Linear transformations are functions between vector spaces that preserve vector addition and scalar multiplication. They form the foundation for understanding linear algebra geometrically.
Definition of Linear Transformation
Definition (Linear Transformation): T : R n → R m T: \mathbb{R}^n \to \mathbb{R}^m T : R n → R m linear transformation if for all vectors u ⃗ , v ⃗ \vec{u}, \vec{v} u , v R n \mathbb{R}^n R n c c c
Additivity: T ( u ⃗ + v ⃗ ) = T ( u ⃗ ) + T ( v ⃗ ) T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) T ( u + v ) = T ( u ) + T ( v ) Homogeneity: T ( c u ⃗ ) = c T ( u ⃗ ) T(c\vec{u}) = cT(\vec{u}) T ( c u ) = c T ( u )
These properties can be combined into one:
Equivalent Definition: T T T u ⃗ , v ⃗ \vec{u}, \vec{v} u , v R n \mathbb{R}^n R n c , d c, d c , d T ( c u ⃗ + d v ⃗ ) = c T ( u ⃗ ) + d T ( v ⃗ ) T(c\vec{u} + d\vec{v}) = cT(\vec{u}) + dT(\vec{v}) T ( c u + d v ) = c T ( u ) + d T ( v )
Intuition: Linear transformations "respect" the structure of vector spaces—they play nicely with addition and scaling.
Example 1: A Linear Transformation
Define T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 T [ x y ] = [ 2 x x + y ] T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x \\ x + y \end{bmatrix} T [ x y ] = [ 2 x x + y ]
Check additivity: Let u ⃗ = [ x 1 y 1 ] \vec{u} = \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} u = [ x 1 y 1 ] v ⃗ = [ x 2 y 2 ] \vec{v} = \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} v = [ x 2 y 2 ]
T ( u ⃗ + v ⃗ ) = T [ x 1 + x 2 y 1 + y 2 ] = [ 2 ( x 1 + x 2 ) ( x 1 + x 2 ) + ( y 1 + y 2 ) ] = [ 2 x 1 + 2 x 2 ( x 1 + y 1 ) + ( x 2 + y 2 ) ] T(\vec{u} + \vec{v}) = T\begin{bmatrix} x_1 + x_2 \\ y_1 + y_2 \end{bmatrix} = \begin{bmatrix} 2(x_1 + x_2) \\ (x_1 + x_2) + (y_1 + y_2) \end{bmatrix} = \begin{bmatrix} 2x_1 + 2x_2 \\ (x_1 + y_1) + (x_2 + y_2) \end{bmatrix} T ( u + v ) = T [ x 1 + x 2 y 1 + y 2 ] = [ 2 ( x 1 + x 2 ) ( x 1 + x 2 ) + ( y 1 + y 2 ) ] = [ 2 x 1 + 2 x 2 ( x 1 + y 1 ) + ( x 2 + y 2 ) ]
T ( u ⃗ ) + T ( v ⃗ ) = [ 2 x 1 x 1 + y 1 ] + [ 2 x 2 x 2 + y 2 ] = [ 2 x 1 + 2 x 2 ( x 1 + y 1 ) + ( x 2 + y 2 ) ] T(\vec{u}) + T(\vec{v}) = \begin{bmatrix} 2x_1 \\ x_1 + y_1 \end{bmatrix} + \begin{bmatrix} 2x_2 \\ x_2 + y_2 \end{bmatrix} = \begin{bmatrix} 2x_1 + 2x_2 \\ (x_1 + y_1) + (x_2 + y_2) \end{bmatrix} T ( u ) + T ( v ) = [ 2 x 1 x 1 + y 1 ] + [ 2 x 2 x 2 + y 2 ] = [ 2 x 1 + 2 x 2 ( x 1 + y 1 ) + ( x 2 + y 2 ) ]
They're equal! ✓
▼ ✏️ Checking Homogeneity T ( c u ⃗ ) = T [ c x 1 c y 1 ] = [ 2 ( c x 1 ) c x 1 + c y 1 ] = [ c ( 2 x 1 ) c ( x 1 + y 1 ) ] = c [ 2 x 1 x 1 + y 1 ] = c T ( u ⃗ ) T(c\vec{u}) = T\begin{bmatrix} cx_1 \\ cy_1 \end{bmatrix} = \begin{bmatrix} 2(cx_1) \\ cx_1 + cy_1 \end{bmatrix} = \begin{bmatrix} c(2x_1) \\ c(x_1 + y_1) \end{bmatrix} = c\begin{bmatrix} 2x_1 \\ x_1 + y_1 \end{bmatrix} = cT(\vec{u}) T ( c u ) = T [ c x 1 c y 1 ] = [ 2 ( c x 1 ) c x 1 + c y 1 ] = [ c ( 2 x 1 ) c ( x 1 + y 1 ) ] = c [ 2 x 1 x 1 + y 1 ] = c T ( u )
Example 2: Not a Linear Transformation
Define T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 T [ x y ] = [ x + 1 y ] T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x + 1 \\ y \end{bmatrix} T [ x y ] = [ x + 1 y ]
Test additivity: Let u ⃗ = [ 1 0 ] \vec{u} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} u = [ 1 0 ] v ⃗ = [ 2 0 ] \vec{v} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} v = [ 2 0 ]
T ( u ⃗ + v ⃗ ) = T [ 3 0 ] = [ 4 0 ] T(\vec{u} + \vec{v}) = T\begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 4 \\ 0 \end{bmatrix} T ( u + v ) = T [ 3 0 ] = [ 4 0 ]
T ( u ⃗ ) + T ( v ⃗ ) = [ 2 0 ] + [ 3 0 ] = [ 5 0 ] T(\vec{u}) + T(\vec{v}) = \begin{bmatrix} 2 \\ 0 \end{bmatrix} + \begin{bmatrix} 3 \\ 0 \end{bmatrix} = \begin{bmatrix} 5 \\ 0 \end{bmatrix} T ( u ) + T ( v ) = [ 2 0 ] + [ 3 0 ] = [ 5 0 ]
Not equal! So T T T not linear .
▼ 📖 Why Adding Constants Breaks Linearity Any transformation of the form T ( x ⃗ ) = A x ⃗ + b ⃗ T(\vec{x}) = A\vec{x} + \vec{b} T ( x ) = A x + b b ⃗ ≠ 0 ⃗ \vec{b} \neq \vec{0} b = 0 not linear . These are called affine transformations .
Why? Check T ( 0 ⃗ ) T(\vec{0}) T ( 0 )
If T T T T ( 0 ⃗ ) = T ( 0 ⋅ v ⃗ ) = 0 ⋅ T ( v ⃗ ) = 0 ⃗ T(\vec{0}) = T(0 \cdot \vec{v}) = 0 \cdot T(\vec{v}) = \vec{0} T ( 0 ) = T ( 0 ⋅ v ) = 0 ⋅ T ( v ) = 0
For affine: T ( 0 ⃗ ) = A 0 ⃗ + b ⃗ = b ⃗ ≠ 0 ⃗ T(\vec{0}) = A\vec{0} + \vec{b} = \vec{b} \neq \vec{0} T ( 0 ) = A 0 + b = b = 0
Key property: Linear transformations always map the zero vector to the zero vector.
Matrix Transformations
Definition (Matrix Transformation): m × n m \times n m × n A A A T : R n → R m T: \mathbb{R}^n \to \mathbb{R}^m T : R n → R m T ( x ⃗ ) = A x ⃗ T(\vec{x}) = A\vec{x} T ( x ) = A x matrix transformation .
Key fact: Every matrix transformation is a linear transformation (we proved this in Note 05).
Example 3: Rotation Matrix
The transformation T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 θ \theta θ
T ( x ⃗ ) = [ cos θ − sin θ sin θ cos θ ] x ⃗ T(\vec{x}) = \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix}\vec{x} T ( x ) = [ cos θ sin θ − sin θ cos θ ] x
For θ = 90 ° \theta = 90° θ = 90°
A = [ 0 − 1 1 0 ] A = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix} A = [ 0 1 − 1 0 ]
T [ 1 0 ] = [ 0 1 ] , T [ 0 1 ] = [ − 1 0 ] T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 \\ 1 \end{bmatrix}, \quad T\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \end{bmatrix} T [ 1 0 ] = [ 0 1 ] , T [ 0 1 ] = [ − 1 0 ]
The unit vectors rotate 90° counterclockwise!
Example 4: Projection Matrix
The transformation that projects vectors in R 2 \mathbb{R}^2 R 2 x x x
T [ x y ] = [ x 0 ] = [ 1 0 0 0 ] [ x y ] T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ 0 \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} T [ x y ] = [ x 0 ] = [ 1 0 0 0 ] [ x y ]
Matrix: A = [ 1 0 0 0 ] A = \begin{bmatrix} 1 & 0 \\ 0 & 0 \end{bmatrix} A = [ 1 0 0 0 ]
Example 5: Scaling Matrix
The transformation that stretches by factor k k k
T ( x ⃗ ) = k x ⃗ = [ k 0 ⋯ 0 0 k ⋯ 0 ⋮ ⋮ ⋱ ⋮ 0 0 ⋯ k ] x ⃗ T(\vec{x}) = k\vec{x} = \begin{bmatrix} k & 0 & \cdots & 0 \\ 0 & k & \cdots & 0 \\ \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & \cdots & k \end{bmatrix}\vec{x} T ( x ) = k x = k 0 ⋮ 0 0 k ⋮ 0 ⋯ ⋯ ⋱ ⋯ 0 0 ⋮ k x
Matrix: A = k I n A = kI_n A = k I n
Standard Matrix Representation
Big question: Is every linear transformation a matrix transformation?
Theorem (Standard Matrix Representation): T : R n → R m T: \mathbb{R}^n \to \mathbb{R}^m T : R n → R m T ( x ⃗ ) = A x ⃗ T(\vec{x}) = A\vec{x} T ( x ) = A x m × n m \times n m × n A A A
The matrix A A A standard matrix for T T T
How to find the standard matrix:
Recipe: e ⃗ 1 , e ⃗ 2 , … , e ⃗ n \vec{e}_1, \vec{e}_2, \ldots, \vec{e}_n e 1 , e 2 , … , e n R n \mathbb{R}^n R n A = [ T ( e ⃗ 1 ) T ( e ⃗ 2 ) ⋯ T ( e ⃗ n ) ] A = \begin{bmatrix} T(\vec{e}_1) & T(\vec{e}_2) & \cdots & T(\vec{e}_n) \end{bmatrix} A = [ T ( e 1 ) T ( e 2 ) ⋯ T ( e n ) ]
The columns of A A A
Why this works: Any vector x ⃗ = [ x 1 x 2 ⋮ x n ] \vec{x} = \begin{bmatrix} x_1 \\ x_2 \\ \vdots \\ x_n \end{bmatrix} x = x 1 x 2 ⋮ x n x ⃗ = x 1 e ⃗ 1 + x 2 e ⃗ 2 + ⋯ + x n e ⃗ n \vec{x} = x_1\vec{e}_1 + x_2\vec{e}_2 + \cdots + x_n\vec{e}_n x = x 1 e 1 + x 2 e 2 + ⋯ + x n e n
By linearity:
T ( x ⃗ ) = T ( x 1 e ⃗ 1 + x 2 e ⃗ 2 + ⋯ + x n e ⃗ n ) = x 1 T ( e ⃗ 1 ) + x 2 T ( e ⃗ 2 ) + ⋯ + x n T ( e ⃗ n ) T(\vec{x}) = T(x_1\vec{e}_1 + x_2\vec{e}_2 + \cdots + x_n\vec{e}_n) = x_1T(\vec{e}_1) + x_2T(\vec{e}_2) + \cdots + x_nT(\vec{e}_n) T ( x ) = T ( x 1 e 1 + x 2 e 2 + ⋯ + x n e n ) = x 1 T ( e 1 ) + x 2 T ( e 2 ) + ⋯ + x n T ( e n )
This is exactly A x ⃗ A\vec{x} A x A A A T ( e ⃗ 1 ) , … , T ( e ⃗ n ) T(\vec{e}_1), \ldots, T(\vec{e}_n) T ( e 1 ) , … , T ( e n )
Example 6: Finding the Standard Matrix
Let T : R 2 → R 3 T: \mathbb{R}^2 \to \mathbb{R}^3 T : R 2 → R 3 T [ x y ] = [ 2 x − y x 3 y ] T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x - y \\ x \\ 3y \end{bmatrix} T [ x y ] = 2 x − y x 3 y
Find the standard matrix for T T T
Solution: Compute T ( e ⃗ 1 ) T(\vec{e}_1) T ( e 1 ) T ( e ⃗ 2 ) T(\vec{e}_2) T ( e 2 )
T [ 1 0 ] = [ 2 ( 1 ) − 0 1 3 ( 0 ) ] = [ 2 1 0 ] T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2(1) - 0 \\ 1 \\ 3(0) \end{bmatrix} = \begin{bmatrix} 2 \\ 1 \\ 0 \end{bmatrix} T [ 1 0 ] = 2 ( 1 ) − 0 1 3 ( 0 ) = 2 1 0
T [ 0 1 ] = [ 2 ( 0 ) − 1 0 3 ( 1 ) ] = [ − 1 0 3 ] T\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 2(0) - 1 \\ 0 \\ 3(1) \end{bmatrix} = \begin{bmatrix} -1 \\ 0 \\ 3 \end{bmatrix} T [ 0 1 ] = 2 ( 0 ) − 1 0 3 ( 1 ) = − 1 0 3
Standard matrix:
A = [ 2 − 1 1 0 0 3 ] A = \begin{bmatrix} 2 & -1 \\ 1 & 0 \\ 0 & 3 \end{bmatrix} A = 2 1 0 − 1 0 3
Verification: A [ x y ] = [ 2 x − y x 3 y ] A\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 2x - y \\ x \\ 3y \end{bmatrix} A [ x y ] = 2 x − y x 3 y
Example 7: Reflection Across a Line
Let T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 y = x y = x y = x
Find the standard matrix.
Solution: Figure out where the standard basis vectors go:
e ⃗ 1 = [ 1 0 ] \vec{e}_1 = \begin{bmatrix} 1 \\ 0 \end{bmatrix} e 1 = [ 1 0 ] x x x [ 0 1 ] \begin{bmatrix} 0 \\ 1 \end{bmatrix} [ 0 1 ] y y y e ⃗ 2 = [ 0 1 ] \vec{e}_2 = \begin{bmatrix} 0 \\ 1 \end{bmatrix} e 2 = [ 0 1 ] y y y [ 1 0 ] \begin{bmatrix} 1 \\ 0 \end{bmatrix} [ 1 0 ] x x x
Standard matrix:
A = [ 0 1 1 0 ] A = \begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} A = [ 0 1 1 0 ]
▼ 📖 Common Geometric Transformations Rotation by angle θ \theta θ
[ cos θ − sin θ sin θ cos θ ] \begin{bmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \end{bmatrix} [ cos θ sin θ − sin θ cos θ ]
Reflection across x x x
[ 1 0 0 − 1 ] \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} [ 1 0 0 − 1 ]
Reflection across y y y
[ − 1 0 0 1 ] \begin{bmatrix} -1 & 0 \\ 0 & 1 \end{bmatrix} [ − 1 0 0 1 ]
Scaling by factor k k k
[ k 0 0 k ] \begin{bmatrix} k & 0 \\ 0 & k \end{bmatrix} [ k 0 0 k ]
Horizontal stretch by a a a b b b
[ a 0 0 b ] \begin{bmatrix} a & 0 \\ 0 & b \end{bmatrix} [ a 0 0 b ]
Shear (horizontal):
[ 1 k 0 1 ] \begin{bmatrix} 1 & k \\ 0 & 1 \end{bmatrix} [ 1 0 k 1 ]
Practice Problems
▼ 🎯 Problem 1: Testing Linearity Determine if T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 T [ x y ] = [ x y x + y ] T\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} xy \\ x + y \end{bmatrix} T [ x y ] = [ x y x + y ]
▼ 💡 Hint Test either additivity or homogeneity with specific examples.
▼ ✅ Solution Test homogeneity: Let u ⃗ = [ 1 1 ] \vec{u} = \begin{bmatrix} 1 \\ 1 \end{bmatrix} u = [ 1 1 ] c = 2 c = 2 c = 2
T ( c u ⃗ ) = T [ 2 2 ] = [ 2 ⋅ 2 2 + 2 ] = [ 4 4 ] T(c\vec{u}) = T\begin{bmatrix} 2 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \cdot 2 \\ 2 + 2 \end{bmatrix} = \begin{bmatrix} 4 \\ 4 \end{bmatrix} T ( c u ) = T [ 2 2 ] = [ 2 ⋅ 2 2 + 2 ] = [ 4 4 ]
c T ( u ⃗ ) = 2 T [ 1 1 ] = 2 [ 1 ⋅ 1 1 + 1 ] = 2 [ 1 2 ] = [ 2 4 ] cT(\vec{u}) = 2T\begin{bmatrix} 1 \\ 1 \end{bmatrix} = 2\begin{bmatrix} 1 \cdot 1 \\ 1 + 1 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 2 \\ 4 \end{bmatrix} c T ( u ) = 2 T [ 1 1 ] = 2 [ 1 ⋅ 1 1 + 1 ] = 2 [ 1 2 ] = [ 2 4 ]
Not equal! Answer: Not linear.
The problem is the x y xy x y
▼ 🎯 Problem 2: Finding Standard Matrix Find the standard matrix for T : R 3 → R 2 T: \mathbb{R}^3 \to \mathbb{R}^2 T : R 3 → R 2 T [ x y z ] = [ x − 2 y + z 3 x + y ] T\begin{bmatrix} x \\ y \\ z \end{bmatrix} = \begin{bmatrix} x - 2y + z \\ 3x + y \end{bmatrix} T x y z = [ x − 2 y + z 3 x + y ]
▼ 💡 Hint Compute T ( e ⃗ 1 ) T(\vec{e}_1) T ( e 1 ) T ( e ⃗ 2 ) T(\vec{e}_2) T ( e 2 ) T ( e ⃗ 3 ) T(\vec{e}_3) T ( e 3 )
▼ ✅ Solution T [ 1 0 0 ] = [ 1 − 0 + 0 3 + 0 ] = [ 1 3 ] T\begin{bmatrix} 1 \\ 0 \\ 0 \end{bmatrix} = \begin{bmatrix} 1 - 0 + 0 \\ 3 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 3 \end{bmatrix} T 1 0 0 = [ 1 − 0 + 0 3 + 0 ] = [ 1 3 ]
T [ 0 1 0 ] = [ 0 − 2 + 0 0 + 1 ] = [ − 2 1 ] T\begin{bmatrix} 0 \\ 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 0 - 2 + 0 \\ 0 + 1 \end{bmatrix} = \begin{bmatrix} -2 \\ 1 \end{bmatrix} T 0 1 0 = [ 0 − 2 + 0 0 + 1 ] = [ − 2 1 ]
T [ 0 0 1 ] = [ 0 − 0 + 1 0 + 0 ] = [ 1 0 ] T\begin{bmatrix} 0 \\ 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 - 0 + 1 \\ 0 + 0 \end{bmatrix} = \begin{bmatrix} 1 \\ 0 \end{bmatrix} T 0 0 1 = [ 0 − 0 + 1 0 + 0 ] = [ 1 0 ]
Standard matrix:
A = [ 1 − 2 1 3 1 0 ] A = \begin{bmatrix} 1 & -2 & 1 \\ 3 & 1 & 0 \end{bmatrix} A = [ 1 3 − 2 1 1 0 ]
▼ 🎯 Problem 3: Geometric Transformation Find the standard matrix for the transformation that reflects vectors in R 2 \mathbb{R}^2 R 2 x x x
▼ 💡 Hint Where do [ 1 0 ] \begin{bmatrix} 1 \\ 0 \end{bmatrix} [ 1 0 ] [ 0 1 ] \begin{bmatrix} 0 \\ 1 \end{bmatrix} [ 0 1 ]
▼ ✅ Solution Reflection across x x x
[ 1 0 ] \begin{bmatrix} 1 \\ 0 \end{bmatrix} [ 1 0 ] [ 1 0 ] \begin{bmatrix} 1 \\ 0 \end{bmatrix} [ 1 0 ] x x x [ 0 1 ] \begin{bmatrix} 0 \\ 1 \end{bmatrix} [ 0 1 ] [ 0 − 1 ] \begin{bmatrix} 0 \\ -1 \end{bmatrix} [ 0 − 1 ] x x x Standard matrix:
A = [ 1 0 0 − 1 ] A = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix} A = [ 1 0 0 − 1 ]
Verification: [ 1 0 0 − 1 ] [ x y ] = [ x − y ] \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}\begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} x \\ -y \end{bmatrix} [ 1 0 0 − 1 ] [ x y ] = [ x − y ] y y y
▼ 🎯 Problem 4: Composition Interpretation If T : R 2 → R 2 T: \mathbb{R}^2 \to \mathbb{R}^2 T : R 2 → R 2 A = [ 2 0 0 3 ] A = \begin{bmatrix} 2 & 0 \\ 0 & 3 \end{bmatrix} A = [ 2 0 0 3 ] T T T
▼ 💡 Hint See what happens to the standard basis vectors.
▼ ✅ Solution T [ 1 0 ] = [ 2 0 ] = 2 [ 1 0 ] T\begin{bmatrix} 1 \\ 0 \end{bmatrix} = \begin{bmatrix} 2 \\ 0 \end{bmatrix} = 2\begin{bmatrix} 1 \\ 0 \end{bmatrix} T [ 1 0 ] = [ 2 0 ] = 2 [ 1 0 ]
T [ 0 1 ] = [ 0 3 ] = 3 [ 0 1 ] T\begin{bmatrix} 0 \\ 1 \end{bmatrix} = \begin{bmatrix} 0 \\ 3 \end{bmatrix} = 3\begin{bmatrix} 0 \\ 1 \end{bmatrix} T [ 0 1 ] = [ 0 3 ] = 3 [ 0 1 ]
Geometric interpretation: T T T x x x y y y non-uniform scaling .
For example: T [ 1 1 ] = [ 2 3 ] T\begin{bmatrix} 1 \\ 1 \end{bmatrix} = \begin{bmatrix} 2 \\ 3 \end{bmatrix} T [ 1 1 ] = [ 2 3 ]
Summary
Linear transformation: Function T : R n → R m T: \mathbb{R}^n \to \mathbb{R}^m T : R n → R m
T ( u ⃗ + v ⃗ ) = T ( u ⃗ ) + T ( v ⃗ ) T(\vec{u} + \vec{v}) = T(\vec{u}) + T(\vec{v}) T ( u + v ) = T ( u ) + T ( v ) T ( c u ⃗ ) = c T ( u ⃗ ) T(c\vec{u}) = cT(\vec{u}) T ( c u ) = c T ( u )
Matrix transformation: T ( x ⃗ ) = A x ⃗ T(\vec{x}) = A\vec{x} T ( x ) = A x A A A
Every matrix transformation is linear
Standard matrix representation: Every linear transformation T : R n → R m T: \mathbb{R}^n \to \mathbb{R}^m T : R n → R m A A A
Columns of A A A T ( e ⃗ 1 ) , T ( e ⃗ 2 ) , … , T ( e ⃗ n ) T(\vec{e}_1), T(\vec{e}_2), \ldots, T(\vec{e}_n) T ( e 1 ) , T ( e 2 ) , … , T ( e n )
T ( x ⃗ ) = A x ⃗ T(\vec{x}) = A\vec{x} T ( x ) = A x x ⃗ \vec{x} x
Key insight: Linear transformations and matrices are two sides of the same coin—every linear transformation is a matrix transformation, and every matrix defines a linear transformation.