Let \( V \) and \( W \) be vector spaces over the same field \( \mathbb{F} \) (typically \( \mathbb{R} \) for control engineering). A mapping \( T: V \to W \) is a linear transformation if it satisfies:

\[ \forall \mathbf{u},\mathbf{v}\in V,\; \forall \alpha\in\mathbb{F}:\quad T(\mathbf{u}+\mathbf{v}) = T(\mathbf{u}) + T(\mathbf{v}),\qquad T(\alpha \mathbf{v}) = \alpha\, T(\mathbf{v}). \]

These two axioms imply several essential identities used repeatedly in control-matrix reasoning (e.g., “superposition”):

\[ \begin{aligned} &T(\mathbf{0})=\mathbf{0}, \\ &T(-\mathbf{v})=-T(\mathbf{v}), \\ &T\!\left(\sum_{k=1}^{m}\alpha_k \mathbf{v}_k\right)=\sum_{k=1}^{m}\alpha_k\,T(\mathbf{v}_k)\quad (\text{finite linear combinations}). \end{aligned} \]

Proof (derivations from axioms):

1) Since \( \mathbf{0} = \mathbf{0} + \mathbf{0} \), linearity gives \( T(\mathbf{0}) = T(\mathbf{0}) + T(\mathbf{0}) \), hence subtracting \( T(\mathbf{0}) \) from both sides yields \( T(\mathbf{0})=\mathbf{0} \).

2) Because \( -\mathbf{v} = (-1)\mathbf{v} \), we have \( T(-\mathbf{v}) = T((-1)\mathbf{v}) = (-1)T(\mathbf{v}) = -T(\mathbf{v}) \).

3) Apply additivity repeatedly and homogeneity term-by-term to obtain the linear-combination property.

flowchart TD
  A["Define a map T: V to W"] --> B["Check linearity axioms"]
  B -->|"T(u+v)=T(u)+T(v)"| C["Additivity holds"]
  B -->|"T(a v)=a T(v)"| D["Homogeneity holds"]
  C --> E["Superposition: T(sum a_k v_k)=sum a_k T(v_k)"]
  D --> E
  E --> F["Choose bases and compute matrix representation"]
  

In modern control, the transformations that map vectors to vectors (e.g., “state to state,” “input to state,” “state to output”) are linear in the models we will build later; this lesson provides the algebraic backbone.

Two canonical subspaces associated with a linear transformation \( T:V\to W \) are the kernel and the image (also called range):

\[ \ker(T) \triangleq \{\mathbf{v}\in V \mid T(\mathbf{v})=\mathbf{0}\},\qquad \operatorname{Im}(T) \triangleq \{\,T(\mathbf{v})\in W \mid \mathbf{v}\in V\,\}. \]

Theorem 2.1: \( \ker(T) \) is a subspace of \( V \) and \( \operatorname{Im}(T) \) is a subspace of \( W \).

Proof:

Kernel. (i) \( \mathbf{0}\in\ker(T) \) since \( T(\mathbf{0})=\mathbf{0} \). (ii) If \( \mathbf{u},\mathbf{v}\in\ker(T) \), then \( T(\mathbf{u}+\mathbf{v})=T(\mathbf{u})+T(\mathbf{v})=\mathbf{0}+\mathbf{0}=\mathbf{0} \), so \( \mathbf{u}+\mathbf{v}\in\ker(T) \). (iii) If \( \mathbf{v}\in\ker(T) \) and \( \alpha\in\mathbb{F} \), then \( T(\alpha \mathbf{v})=\alpha T(\mathbf{v})=\alpha\mathbf{0}=\mathbf{0} \), so \( \alpha\mathbf{v}\in\ker(T) \). Hence \( \ker(T) \) is a subspace.

Image. (i) \( \mathbf{0}\in\operatorname{Im}(T) \) since \( \mathbf{0}=T(\mathbf{0}) \). (ii) If \( \mathbf{w}_1=T(\mathbf{u}) \) and \( \mathbf{w}_2=T(\mathbf{v}) \) are in the image, then \( \mathbf{w}_1+\mathbf{w}_2=T(\mathbf{u})+T(\mathbf{v})=T(\mathbf{u}+\mathbf{v}) \), so it is also in the image. (iii) For \( \alpha\in\mathbb{F} \), \( \alpha \mathbf{w}_1 = \alpha T(\mathbf{u}) = T(\alpha \mathbf{u}) \), thus the image is a subspace.

The kernel characterizes “directions” in \( V \) that are annihilated by the transformation; in later control lessons, such subspaces will encode “invisible” or “ineffective” components under certain mappings.

Let \( \mathcal{B}=\{\mathbf{b}_1,\dots,\mathbf{b}_n\} \) be a basis of a finite-dimensional vector space \( V \) (from Lesson 1). Every vector \( \mathbf{v}\in V \) can be written uniquely as:

\[ \mathbf{v} = \sum_{i=1}^{n} x_i \mathbf{b}_i,\quad \text{with unique scalars } x_1,\dots,x_n\in\mathbb{F}. \]

Theorem 3.1 (Basis-determination theorem): If \( T:V\to W \) is linear, then \( T \) is completely determined by the values \( T(\mathbf{b}_1),\dots,T(\mathbf{b}_n) \).

Proof: For any \( \mathbf{v}=\sum_{i=1}^{n} x_i \mathbf{b}_i \), linearity implies

\[ T(\mathbf{v}) = T\!\left(\sum_{i=1}^{n} x_i \mathbf{b}_i\right) = \sum_{i=1}^{n} x_i\, T(\mathbf{b}_i). \]

Therefore, once the basis images are fixed, \( T(\mathbf{v}) \) is fixed for every \( \mathbf{v}\in V \).

This theorem is the conceptual bridge to matrices: the “data” of a linear map in finite dimensions is a finite set of vectors (the basis images), which will become the columns of a matrix.

For the most common control-engineering setting, take \( V=\mathbb{R}^n \) and \( W=\mathbb{R}^m \) with the standard bases \( \mathbf{e}_1,\dots,\mathbf{e}_n \) and \( \mathbf{f}_1,\dots,\mathbf{f}_m \). A linear transformation \( T:\mathbb{R}^n\to\mathbb{R}^m \) can be represented by a matrix \( \mathbf{A}\in\mathbb{R}^{m\times n} \).

Theorem 4.1 (Existence and uniqueness of a matrix representation): For every linear transformation \( T:\mathbb{R}^n\to\mathbb{R}^m \), there exists a unique matrix \( \mathbf{A}\in\mathbb{R}^{m\times n} \) such that for all \( \mathbf{x}\in\mathbb{R}^n \),

\[ T(\mathbf{x}) = \mathbf{A}\mathbf{x}. \]

Proof (constructive): Define the columns of \( \mathbf{A} \) by the images of the standard basis vectors:

\[ \mathbf{A} \triangleq \begin{bmatrix} T(\mathbf{e}_1) & T(\mathbf{e}_2) & \cdots & T(\mathbf{e}_n)\end{bmatrix}. \]

Any \( \mathbf{x}\in\mathbb{R}^n \) can be written as \( \mathbf{x}=\sum_{i=1}^{n} x_i \mathbf{e}_i \). By linearity,

\[ T(\mathbf{x}) = \sum_{i=1}^{n} x_i\, T(\mathbf{e}_i) = \begin{bmatrix} T(\mathbf{e}_1) & \cdots & T(\mathbf{e}_n)\end{bmatrix} \begin{bmatrix} x_1 \\ \vdots \\ x_n \end{bmatrix} = \mathbf{A}\mathbf{x}. \]

Uniqueness: if \( \mathbf{A}\mathbf{x}= \mathbf{B}\mathbf{x} \) for all \( \mathbf{x} \), then in particular for \( \mathbf{x}=\mathbf{e}_i \) we get \( \mathbf{A}\mathbf{e}_i=\mathbf{B}\mathbf{e}_i \), meaning the \( i \)-th columns are equal for all \( i \), so \( \mathbf{A}=\mathbf{B} \).

Practical interpretation: the \( i \)-th column of \( \mathbf{A} \) is the output produced by feeding the transformation the \( i \)-th basis direction.

Because Lesson 1 established bases and coordinates, we can represent vectors by their coordinate vectors. Let \( \mathcal{B}=\{\mathbf{b}_1,\dots,\mathbf{b}_n\} \) be a basis of \( V \). The coordinate vector of \( \mathbf{v}\in V \) in basis \( \mathcal{B} \) is \( [\mathbf{v}]_{\mathcal{B}} \in \mathbb{F}^n \):

\[ \mathbf{v}=\sum_{i=1}^{n} x_i \mathbf{b}_i \quad \Longleftrightarrow \quad [\mathbf{v}]_{\mathcal{B}} \triangleq \begin{bmatrix} x_1\\ \vdots\\ x_n \end{bmatrix}. \]

Now let \( T:V\to W \) be linear, and let \( \mathcal{C}=\{\mathbf{c}_1,\dots,\mathbf{c}_m\} \) be a basis of \( W \). The matrix of \( T \) relative to \( (\mathcal{B},\mathcal{C}) \) is the unique matrix \( [T]_{\mathcal{C}\leftarrow \mathcal{B}}\in\mathbb{F}^{m\times n} \) satisfying:

\[ [\,T(\mathbf{v})\,]_{\mathcal{C}} = [T]_{\mathcal{C}\leftarrow \mathcal{B}}\; [\mathbf{v}]_{\mathcal{B}},\qquad \forall \mathbf{v}\in V. \]

Construction: write each basis image \( T(\mathbf{b}_i) \) in the output basis \( \mathcal{C} \):

\[ [\,T(\mathbf{b}_i)\,]_{\mathcal{C}} = \begin{bmatrix} a_{1i}\\ \vdots\\ a_{mi}\end{bmatrix} \quad \Longrightarrow \quad [T]_{\mathcal{C}\leftarrow \mathcal{B}} = \begin{bmatrix} a_{11} & a_{12} & \cdots & a_{1n}\\ \vdots & \vdots & \ddots & \vdots\\ a_{m1} & a_{m2} & \cdots & a_{mn} \end{bmatrix}. \]

We will study how matrix representations change when bases change in a later lesson (change of coordinates). Here, the key point is: once bases are fixed, linear transformations become matrices and vector mapping becomes matrix-vector multiplication.

Let \( T:V\to W \) and \( S:W\to U \) be linear transformations. Their composition \( S\circ T:V\to U \) is defined by \( (S\circ T)(\mathbf{v}) \triangleq S(T(\mathbf{v})) \).

Theorem 6.1: If \( V=\mathbb{F}^n \), \( W=\mathbb{F}^m \), \( U=\mathbb{F}^p \), and \( T(\mathbf{x})=\mathbf{A}\mathbf{x} \), \( S(\mathbf{y})=\mathbf{B}\mathbf{y} \), then

\[ (S\circ T)(\mathbf{x}) = \mathbf{B}\mathbf{A}\mathbf{x}. \]

Proof: Substitute \( T(\mathbf{x})=\mathbf{A}\mathbf{x} \) into \( S \):

\[ (S\circ T)(\mathbf{x}) = S(T(\mathbf{x})) = S(\mathbf{A}\mathbf{x}) = \mathbf{B}(\mathbf{A}\mathbf{x}) = (\mathbf{B}\mathbf{A})\mathbf{x}. \]

Therefore, matrix multiplication is not an arbitrary rule: it is the algebraic encoding of “apply one linear transformation, then another.” This becomes central in control when multiple linear mappings are chained (e.g., coordinate extraction, actuator mixing, sensor mixing).

A linear transformation \( T:V\to V \) is invertible if there exists a map \( T^{-1}:V\to V \) such that \( T^{-1}(T(\mathbf{v}))=\mathbf{v} \) and \( T(T^{-1}(\mathbf{v}))=\mathbf{v} \) for all \( \mathbf{v}\in V \).

Theorem 7.1: If \( T(\mathbf{x})=\mathbf{A}\mathbf{x} \) for a square matrix \( \mathbf{A}\in\mathbb{F}^{n\times n} \), then \( T \) is invertible if and only if \( \mathbf{A} \) is invertible, and in that case \( T^{-1}(\mathbf{x})=\mathbf{A}^{-1}\mathbf{x} \).

Proof:

(Only if) Assume \( T \) invertible with inverse \( T^{-1} \). Since \( T^{-1} \) is also linear (can be proven by applying \( T \) to both sides of the linearity condition), its matrix representation is some \( \mathbf{B} \) such that \( T^{-1}(\mathbf{x})=\mathbf{B}\mathbf{x} \). Then for all \( \mathbf{x} \):

\[ \mathbf{x} = T^{-1}(T(\mathbf{x})) = \mathbf{B}(\mathbf{A}\mathbf{x}) = (\mathbf{B}\mathbf{A})\mathbf{x} \quad \Longrightarrow \quad \mathbf{B}\mathbf{A}=\mathbf{I}. \]

Similarly, \( T(T^{-1}(\mathbf{x}))=\mathbf{x} \) implies \( \mathbf{A}\mathbf{B}=\mathbf{I} \). Hence \( \mathbf{B}=\mathbf{A}^{-1} \), so \( \mathbf{A} \) is invertible.

(If) If \( \mathbf{A} \) is invertible, define \( T^{-1}(\mathbf{x})=\mathbf{A}^{-1}\mathbf{x} \). Then \( T^{-1}(T(\mathbf{x}))=\mathbf{A}^{-1}\mathbf{A}\mathbf{x}=\mathbf{x} \) and \( T(T^{-1}(\mathbf{x}))=\mathbf{A}\mathbf{A}^{-1}\mathbf{x}=\mathbf{x} \). Thus \( T \) is invertible.

A useful characterization connecting to Section 2 is: \( T \) is injective iff \( \ker(T)=\{\mathbf{0}\} \). (This follows immediately: if \( T(\mathbf{u})=T(\mathbf{v}) \) then \( T(\mathbf{u}-\mathbf{v})=\mathbf{0} \).)

Even before formal state-space modeling, many control setups rely on linear operators that “select,” “mix,” or “rotate” vector quantities.

(a) Selection (pick components): For \( \mathbf{x}\in\mathbb{R}^n \), selecting components \( i_1,\dots,i_k \) is linear:

\[ \mathbf{y} = \mathbf{S}\mathbf{x},\quad \mathbf{S}\in\mathbb{R}^{k\times n}\ \text{has rows chosen from}\ \mathbf{I}_n. \]

(b) Mixing (actuator/sensor combination): A mixing matrix \( \mathbf{M} \) forms linear combinations:

\[ \mathbf{y} = \mathbf{M}\mathbf{u},\quad y_i = \sum_{j=1}^{m} m_{ij} u_j. \]

(c) Planar rotation: A rotation by angle \( \theta \) is linear and represented by

\[ \mathbf{R}(\theta)= \begin{bmatrix} \cos\theta & -\sin\theta\\ \sin\theta & \cos\theta \end{bmatrix},\qquad \mathbf{x}_{\text{rot}}=\mathbf{R}(\theta)\mathbf{x}. \]

These matrices will later appear as parts of larger block structures in modern control models; the key is that they are linear transformations and therefore behave predictably under composition.

We implement: (i) building \( \mathbf{A} \) from basis images, (ii) verifying linearity numerically, and (iii) composing transformations via matrix multiplication. In later chapters, the same matrix objects will feed directly into state-space computations (common packages include numpy, scipy, sympy, and control-oriented tools such as python-control).

import numpy as np

def build_matrix_from_images(images):
    """
    images: list of m-vectors [T(e1), T(e2), ..., T(en)]
    returns A in R^{m x n} with columns equal to images
    """
    return np.column_stack(images)

# Example: T: R^3 -> R^2 given by
# T(e1) = [1, 0]^T, T(e2) = [2, -1]^T, T(e3) = [0, 3]^T
Te1 = np.array([1.0, 0.0])
Te2 = np.array([2.0, -1.0])
Te3 = np.array([0.0, 3.0])

A = build_matrix_from_images([Te1, Te2, Te3])

x = np.array([1.0, -2.0, 0.5])
Tx = A @ x

print("A=\n", A)
print("x=", x)
print("T(x)=A x =", Tx)

# Numerical linearity check on random vectors
rng = np.random.default_rng(0)
u = rng.standard_normal(3)
v = rng.standard_normal(3)
alpha = 1.7

left = A @ (u + v)
right = (A @ u) + (A @ v)
print("Additivity error norm:", np.linalg.norm(left - right))

left2 = A @ (alpha * v)
right2 = alpha * (A @ v)
print("Homogeneity error norm:", np.linalg.norm(left2 - right2))

# Composition: S: R^2 -> R^2 with matrix B
B = np.array([[0.0, 1.0],
              [-1.0, 0.0]])  # 90-degree rotation in the plane

# (S o T)(x) = B (A x) = (B A) x
BA = B @ A
print("Composite matrix BA=\n", BA)
print("Composite output:", (BA @ x))
      

For exact algebraic verification (symbolic matrices), use sympy to prove identities without floating-point error.

import sympy as sp

A = sp.Matrix([[1, 2, 0],
               [0, -1, 3]])
B = sp.Matrix([[0, 1],
               [-1, 0]])

x1, x2, x3 = sp.symbols('x1 x2 x3')
x = sp.Matrix([x1, x2, x3])

expr1 = B*(A*x)
expr2 = (B*A)*x
print("Symbolic equality check:", sp.simplify(expr1 - expr2) == sp.Matrix([0, 0]))
      

The lesson’s concepts map directly to standard linear-algebra tooling used throughout modern control implementations. Below are minimal, idiomatic examples for each environment.

10.1 C++ (Eigen)

#include <iostream>
#include <Eigen/Dense>

int main() {
  // T: R^3 -> R^2 with matrix A
  Eigen::Matrix<double, 2, 3> A;
  A << 1, 2, 0,
       0,-1, 3;

  Eigen::Vector3d x;
  x << 1.0, -2.0, 0.5;

  Eigen::Vector2d Tx = A * x;

  std::cout << "A=\n" << A << "\n";
  std::cout << "x=\n" << x << "\n";
  std::cout << "T(x)=A x=\n" << Tx << "\n";

  // Composition with S: R^2 -> R^2
  Eigen::Matrix2d B;
  B << 0, 1,
      -1, 0;

  Eigen::Matrix<double, 2, 3> BA = B * A;
  std::cout << "Composite BA=\n" << BA << "\n";
  std::cout << "(S o T)(x)=BA x=\n" << (BA * x) << "\n";

  return 0;
}
      

10.2 Java (EJML)

import org.ejml.simple.SimpleMatrix;

public class LinearMapDemo {
    public static void main(String[] args) {
        // A in R^{2x3}
        SimpleMatrix A = new SimpleMatrix(new double[][]{
                {1, 2, 0},
                {0, -1, 3}
        });

        SimpleMatrix x = new SimpleMatrix(new double[][]{
                {1.0},
                {-2.0},
                {0.5}
        });

        SimpleMatrix Tx = A.mult(x);

        System.out.println("A=\n" + A);
        System.out.println("x=\n" + x);
        System.out.println("T(x)=A x=\n" + Tx);

        // Composition with B in R^{2x2}
        SimpleMatrix B = new SimpleMatrix(new double[][]{
                {0, 1},
                {-1, 0}
        });

        SimpleMatrix BA = B.mult(A);
        System.out.println("BA=\n" + BA);
        System.out.println("(S o T)(x)=BA x=\n" + BA.mult(x));
    }
}
      

10.3 MATLAB (Matrix Representation) and Simulink (Matrix Gain Block)

% T: R^3 -> R^2
A = [1 2 0; 0 -1 3];
x = [1; -2; 0.5];
Tx = A*x;

disp('A='); disp(A);
disp('x='); disp(x);
disp('T(x)=A*x='); disp(Tx);

% Composition with S: R^2 -> R^2
B = [0 1; -1 0];
BA = B*A;
disp('BA='); disp(BA);
disp('(S o T)(x)=BA*x='); disp(BA*x);
      

In Simulink, matrix-based linear transformations are implemented with blocks such as Matrix Gain, Gain, Sum, and Selector. The script below creates a small model that computes \( \mathbf{y}=\mathbf{A}\mathbf{x} \).

% Programmatic Simulink model: y = A x using Matrix Gain
model = 'LinearMap_Model';
new_system(model); open_system(model);

A = [1 2 0; 0 -1 3];

add_block('simulink/Sources/Constant', [model '/x'], 'Position', [50 80 120 120]);
set_param([model '/x'], 'Value', '[1; -2; 0.5]');

add_block('simulink/Math Operations/Matrix Gain', [model '/A'], 'Position', [180 75 260 125]);
set_param([model '/A'], 'Gain', 'A');

add_block('simulink/Sinks/Display', [model '/y'], 'Position', [320 80 380 120]);

add_line(model, 'x/1', 'A/1');
add_line(model, 'A/1', 'y/1');

save_system(model);
% Run simulation (Display block will show y)
sim(model);
      

10.4 Wolfram Mathematica

(* T: R^3 -> R^2 *)
A = { {1, 2, 0}, {0, -1, 3} };
x = {1, -2, 1/2};
Tx = A.x;

Print["A = ", MatrixForm[A]];
Print["x = ", x];
Print["T(x) = A.x = ", Tx];

(* Composition with B: R^2 -> R^2 *)
B = { {0, 1}, {-1, 0} };
BA = B.A;
Print["BA = ", MatrixForm[BA]];
Print["(S o T)(x) = BA.x = ", BA.x];
      

The problems below reinforce the core workflow: verify linearity, compute basis images, build the matrix, and use matrix algebra to reason about composition and invertibility.

Problem 1 (Linearity test): Define \( T:\mathbb{R}^2\to\mathbb{R}^2 \) by \( T(x_1,x_2) = (x_1 + 2x_2,\; 3x_1 - x_2) \). Prove that \( T \) is linear and find its matrix \( \mathbf{A} \) in the standard basis.

Solution: Let \( \mathbf{u}=(u_1,u_2) \), \( \mathbf{v}=(v_1,v_2) \), and \( \alpha\in\mathbb{R} \). Compute additivity:

\[ \begin{aligned} T(\mathbf{u}+\mathbf{v}) &= T(u_1+v_1,\;u_2+v_2) \\ &= \big((u_1+v_1)+2(u_2+v_2),\; 3(u_1+v_1)-(u_2+v_2)\big)\\ &= (u_1+2u_2,\;3u_1-u_2) + (v_1+2v_2,\;3v_1-v_2)\\ &= T(\mathbf{u}) + T(\mathbf{v}). \end{aligned} \]

Homogeneity:

\[ \begin{aligned} T(\alpha \mathbf{v}) &= T(\alpha v_1,\;\alpha v_2) = (\alpha v_1 + 2\alpha v_2,\; 3\alpha v_1 - \alpha v_2) = \alpha (v_1+2v_2,\;3v_1-v_2) = \alpha T(\mathbf{v}). \end{aligned} \]

Hence \( T \) is linear. The matrix is obtained from columns \( T(\mathbf{e}_1) \) and \( T(\mathbf{e}_2) \):

\[ T(\mathbf{e}_1)=T(1,0)=(1,3),\qquad T(\mathbf{e}_2)=T(0,1)=(2,-1), \]

\[ \mathbf{A}= \begin{bmatrix} 1 & 2\\ 3 & -1 \end{bmatrix} \quad \text{so that} \quad T(\mathbf{x})=\mathbf{A}\mathbf{x}. \]

Problem 2 (Build the matrix from basis images): Let \( T:\mathbb{R}^3\to\mathbb{R}^2 \) satisfy \( T(\mathbf{e}_1)=(1,0) \), \( T(\mathbf{e}_2)=(2,-1) \), \( T(\mathbf{e}_3)=(0,3) \). Compute \( T(1,-2,\tfrac{1}{2}) \).

Solution: The matrix is \( \mathbf{A}=\begin{bmatrix} 1 & 2 & 0\\ 0 & -1 & 3\end{bmatrix} \). Then

\[ T\!\left(\begin{bmatrix} 1\\ -2\\ \tfrac{1}{2}\end{bmatrix}\right) = \mathbf{A}\begin{bmatrix} 1\\ -2\\ \tfrac{1}{2}\end{bmatrix} = \begin{bmatrix} 1\cdot 1 + 2\cdot (-2) + 0\cdot \tfrac{1}{2}\\ 0\cdot 1 + (-1)\cdot (-2) + 3\cdot \tfrac{1}{2}\end{bmatrix} = \begin{bmatrix} -3\\ \tfrac{7}{2}\end{bmatrix}. \]

Problem 3 (Composition becomes multiplication): Let \( T(\mathbf{x})=\mathbf{A}\mathbf{x} \) with \( \mathbf{A}\in\mathbb{R}^{2\times 3} \) and \( S(\mathbf{y})=\mathbf{B}\mathbf{y} \) with \( \mathbf{B}\in\mathbb{R}^{2\times 2} \). Prove that \( S\circ T \) is linear and its matrix is \( \mathbf{B}\mathbf{A} \).

Solution: Linearity follows from the algebra of matrices. For additivity: \( (S\circ T)(\mathbf{u}+\mathbf{v})=\mathbf{B}\mathbf{A}(\mathbf{u}+\mathbf{v})=\mathbf{B}\mathbf{A}\mathbf{u}+\mathbf{B}\mathbf{A}\mathbf{v} \). For homogeneity: \( (S\circ T)(\alpha\mathbf{v})=\mathbf{B}\mathbf{A}(\alpha\mathbf{v})=\alpha(\mathbf{B}\mathbf{A}\mathbf{v}) \). Hence \( S\circ T \) is linear with matrix \( \mathbf{B}\mathbf{A} \).

flowchart LR
  X["x in R^n"] -->|"apply T: A x"| Y["y in R^m"]
  Y -->|"apply S: B y"| Z["z in R^p"]
  X -->|"single step: (B A) x"| Z
  

Problem 4 (Kernel and injectivity): Consider \( T:\mathbb{R}^2\to\mathbb{R}^2 \) with \( \mathbf{A}=\begin{bmatrix} 1 & 2\\ 3 & 6\end{bmatrix} \). Find a nonzero vector in \( \ker(T) \) and conclude that \( T \) is not injective.

Solution: Solve \( \mathbf{A}\mathbf{x}=\mathbf{0} \):

\[ \begin{aligned} x_1 + 2x_2 &= 0,\\ 3x_1 + 6x_2 &= 0. \end{aligned} \]

The second equation is three times the first, so pick \( x_2=1 \) giving \( x_1=-2 \). Thus \( \mathbf{x}=\begin{bmatrix}-2\\ 1\end{bmatrix}\neq \mathbf{0} \) and \( T(\mathbf{x})=\mathbf{A}\mathbf{x}=\mathbf{0} \), hence \( \ker(T)\neq\{\mathbf{0}\} \) and \( T \) is not injective.

Problem 5 (Uniqueness of matrix representation): Suppose \( \mathbf{A},\mathbf{B}\in\mathbb{R}^{m\times n} \) satisfy \( \mathbf{A}\mathbf{x}=\mathbf{B}\mathbf{x} \) for all \( \mathbf{x}\in\mathbb{R}^n \). Prove \( \mathbf{A}=\mathbf{B} \).

Solution: For each standard basis vector \( \mathbf{e}_i \), the assumption gives \( \mathbf{A}\mathbf{e}_i=\mathbf{B}\mathbf{e}_i \). But \( \mathbf{A}\mathbf{e}_i \) is exactly the \( i \)-th column of \( \mathbf{A} \), and similarly for \( \mathbf{B} \). Therefore every column matches, so \( \mathbf{A}=\mathbf{B} \).

We defined linear transformations and proved fundamental consequences of linearity, introduced kernel and image as subspaces, and established that a linear map is fully determined by its action on a basis. For finite-dimensional spaces, fixing bases yields a unique matrix representation, with the columns equal to the coordinate vectors of basis images. Composition of transformations becomes matrix multiplication, and invertibility of a linear map corresponds exactly to invertibility of its representing matrix. These results are the algebraic substrate for later state-space constructions in modern control.