Before we dive into the ocean by the name of linear transformations and matrices, it makes sense to remind some definitions and concepts that will be used later.
A functionf : X ↦ Y is a rule that associates with an element x (input) from set X one and only one element y (output) of set Y. We say that f maps element x to the element y, which we denote as y = f(x). The set X is known as domain of f and Y is called codomain.
Codomain (in blue)
For any subset A ⊆ X, we let f(A) = {f(𝑎) ∣ 𝑎 ∈ A}; the set f(A) is called the image of A under f. In particular, y = f(x) is the image of the element x. The image of the domain of f is called the range or image. We denote this set as Im(f) or just Imf.
As fields of scalars we consider one of the following three one-dimensional vector spaces: ℚ, rational numbers, ℝ, real numbers, or ℂ, complex numbers. When it does not matter which of these fields is in use, we denote it by 𝔽 each of them.
Next, we form from each of the scalar spaces a larger vector space---the direct product of n copies of them:
Any two ordered n-tuples (which are lists or finite sequences of numbers) to be regarded as the same, they must list the same numbers in the same order. Recall also that if n = 2, then the n-tuple is called an "ordered pair," and if n = 3, it is called an "ordered triple."
Note that independently which particular realisation is used, the direct product 𝔽n has the same basis:
When 𝔽 is ℝ, the set of real numbers, and n is eithher 2 or 3, we can provide a geometric interpretation to a function. However, when n is larger than 3, or complex numbers are involved, the only practical implementation of a function is to define it algebraically.
An injective function (also known as injection, or one-to-one function) is a function f that maps distinct elements of its domain to distinct elements; that is, f(𝑎) = f(b) implies 𝑎 = b.
A surjective function (also known as surjection, or onto function) is a function f such that for every element y from the codomain of
f there exists an input element x such that f(x) = y.
A bijective functionf : X → Y is a one-to-one (injective) and onto (surjective) mapping of a set X to a set Y
Linear Transformations
Now we turn our attention to functions or transformations of vector spaces that do not “mess up” vector addition and scalar multiplication.
Let V = 𝔽n and U = 𝔽m be vector spaces over the same field 𝔽. We call a function T : V ⇾ U a linear transformation (or operator) from V into U if for all vectors x, y ∈ V
and any scalar α ∈ 𝔽, we have
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}
Example 1:
We consider several examples of linear transformations.
Let V = ℝ≤n[x] be a space of polynomials in variable x with real coefficients of degree no larger than n. Let U = ℝ≤n−1[x] be a similar space of polynomials but with degree up to n−1. We consider the differential operator \( \displaystyle \texttt{D} :\, V \mapsto U \) that is defined by \( \displaystyle \texttt{D}\,p(x) = p'(x) . \) As you know from calculus, this differential operator is linear.
Let V = ℝ≤1[x] be a space of polynomials in variable x with real coefficients of degree 1, and U = ℝ≤2[x]. For any polynomial p(x) = 𝑎x + b ∈ V, we assign another polynomial T(𝑎x + b) = (½)𝑎x² + bx ∈ U.
This transformation T : V ⇾ U is linear because
\begin{align*}
T \left( ax +b + cx +d \right) &= T \left( (a+c)\,x +b + d \right)
\\
&= \frac{1}{2} \left( a + b \right) x^2 + \left( b + d \right) x
\\
&= T \left( ax +b \right) + T \left( cx +d \right) ,
\end{align*}
and for any real number λ, we have
\begin{align*}
T \left( \lambda \left( ax +b \right) \right) &= T \left( \lambda ax + \lambda b \right)
\\
&= \frac{1}{2}\, \lambda a\,x^2 + \lambda b\,x = \lambda T \left( ax +b \right) .
\end{align*}
Let V = ℭ[0, 1] be a space of all continuous real-valued functions on interval [0, 1]. We define a linear transformation φ : V ⇾ ℝ by
\[
\varphi (f) = \int_0^1 f(x)\,{\text d} x . \qquad \forall f \in V.
\]
This linear transformation is known as a functional on V.
Let V = ℝ3,2 be the space of all 3×2 matrices with real entries. We define a linear operator A : V ⇾ V by
\[
A\,\begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix} = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \, \begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix} ,
\]
where it is understood that two matrices are multiplied:
\[
\begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix} \, \begin{bmatrix} a & b \\ c & d \\ e & f \end{bmatrix} = \begin{bmatrix} a + 2c + 3 e & b + 2 d + 3 f \\ 4a + 5 c + 6 e & 4 b + 5 d + 6 f \\ 7a + 8c + 9e & 7b +8 d + 9f \end{bmatrix}
\]
So multiplication by a 3 × 3 mutrix we generate a linear transformation in the vector space of matrices ℝ3,2.
Let us consider a vector space of square matrices with real coefficients: ℝn,n. For any matrix A ∈ ℝn,n, we define a linear transformation
\[
T \left( {\bf A} \right) = {\bf A}^{\mathrm{T}} ,
\]
where "T" means transposition (swap rows and columns).
End of Example 1
We often simply call Tlinear. The space V is referred to as the domain
of the linear transformation and the space U is called codomain of T. For example, if f : ℝn ⇾ ℝm is a real-valued functionthat preserves linearity, we call it real linear transformation because scalars are real numbers. Geometrically, real linear transformations can be thought of as the functions that rotate, stretch, shrink, and/or reflect ℝn, but do so somewhat uniformly. Eq.\eqref{EqTransform.1} tells us that a linear transformation maps straight lines into straight lines because tx + y is a parametric equation of line.
Let T : V ⇾ W be a linear transformation. The kernel (also known as the null space or nullspace) of T, denoted ker(T), is
the vector space of all elements v of V such that
\[
\mbox{ker}(T) = \left\{ {\bf v} \in V \mid T({\bf v}) = 0 \right\} .
\]
Example 2:
Let us consider a linear transformation from ℝ³ into ℝ² defined by
\[
T({\bf x}) = (x_1 + x_2 , x_1 -2 x_3 ) , \qquad {\bf x} = (x_1 , x_2 , x_3 ) \in \mathbb{R}^3 .
\]
If x ∈ ker(T), then
\[
x_1 + x_2 = 0 \qquad \mbox{and} \qquad x_1 -2 x_3 = 0.
\]
Setting the free variable x₃ = t, we get
\[
x_1 = 2\,t \qquad \Longrightarrow \qquad x_2 = -x_1 = -2\,t .
\]
Hence ker(T) is the one-dimensional subspace of ℝ³ consisting of all vectors of the form
\[
\mbox{ker}(T) = \left\{ (2\,t, -2\,t , t \mid t \in \mathbb{R} \right\} .
\]
So the nullspace of T is spanned on the vector (2, −2, 1).
End of Example 2
We summarize
the almost obvious statements about linear transformation in the following proposition.
Theorem 1: Let V and U be a vector spaces and
\( T\,:\, V\to U \) be linear transformation.
If T linear, then \( T(0) =0. \)
T is linear if and only if for \( {\bf x}_1 , \ldots , {\bf x}_n \in V \) and
any real or complex scalars \( a_1 , \ldots , a_n \)
\[
T \left( \sum_{i=1}^r a_i {\bf x}_i \right) = \sum_{i=1}^r a_i T \left( {\bf x}_i \right) .
\]
Example 3:
Let us construct a linear transformation from ℝ² into ℝ³. We choose two basis vectors in ℝ² and corresponding vectors in ℝ³:
\[
T \left( x \begin{bmatrix} -1 \\ \phantom{-}2 \end{bmatrix} + y \begin{bmatrix} 1 \\ 1 \end{bmatrix} \right) = x\begin{bmatrix} -1 \\ \phantom{-}1 \\ \phantom{-}2 \end{bmatrix} + y \begin{bmatrix} 2 \\ 1 \\ 3 \end{bmatrix} =
\]
To prove that T is a linear transformation, all we need to
do is say: by Theorem 1, T is a linear transformation.
An alternative way to write T is as follows:
Now we come a main issue of linear transformations: how practically define a linear map between two vector spaces. To achieve this, we need to transfer direct product form of 𝔽n into matrix form (for any positive integer n ∈ ℤ). Now we treat n-tuples (or lists of size n) as column vectors:
where we denote by 𝔽n×1 = 𝔽n,1 the vector space of column vectors with entries from field 𝔽 (which is either ℚ or ℝ or ℂ). We can also transfer 𝔽n into vector space of row vectors:
where 𝔽1×n = 𝔽1,n the vector space of row vectors.
Of course, 𝔽n×1 and 𝔽1×n, are isomorphic to the direct product, 𝔽n ≌ 𝔽n,1 ≌ 𝔽1,n, but these vector spaces have different structure. Your software package distinguishes an n-tuple (or list of size n) from a column vector or row vector that have dimensions n×1 and 1×n, respectively. In Mathematica, a list of n elements is defined as
v = {1, 2, 3}
{1, 2, 3}
Dimensions[v]
{3}
As you see, v is just a list of three elements or 3-tuple of dimension 3.
A column vector is putting in as
v1 = {{1}, {2}, {3}}
{{1}, {2}, {3}}
Dimensions[v1]
{3, 1}
SameQ[v, v1]
False
We can convert an n-tuple into a row vector as
rowVector = ArrayReshape[v, {1, Length[v]}]
{{1, 2, 3}}
Dimensions[%]
{1, 3}
rowVector1 = {v}
{{1, 2, 3}}
Converting a list into column vector can be acomplished as follows:
colVector = ArrayReshape[v, {Length[v], 1}]
{{1}, {2}, {3}}
Dimensions[colVector]
{3, 1}
colVecto1r = List /@ v
{{1}, {2}, {3}}
As it is seen from Mathematica's codes, this computer algebra system distinguishes lists from column vectors and row vectors. Recall that we denote by 𝔽1,n = 𝔽1×n the vector space of row vectors with entries from filed 𝔽.
Example: The span of the empty set \( \varnothing \) consists of a unique element 0.
Therefore, \( \varnothing \) is linearly independent and it is a basis for the trivial vector space
consisting of the unique element---zero. Its dimension is zero.
Example: Recall from section on Vector Spaces that the set of
all ordered n-tuples or real numbers is denoted by the symbol \( \mathbb{R}^n . \) It is a custom
to represent ordered n-tuples in matrix notation as column vectors. For example, the matrix
can be used as an alternative to \( {\bf v} = \left[ v_1 , v_2 , \ldots , v_n \right] \quad\mbox{or}\quad
{\bf v} = \left( v_1 , v_2 , \ldots , v_n \right) . \) The latter is called the comma-delimited form of a
vector and former is called the column-vector form.
In \( \mathbb{R}^n , \) the vectors
\( e_1 [1,0,0,\ldots , 0] , \quad e_2 =[0,1,0,\ldots , 0], \quad \ldots , e_n =[0,0,\ldots , 0,1] \)
form a basis for n-dimensional real space, and it is called the standard basis. Its dimension is n.
For example, the vectors
If f is a function with domain \( \mathbb{R}^m \) and codomain
\( \mathbb{R}^n , \) then we say that f is a transformation from
\( \mathbb{R}^m \) to \( \mathbb{R}^n \) or that f
maps \( \mathbb{R}^m \) into \( \mathbb{R}^n , \) which we
denote by writing \( f\,:\, \mathbb{R}^m \to \mathbb{R}^n . \) In the special case where
n = m, a transformation is sometimes called an operator on \( \mathbb{R}^n . \)
Although the latter represents a linear system of equations, we could view it instead as a transformation that maps a
vector x from \( \mathbb{R}^m \) into the vector from
\( \mathbb{R}^n \) by multiplying x on the left by A.
We call this a matrix transformation and denote by \( T_{\bf A}:\, \mathbb{R}^m \to \mathbb{R}^n \)
(in case where m=n, it is called matrix operator). This transformation is generated by
matrix multiplication.
Theorem: Let \( T:\, \mathbb{R}^m \to \mathbb{R}^n \) be a
linear transformation. Then there exists a unique matrix A such that
\[
T \left( {\bf x} \right) = {\bf A}\, {\bf x} \qquad\mbox{for all } {\bf x} \in \mathbb{R}^m .
\]
In fact, A is the \( n \times m \) matrix whose j-th column is the
vector \( T\left( {\bf e}_j \right) , \) where \( {\bf e}_j \) is the
j-th column of the identity matrix in \( \mathbb{R}^m : \)
\[
{\bf A} = \left[ T \left( {\bf e}_1 \right) , T \left( {\bf e}_2 \right) , \cdots , T \left( {\bf e}_m \right) \right] .
\]
Write \( {\bf x} = {\bf I}_m {\bf x} = \left[ {\bf e}_1 \ \cdots \ {\bf e}_m \right] {\bf x} =
x_1 {\bf e}_1 + \cdots + x_m {\bf e}_m , \) and use the linearity of T to compute
Such representation is unique, which could be proved by showing that for any other matrix representation Bx
of transformation T, it follows that A = B.
Example: The transformation T from \( \mathbb{R}^4 \) to \( \mathbb{R}^3 \)
defined by the equations
so multiplication by zero maps every vector from \( \mathbb{R}^m \) into the zero vector
in \( \mathbb{R}^n . \) Such transformation is called the zero transformation from
\( \mathbb{R}^m \) to \( \mathbb{R}^n . \)
■
Example: If I is the \( n \times n \)
identity matrix, then
■
Matrices of Linear Transformations
Theorem: For every matrix A the matrix transformation
\( T_{\bf A}:\, \mathbb{R}^m \to \mathbb{R}^n \) has the following properties for all vectors
v and u and for every scalar k:
Theorem: \( T\,:\, \mathbb{R}^m \to \mathbb{R}^n \)
is a matrix transformation if and only if the following relationships hold for all vectors v and u
in \( \mathbb{R}^m \) and for every scalar k:
\( T \left( {\bf v} + {\bf u} \right) = T \left( {\bf v} \right) + T \left( {\bf u} \right) \) (Additivity property),
where \( {\bf e}_i , \quad i=1,2,\ldots , m , \) are the standard basis vectors for
\( \mathbb{R}^m . \) We know that Ax is a
linear combination of the columns of A in which the successive coefficients are the entries
\( x_1 , x_2 , \ldots , x_m \) of x. That is,
\[
{\bf A}\,{\bf x} = x_1 T \left( {\bf e}_1 \right) + x_2 T \left( {\bf e}_2 \right) + \cdots + x_m T \left( {\bf e}_m \right) .
\]
Theorem: Every linear transformation from \( \mathbb{R}^m \) to
\( \mathbb{R}^n \) is a matrix transformation, and conversely, every matrix transformation from
\( \mathbb{R}^m \) to
\( \mathbb{R}^n \) is a linear transformation.
Theorem: If \( T_{\bf A}\,:\, \mathbb{R}^m \to \mathbb{R}^n \) and
\( T_{\bf B}\,:\, \mathbb{R}^m \to \mathbb{R}^n \) are matrix transformations,
and if \( T_{\bf A} \left( {\bf v} \right) = T_{\bf B} \left( {\bf v} \right) \)
for every vector \( {\bf v} \in \mathbb{R}^m , \) then A = B.
To say that \( T_{\bf A} \left( {\bf v} \right) = T_{\bf B} \left( {\bf v} \right) \) for every
vector in \( \mathbb{R}^m \) is the same as saying that
\[
{\bf A}\,{\bf v} = {\bf B}\,{\bf v}
\]
for every vector v in \( \mathbb{R}^m . \) This will be true, in particular, if v
is any of the standard basis vectors \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_m \) for
\( \mathbb{R}^m ; \) that is,
Since every entry of ej is 0 except for the j-th, which is 1, it follows
that Aej is the j-th column of A and
Bej is the j-th column of B. Thus,
\( {\bf A}\,{\bf e}_j = {\bf B}\,{\bf e}_j \) implies that corresponding columns of A
and B are the same, and hence A = B.
The above theorem tells us that there is a one-to-one correspondence
between n-by-m matrices and matrix transformations from
ℝm to ℝn in the sense that every
n×m\( n \times m \)
matrix A generates exactly one matrix transformation
(multiplication by A) and every matrix transformation from ℝm to
ℝn arises from exactly one \( n \times m \)
matrix: we call that matrix the standard matrix for the transformation, which is given by the formula:
\[
{\bf A} = \left[ T \left( {\bf e}_1 \right) \,|\, T \left( {\bf e}_2 \right) \,|\, \cdots \,| T \left( {\bf e}_m \right) \right] .
\]
This suggests the following procedure for finding standard matrices.
Algorithm for finding the standard matrix of a linear transformation:Step 1: Find the images of the standard basis vectors \( {\bf e}_1 , {\bf e}_2 , \ldots , {\bf e}_m \) for
\( \mathbb{R}^m . \) Step 2: Construct the matrix that has the images obtained in Step 1 as its successive
columns. This matrix is the standard matrix for the transformation.
Example: Find the standard matrix A for the linear transformation:
Example: Over the field of complex numbers, the vector space \( \mathbb{C} \)
of all complex numbers has dimension 1 because its basis consists of one element \( \{ 1 \} . \)
Over the field of real numbers, the vector space \( \mathbb{C} \) of all complex numbers
has dimension 2 because its basis consists of two elements \( \{ 1, {\bf j} \} . \)
Does the transformation satisfy the property T(cA) =
cT(A) for a constant c ?
\[
\mbox{(a)}\quad \begin{bmatrix}
x_{1} \\
x_{2} \\
\end{bmatrix} \,\mapsto \,
\begin{bmatrix}
x_{1}+2x_{2} \\
3x_{2}
\end{bmatrix}; \qquad \mbox{(b)}\quad \begin{bmatrix}
x_{1} \\
x_{2}
\end{bmatrix} \,\mapsto \,
\begin{bmatrix}
x_{1}^2 \\
x_{2}
\end{bmatrix}.
\]
Is the following transformation linear? Check that both properties
T(A + B)
= T(A) + T(B) and T(cA) =
cT(A) are satisfied.
\[
\mbox{(a)}\quad \begin{bmatrix}
x_{1} \\
x_{2} \\
x_{3} \\
\end{bmatrix} \,\mapsto \,
\begin{bmatrix}
4x_{1}+x_{3}\\
-2x_{1}+3x_{2}
\end{bmatrix}\qquad \mbox{(b)}\quad
\]
Find the composition of transformations S with T, i.e.
S∘T.
