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.
Let V and U be vector spaces over a scalar field 𝔽 (which is either ℂ or ℝ or ℚ). 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 E into F, then T(O) 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.
Linear maps transform linear combinations into linear combinations having the same coeficients.
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
Theorem 2: Let E be a finite-dimensional vector space over the field 𝔽 and let [e₁, e₂, … , en] be an ordered basis for E. Let F be a vector space over the same field 𝔽 and let b₁, b₂, … , bn be any vectors in F. Then there is precisely one linear transformation T from E into F 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 span class="math">\( \displaystyle \quad {\bf v} = \sum_{1 \le i \le n} x_i {\bf e}_i \quad \) 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.

This theorem provides a really nice way to create linear transformations. Let V and U be vector spaces over the same field 𝔽 (which is either ℂ or ℝ or ℚ).
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.

 

Image and Kernel of a Linear Map

For any map f : E ↦ F between sets, and any subset X ⊂ V, its image is denoted as f(X) = { f(x) : x ∈ X} ⊂ Y. The image of f is
\[ \mbox{im}(f) = \left\{ f(x) \,:\ x \in X \right\} \subset F. \]
The inverse image of a subset Y ⊂ F is
\[ f^{-1} (Y) = \left\{ x \in E \, : \ f(x) \in Y \right\} \subset E . \]
Note that this definition does not require f to be invertible, but when it is, the inverse image of Y is also the image of Y by the inverse of f so that the notation is coherent.
Theorem 2: Let T : U ⇾ V be linear transformation.
  1. The inverse image T−1(Y) of any vector subspace Y ⊂ V is a vector subspace of U.
  2. The image T(X) of any vector subspace X ⊂ U is a subspace of V.
  1. Let us consider a subspace Y of V. If x, u ∈ T-1, namely, T(x), T(u) ∈ Y. Then for k ∈ 𝔽, we have \[ T(k{\bf x} + {\bf u}) = k\,T({\bf x}) + T({\bf u}) \in Y. \] This means that kx + u ∈ T-1(Y). This proves the first assertion.
  2. Let now X be any subspace of U. Take two elements of T(X). They can be written in the form T(x) and T(u) for some x, u ∈ X. Then \[ T(k{\bf x} + {\bf u}) = k\,T({\bf x}) + T({\bf u}) \in T(X) \] is in the image of X. Hence, T(X) is a subspace of V.
Example 4: In order to create a linear transformation from ℳ2×2 , we need a basis for M 2,2 . The standard basis
End of Example 4
Corollary 1: For any linear transformation T : U ⇾ V, T−1(0V) is a subspace of U and im(T) = T(U) is a subspace of V.
Corollary 2: For any two linear transformations of 𝔽-vector space U into another 𝔽-vector space V, the coincidence subset \[ \left\{ {\bf x} \in U \ : \ f({\bf x}) = g({\bf x}) \right\} \] is a vector subspace of U.
The coincidence subset of f and g consists precisely of the x ∈ U such that \[ \left( f - g \right) ({\bf x}) = f({\bf x}) - g ({\bf x}) = {\bf 0}_V . \] So it is the inverse image of the subspace {0} by the linear map (f − g ).
Example 4: In order to create a linear transformation from ℳ2×2 , we need a basis for M 2,2 . The standard basis
End of Example 5
Let T : U ⇾ V be a linear transformation of two vector spaces over the same scalar field. The kernel of T is the subspace ker(T) = T−1(0) of U. The image of T is the subspace T(U) of V. The nullity of T is the dimension (when finite) of its kernel. The rank of T is the dimension (when finite) of its image.
If a linear map T : U ⇾ V is injective (one-to-one), then 0 ∈ U is the only element having image 0 ∈ V, hence ker(T) = {0}. There is a converse statement.
Theorem 3: For a linear transformation T : U ⇾ V between two vector spaces, \[ \mbox{ker}(T) = \left\{ {\bf 0} \right\} \qquad \iff \qquad T \mbox{ is injective}. \]
For any linear mao \[ T({\bf x}) = T({\bf y}) \qquad \ff \qquad T({\bf x}) - T({\bf y}) = {\bf 0} \] If ker(T) = {0}, we have \[ T({\bf x}) = T({\bf y}) \qquad \iff \qquad {\bf x} - {\bf y} = {\bf 0} \qquad \iff \qquad {\bf x} = {\bf y} . \] This proves that T is injective.
Example 6: In order to create a linear transformation from ℳ2×2 , we need a basis for M 2,2 . The standard basis
End of Example 6

 

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 ∈ &Fop; and a linear transformation T ∈ ℒ(X, Y), their product is defined by \[ \left( \lambda\,T \right) ({\bf x}) = \lambda\,T({\bf x}) , \qquad \forall {\bf x} \in X . \]
Theorem 4: The set of all linear transformations ℒ(X, Y) is a vector space subject that X and Y are vector spaces over field 𝔽.
Example 7: In order to create a linear transformation from ℳ2×2 , we need a basis for M 2,2 . The standard basis
End of Example 7

 

How to Construct Linear Maps?

Here is a general method for the construction of linear transformations.
Theorem 4: Let S = {u1, u2, … , un} be a linearly independent subset of the vector space U. For any family of vectors v1, v2, … , vn of the same size as S from vector space V, there is a linear transformation T : U ⇾ V such that T(uk) = vk 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 decomposion \( \displaystyle {\bf x} = \sum_{j\in J} c_j {\bf u}_j \) 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 1:
End of Example 1
■
Theorem 4: Let f,g : E ⇾ F be two linear maps. If f and g agree on a subset S c E, then they agree on the linear span of S. If f and g agree on a set of generators of E, then f = g.
The coincidence set of f and g is equal to the subspace ker(f - g) so that both statements follow.
Due to its fundamental importance, we formulate explicitly a statement based on the preceding two propositions.
Corollary 4: Let {e₁, e₂, … , en} be a basis of the vector space <>E. For any choice of family of vectors {b₁, b₂, … , bn} in a vector space F, there is one and only one linear map T : E ⇾ F such that T(ei) = bi for every i = 1, 2, … n.
  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.