Z). It is an isomorphism if T is an isomorphism.
The algebraic properties of composition stated in the Corollary can be combined as follows: \begin{align*} \left( c_1 S_1 + c_2 S_2 \right) \circ T &= c_1 \left( S_1 \circ T \right) + c_2 \left( S_2 \circ T \right) ; \\ S \circ \left( c_1 T_1 + c_2 T_2 \right) &= c_1 \left( S \circ T_1 \right) + c_2 \left( S \circ T_2 \right) . \end{align*} The first equation says exactly that composition on the right by a fixed T is a linear transformation. (Write S ∘ T as Image(S) if the equations still don't look right.) If T is an isomorphism, then composition by T-1 "undoes" composition by T, and so is its inverse.

The second equation implies a similar corollary about composition on the left by a fixed S.

   
Example 8:    ■
End of Example 8
Theorem 7: If U is a product vector space, U = ΠkUk, then a mapping T from a vector space V to U is linear if and only if πk ∘ T is a linear for each coordinate projection πk : U ↦ Uk.
If T is linear, then composition πk ∘ T is linear by Corollary 1. Now suppose, conversely, that all the maps πk ∘ T are linear. Then \begin{align*} \pi_k \left( T \left( \lambda\,\mathbf{x} + \beta\,\mathbf{y} \right) \right) &= \pi_k \circ T \left( \lambda\,\mathbf{x} + \beta\,\mathbf{y} \right) \\ &= \lambda\left( \pi_k \circ T \right) (\mathbf{x}) + \beta \left( \pi_k \circ T \right) (\mathbf{y}) \\ &= \lambda\,\pi_k \left( T(\mathbf{x}) \right) + \beta\,\pi_k \left( T(\mathbf{y}) \right) = \pi_k \left( \lambda\,T(\mathbf{x}) + \beta\,T(\mathbf{y}) \right) . \end{align*} But if πk(f) = πk(g) for all k, then f = g. Therefore, T(λx + βy) = λ T(x) + β T(y), so T is linear.
   
Example 9: Let T be a linear mapping from a vector space V into the Cartesian product U = ℝn. Then k-th projection πk on ℝ is just the k-th component, so π₁(𝑎₁, 𝑎₂, … , 𝑎n) = (𝑎₁, 0, … , 0). Therefore, T : V ⇾ ℝn is linear if and only if its every component is linear.    ■
End of Example 9
  1. Determine whether the following are linear transformations from ℝ≤2[x] to ℝ≤3[x] :
    1. T(p(x)) = x p(x);
    2. T(p(x)) = x² + p(x);
    3. T(p(x)) = x²p(0);
    4. T(p(x)) = p(x) + x p(x) + x²p′(x).
  2. For any continuous function f ∈ ℭ[0, 1] on interval [0, 1], show that the following transformation is linear: \[ f \,\mapsto \,\int_0^x f(t)\,{\text d}t , \qquad 0 \le x \le 1. \]
  3. Determine whether the following are linear transformations from the space of continuous functions ℭ[0, 1] into ℝ:
    1. T(f) = |f(0)|;
    2. T(f) = f(0);
    3. T(f) = [f(0) +f(1) ]/2;
    4. T(f) = \( \displaystyle \left\{ \int_0^1 \left\vert f(t) \right\vert^2 {\text d} t \right\}^{1/2} . \)
  4. Let β = {v1, v2, … , vn} be a basis for a vector space V, and let T₁ and T₂ be two linear transformations mapping V into a vector space W. Show that if \[ T_1 ({\bf v}_i ) = T_2 ({\bf v}_i ) , \qquad i = 1, 2, \ldots , n, \] then T₁ = T₂ [i.e., show that T₁(v) = T₂(v) for all v ∈ V.
  5. Let T be a linear operator on a vector space V, that is, T : V ⇾ V. Define Tn, n ≥ 1, recursively by \[ T^n ({\bf v}) = T \left( T^{n-1} ({\bf v})\right), \qquad \forall {\bf v} \in V. \] Show that Tn is a linear operator on V for each n ≥ 1.
  6. Assume that T : ℝn ⇾ ℝn, is a linear transformation, and that u1, u2, … , un is a basis of ℝn. For every i = 1, 2, … , n, let vi = T(ui). Let A = [T] be the matrix that has u1, u2, … , un as its columns, and let B be the matrix that has v1, v2, … , vn as its columns. Show that A is invertible and the matrix of T is BA−1.
  7. Let T : V ⇾ V be a linear operator and define Uj = ker(Tj) and Wj = image(Tj), j = 1, 2, …. .. Show that for all j, Uj ⊆ Uj+1 and Wj ⊇ Wj+1.

 

  1. Anton, Howard (2005), Elementary Linear Algebra (Applications Version) (9th ed.), Wiley International
  2. Beezer, R.A., A First Course in Linear Algebra, 2017.
  3. Fitzpatrick, S., Linear Algebra: A second course, featuring proofs and Python.