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# Matrix Transformations

Let V and U be vector spaces. We call a function $$T\,:\,V \to U$$ a linear transformation (or operator) from V into U if for all $${\bf x} , {\bf y} \in V$$ any scalar α, we have
• $$T({\bf x} + {\bf y}) = T({\bf x}) + T({\bf y})$$  (Additivity property),
• $$T( \alpha\,{\bf x} ) = \alpha\,T( {\bf x} )$$          (Homogeneity property).
We often simply call T linear. The space V is referred to as the domain of the lienar transformation and the space U is called codomain of T. We summarize the almost obvious statements about linear transformation in the following proposition.

Theorem: Let V and U be a vector spaces and $$T\,:\, V\to U$$ be linear transformation.

1. If T linear, then $$T(0) =0.$$
2. T is linear if and only if $$T(c{\bf x} + {\bf y} ) = c\,T({\bf x}) + T({\bf y})$$ for all $${\bf x}, {\bf y} \in V$$ and any scalar c.
3. T is linear if and only if for $${\bf x}_1 , \ldots , {\bf x}_n \in V$$ and any real or complex scalars $$a_1 , \ldots , a_n$$
$T \left( \sum_{i=1}^n a_i {\bf x}_i \right) = \sum_{i=1}^n a_i T \left( {\bf x}_i \right) .$

Recall that a set of vectors β is said to generate or span a vector space V if every element from V can be represented as a linear combination of vectors from β.

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 vectorsd. For example, the matrix
${\bf v} = \begin{bmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{bmatrix} \qquad\mbox{or}\qquad {\bf v} = \begin{pmatrix} v_1 \\ v_2 \\ \vdots \\ v_n \end{pmatrix}$
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
${\bf e}_1 = {\bf i} = \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} \qquad {\bf e}_2 = {\bf j} = \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} \qquad {\bf e}_3 = {\bf k} = \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix}$
form the standard basis vectors for $$\mathbb{R}^3 .$$ Any three-dimensional vecor could expressed through these basic vectors:
${\bf x} = \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} = \begin{bmatrix} x_1 \\ 0 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ x_2 \\ 0 \end{bmatrix} + \begin{bmatrix} 0 \\ 0 \\ x_3 \end{bmatrix} = x_1 \begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} + x_2 \begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} + x_3 \begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = x_1 {\bf e}_1 + x_2 {\bf e}_2 + x_3 {\bf e}_3 = x_1 {\bf i} + x_2 {\bf j} + x_3 {\bf k} .$
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 wroting $$f\,:\, \mathbb{R}^m \to \mathbb{R}^n .$$ In the special case where n = m, a transformation is sometimes called an operator on $$\mathbb{R}^n .$$
Suppose we have the system of linear equations
\begin{eqnarray*} a_{11} x_1 + a_{12} x_2 + \cdots + a_{1m} x_m &=& b_1 , \\ a_{21} x_1 + a_{22} x_2 + \cdots + a_{2m} x_m &=& b_2 , \\ \vdots &=& \vdots \\ a_{n1} x_1 + a_{12} x_2 + \cdots + a_{1m} x_m &=& b_n , \\ \end{eqnarray*}
which can be written in matrix notation as
$\begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1m} \\ a_{21} & a_{22} & \cdots & a_{2m} \\ \vdots & \vdots & \ddots & \vdots \\ a_{n1} & a_{n2} & \cdots & a_{nm} \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_m \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ \vdots \\ b_n \end{bmatrix} ,$
or more briefly as
${\bf A}\, {\bf x} = {\bf b} .$
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 usethe linearlity of T to compute
\begin{align*} T \left( {\bf x} \right) &= T \left( x_1 {\bf e}_1 + \cdots + x_m {\bf e}_m \right) = x_1 T \left( {\bf e}_1 \right) + \cdots + x_m T \left( {\bf e}_m \right) \\ &= \left[ T \left( {\bf e}_1 \right) \ \cdots \ T \left( {\bf e}_m \right) \right] \begin{bmatrix} x_1 \\ \vdots \\ x_m \end{bmatrix} = {\bf A} \, {\bf x} . \end{align*}
Such representation is unique, which could be proved by showing that for any other matrix representation B x of transformation T, it follows that A = B.

Example: The transformation T from $$\mathbb{R}^4$$ to $$\mathbb{R}^3$$ defined by the equations
\begin{eqnarray*} 3\, x_1 -2\, x_2 + 5\, x_3 - 7\, x_4 &=& b_1 , \\ x_1 + 7\, x_2 -3\, x_3 + 5\, x_4 &=& b_2 , \\ 4\, x_1 -3\, x_2 + x_3 -6\, x_4 &=& b_3 , \\ \end{eqnarray*}
can be represented in matrix form as
$\begin{bmatrix} 3 & -2 & 5 & -7 \\ 1 & 7 & -3 & 5 \\ 4 & -3 & 1 & -6 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \\ x_4 \end{bmatrix} = \begin{bmatrix} b_1 \\ b_2 \\ b_3 \end{bmatrix} ,$
from which we see that the transformation can be interpreted as a multiplication by
${\bf A} = \begin{bmatrix} 3 & -2 & 5 & -7 \\ 1 & 7 & -3 & 5 \\ 4 & -3 & 1 & -6 \end{bmatrix} .$
Therefore, the transformation T is generated by matrix A. For example, if
${\bf x} = \begin{bmatrix} -1 \\ 2 \\ 3 \\ -4 \end{bmatrix} ,$
then
$T_{\bf A} \left( {\bf x} \right) = {\bf A}\, {\bf x} = \begin{bmatrix} 3 & -2 & 5 & -7 \\ 1 & 7 & -3 & 5 \\ 4 & -3 & 1 & -6 \end{bmatrix} \begin{bmatrix} -1 \\ 2 \\ 3 \\ -4 \end{bmatrix} = \begin{bmatrix} 36 \\ -16 \\ 17 \end{bmatrix} .$

Example: If 0 is the $$n \times m$$ zero matrix, then
$T_{\bf 0} \left( {\bf x} \right) = {\bf 0}\,{\bf x} = {\bf 0} ,$
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 ■

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 anf u and for every scalar k:

1. $$T_{\bf A} \left( {\bf 0} \right) = {\bf 0} .$$
2. $$T_{\bf A} \left( k{\bf u} \right) = k\,T_{\bf A} \left( {\bf u} \right) .$$
3. $$T_{\bf A} \left( {\bf v} \pm {\bf u} \right) = T_{\bf A} \left( {\bf v} \right) \pm T_{\bf A} \left( {\bf u} \right) .$$

All parts are restatements of the following properties of matrix arithmetics:
${\bf A}{\bf 0} = {\bf 0}, \qquad {\bf A}\left( k{\bf u} \right) = k \left( {\bf A}\, {\bf u} \right) , \qquad {\bf A} \left( {\bf v} \pm {\bf u} \right) = {\bf A} \left( {\bf v} \right) \pm {\bf A} \left( {\bf u} \right) .$

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:

1. $$T \left( {\bf v} + {\bf u} \right) = T \left( {\bf v} \right) + T \left( {\bf u} \right)$$  (Additivity property),
2. $$T \left( k{\bf v} \right) = k\,T \left( {\bf v} \right)$$          (Homogeneity property).

If T is a matrix transformation, then properties (i) and (ii) follow respectively from previous theorem.

Conversely, assume that properties (i) and (ii) are valid. We must show that there exists an n-by-m matrix A such that

$T \left( {\bf x} \right) = {\bf A}\, {\bf x}$
for every $${\bf x} \in \mathbb{R}^m .$$ Since T is a linear transformation, we have
$T \left( k_1 {\bf x}_1 + k_2 {\bf x}_2 + \cdots + k_m {\bf x}_m \right) = k_1 T \left( {\bf x}_1 \right) + k_2 T \left( {\bf x}_2 \right) + \cdots + k_m T \left( {\bf x}_m \right)$
for all vectors $${\bf x}_i , \quad i=1,2,\ldots , m ,$$ and for all scalars $$k_i , \quad i=1,2,\ldots , m .$$ Let A be the matrix
${\bf A} = \left[ T \left( {\bf e}_1 \right) \, | \, T \left( {\bf e}_2 \right) \, | \, \cdots \, T \left( {\bf e}_m \right) \right] ,$
where $${\bf e}_i , \quad i=1,2,\ldots , m ,$$ are the staaaaaandard basis vectors for $$\mathbb{R}^m .$$ We know that A x 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,
${\bf A}\,{\bf x} = x_1 T \left( {\bf e}_1 \right) + x_2 T \left( {\bf e}_2 \right) + \cdots + x_m T \left( {\bf e}_m \right) .$
Using linearlity of T, we have
${\bf A}\,{\bf x} = T \left( x_1 {\bf e}_1 + x_2 {\bf e}_2 + \cdots + x_m {\bf e}_m \right) = T \left( {\bf x} \right) ,$
which completes the proof.

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,
${\bf A}\,{\bf e}_j = {\bf B}\,{\bf e}_j \qquad (j=1,2,\ldots , m) .$
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 coerresponding 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 $$\mathbb{R}^m$$ to $$\mathbb{R}^n$$ in teh sense that every $$n \times m$$ matrix A generates exactly one matrix transformation (multiplication by A) and every matrix transformation from $$\mathbb{R}^m$$ to $$\mathbb{R}^n$$ 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:
$T \left( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \right) = \begin{bmatrix} 3\,x_1 -2\,x_3 \\ 2\,x_2 + 5\,x_3 \end{bmatrix} .$

To answer the question, we apply the linear transformation T to every basic vector:

$T \left( {\bf e}_1 \right) = 3\, {\bf i} = \begin{bmatrix} 3 \\ 0 \end{bmatrix} , \quad T \left( {\bf e}_2 \right) = 2\, {\bf j} = \begin{bmatrix} 0 \\ 2 \end{bmatrix} , \quad T \left( {\bf e}_3 \right) = -2\, {\bf i} + 5\,{\bf j} = \begin{bmatrix} -2 \\ 5 \end{bmatrix} .$
Therefore, the standard matrix becomes
$T \left( \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} \right) = {\bf A}\,{\bf x} = \begin{bmatrix} 3&0&-2 \\ 0&2&5 \end{bmatrix} \begin{bmatrix} x_1 \\ x_2 \\ x_3 \end{bmatrix} .$

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 $$\{ 1, {\bf j} \} .$$