📐 Concept diagram

08-03 — Linear Transformations

Phase: 8 — Linear Algebra (Rigorous) Subject: 08-03 Prerequisites: 08-02 — Linear Independence and Basis Next subject: 08-04 — Matrices as Linear Transformations

Learning Objectives

By the end of this subject, you will be able to:

1. Define a linear transformation and verify linearity using T(u+v) = T(u) + T(v) and T(cv) = cT(v)
2. Compute the kernel and range of a linear transformation, and use them to determine injectivity and surjectivity
3. Construct the matrix representation of a linear transformation relative to chosen bases, and apply it to compute T(v)
4. Compose linear transformations and compute their matrix products
5. Determine whether a linear transformation is invertible, and identify isomorphisms between vector spaces

Core Content

1. Definition and Basic Properties

Definition: A function T : V → W between vector spaces over the same field F is a linear transformation if for all u, v ∈ V and all scalars c ∈ F:

1. Additivity: T(u + v) = T(u) + T(v)
2. Homogeneity: T(cv) = c T(v)

CRITICAL — Foundational: These two properties define linearity. EVERYTHING — kernel, range, rank-nullity, matrix representation — follows from them. Always check BOTH.

Equivalently (and often more convenient): T(cu + v) = c T(u) + T(v) for all u, v, c.

Immediate consequences: - T(0_V) = 0_W (take c = 0, v arbitrary) - T(−v) = −T(v) (take c = −1) - T(Σ c_i v_i) = Σ c_i T(v_i) — LT preserves linear combinations

Examples of linear transformations: - T : R^n → R^m defined by T(x) = Ax (matrix multiplication) - T : P_n → P_{n-1} defined by T(p) = p′ (differentiation) - T : C[a, b] → R defined by T(f) = ∫_a^b f(x) dx (integration) - T : V → V defined by T(v) = 0 for all v (the zero transformation) - T : V → V defined by T(v) = v (the identity transformation)

Non-examples: - T(x) = x + b (b ≠ 0) is NOT linear: T(0) = b ≠ 0. This is an affine transformation. - T(x) = ‖x‖ (norm) is NOT linear: T(−x) = ‖x‖ ≠ −‖x‖. - T(x) = x² (componentwise) is NOT linear: T(x+y) ≠ T(x) + T(y) in general.

2. Kernel and Range

Definition: For T : V → W linear: - Kernel (nullspace): ker(T) = {v ∈ V : T(v) = 0_W} — a subspace of V - Range (image): range(T) = {T(v) : v ∈ V} — a subspace of W

Theorem: T is injective (one-to-one) iff ker(T) = {0}.

Common Pitfall: Injectivity and surjectivity are INDEPENDENT — a transformation can be one, both, or neither. Always check separately.

Proof: If ker(T) = {0} and T(u) = T(v), then T(u−v) = T(u) − T(v) = 0, so u−v ∈ ker(T) → u−v = 0 → u = v. Conversely, if T is injective and T(v) = 0 = T(0), then v = 0. ∎

Theorem: T is surjective (onto) iff range(T) = W.

Rank-Nullity Theorem (for linear transformations):

$dim(V) = dim(ker(T)) + dim(range(T))
$

This is a fundamental structural result — the dimension of the domain decomposes into what gets "crushed to zero" (kernel) and what "survives" (range).

Example: T : R^3 → R^2 defined by T(x, y, z) = (x + y, y + z).

Kernel: Solve x+y = 0, y+z = 0 → z = −y, x = −y. So ker(T) = {t(−1, 1, −1) : t ∈ R}, dim = 1.

Range: T(x, y, z) = x(1, 0) + y(1, 1) + z(0, 1) = span{(1, 0), (1, 1)}. These are independent → dim = 2 = dim(R^2), so T is surjective.

Check rank-nullity: dim(V) = 3, dim(ker) = 1, dim(range) = 2, and 3 = 1 + 2. ✓

3. Matrix Representation

If V has basis B = {b₁, ..., b_n} and W has basis C = {c₁, ..., c_m}, any linear transformation T : V → W corresponds to a unique m × n matrix [T]_{C←B} such that:

$[T(v)]_C = [T]_{C←B} [v]_B
$

Construction: The j-th column of [T]_{C←B} is [T(b_j)]_C — the coordinates of T(b_j) expressed in the C basis.

Example: T : R^2 → R^2, T(x, y) = (2x + y, x − y). With standard basis E on both sides:

T(e₁) = T(1, 0) = (2, 1) → column 1 = (2, 1) T(e₂) = T(0, 1) = (1, −1) → column 2 = (1, −1)

$[T]_E = [2   1]
        [1  -1]
$

Check: [T]_E [x; y] = [2x+y; x−y] = T(x, y). ✓

With non-standard basis: Same T, but now B = {(1, 1), (1, −1)} on domain, C = {e₁, e₂} on codomain: T(b₁) = T(1,1) = (3, 0) = 3e₁ + 0e₂ → column 1 = (3, 0) T(b₂) = T(1,−1) = (1, 2) = 1e₁ + 2e₂ → column 2 = (1, 2)

$[T]_{C←B} = [3  1]
            [0  2]
$

Key insight: The matrix depends on the choice of bases! The transformation itself is basis-independent; the matrix is its coordinate-dependent representation.

4. Composition of Linear Transformations

If T : U → V and S : V → W are linear, then S ∘ T : U → W defined by (S ∘ T)(u) = S(T(u)) is linear.

Matrix multiplication corresponds to composition:

[S ∘ T]_{D←B} = [S]_{D←C} [T]_{C←B}

when the intermediate basis C matches. This is WHY matrix multiplication is defined the way it is — it's the coordinate representation of composition.

5. Isomorphisms and Inverses

Definition: A linear transformation T : V → W is an isomorphism if it is both injective and surjective. V and W are isomorphic if such a T exists, denoted V ≅ W.

Theorem: Two finite-dimensional vector spaces over F are isomorphic iff they have the same dimension. In particular, every n-dimensional vector space over R is isomorphic to R^n — this is why we can always "work in coordinates."

Inverse: If T : V → W is an isomorphism, there exists a unique T⁻¹ : W → V such that T⁻¹ ∘ T = I_V and T ∘ T⁻¹ = I_W. T⁻¹ is also linear.

For matrices: T⁻¹ exists iff [T] is invertible (square and det ≠ 0, equivalently, rank = dim).

Example: T : P₂ → R^3 defined by T(ax² + bx + c) = (a, b, c) is an isomorphism. Both have dimension 3.

Key Terms

• 08 03 Linear Transformations
• Composition of Linear Transformations
• Correct: A) It must also be linearly dependent
• Correct: A) S must be injective
• Correct: B) 1
• Correct: B) Injective
• Correct: B) T is injective
• Correct: B) V = R^2, W = R^3
• Correct: C) 2
• Correct: C) 5
• Correct: C) T(x, y) = (2x − y, x + 3y)
• Definition and Basic Properties

Worked Examples

Example 1: Verifying Linearity

Problem: Is T : R^2 → R^2, T(x, y) = (x + 2y, 3x − y) linear?

Solution: Check the two conditions.

Let u = (x₁, y₁), v = (x₂, y₂).

Additivity: T(u+v) = T(x₁+x₂, y₁+y₂) = ((x₁+x₂) + 2(y₁+y₂), 3(x₁+x₂) − (y₁+y₂)) = (x₁+2y₁ + x₂+2y₂, 3x₁−y₁ + 3x₂−y₂) = (x₁+2y₁, 3x₁−y₁) + (x₂+2y₂, 3x₂−y₂) = T(u) + T(v). ✓

Homogeneity: T(cu) = T(cx₁, cy₁) = (cx₁ + 2cy₁, 3cx₁ − cy₁) = c(x₁+2y₁, 3x₁−y₁) = c T(u). ✓

Yes, T is linear. In fact, T(x) = Ax with A = [1 2; 3 −1].

Example 2: Kernel, Range, Injectivity, Surjectivity

Problem: T : R^3 → R^2, T(x, y, z) = (x − y + z, 2x + y − z). Find ker(T), range(T), and determine injectivity/surjectivity.

Solution:

Kernel: Solve T(x, y, z) = (0, 0):

$x − y + z = 0
2x + y − z = 0
$

Add: 3x = 0 → x = 0. Then −y + z = 0 and y − z = 0 → y = z. So ker(T) = {(0, t, t) : t ∈ R} = span{(0, 1, 1)}. dim(ker) = 1 ≠ 0, so T is NOT injective.

Range: T(x,y,z) = x(1,2) + y(−1,1) + z(1,−1). The vectors are: v₁ = (1,2), v₂ = (−1,1), v₃ = (1,−1). Note v₃ = v₁ − v₂, so range(T) = span{v₁, v₂}. v₁ and v₂ are independent, so dim(range) = 2. Since dim(R^2) = 2, range(T) = R^2, so T IS surjective.

Rank-nullity check: dim(domain) = 3, dim(ker) = 1, dim(range) = 2. 3 = 1+2. ✓

Example 3: Matrix Representation with Non-Standard Bases

Problem: T : R^2 → R^2, T(x, y) = (x + y, 2x). Find [T]_{B←B} where B = {(1, 2), (2, 1)}.

Solution:

Step 1: Compute T on basis vectors. T(1, 2) = (1+2, 2·1) = (3, 2) T(2, 1) = (2+1, 2·2) = (3, 4)

Step 2: Express T(b₁) and T(b₂) in basis B. For T(b₁) = (3, 2): solve α(1,2) + β(2,1) = (3,2).

$α + 2β = 3
2α + β = 2
$

From eq2: β = 2 − 2α. Substitute: α + 2(2−2α) = α + 4 − 4α = 4 − 3α = 3 → α = 1/3. β = 2 − 2/3 = 4/3. So [T(b₁)]_B = (1/3, 4/3).

For T(b₂) = (3, 4): solve α(1,2) + β(2,1) = (3,4).

$α + 2β = 3
2α + β = 4
$

β = 4 − 2α. Substitute: α + 2(4−2α) = α + 8 − 4α = 8 − 3α = 3 → α = 5/3. β = 4 − 10/3 = 2/3. So [T(b₂)]_B = (5/3, 2/3).

Step 3: Assemble.

$[T]_{B←B} = [1/3  5/3]
            [4/3  2/3]
$

Verification: Take v = (1, 2) = 1·b₁ + 0·b₂, so [v]B = (1, 0). [T(v)]_B = [T]{B←B} [v]_B = (1/3, 4/3). This means T(v) = (1/3)b₁ + (4/3)b₂ = (1/3)(1,2) + (4/3)(2,1) = (1/3+8/3, 2/3+4/3) = (3, 2) = T(1,2). ✓

Example 4: Composition

Problem: S : R^2 → R^3 defined by S(x, y) = (x+y, x−y, 2x), and T : R^3 → R^2 defined by T(x, y, z) = (x+z, y−z). Compute (T ∘ S)(x, y) and its matrix.

Solution:

Direct computation: (T ∘ S)(x, y) = T(S(x,y)) = T(x+y, x−y, 2x) = ((x+y) + 2x, (x−y) − 2x) = (3x + y, −x − y)

Matrix approach (standard bases everywhere):

$[S] = [1   1]      [T] = [1  0  1]
      [1  -1]            [0  1 -1]
      [2   0]
$

$[T ∘ S] = [T][S] = [1 0 1] [1  1]   = [1·1+0·1+1·2   1·1+0·(−1)+1·0]   = [3   1]
         [0 1 -1] [1 -1]     [0·1+1·1+(−1)·2   0·1+1·(−1)+(−1)·0]     [-1  -1]
                    [2  0]
$

Matches direct computation: (3x+y, −x−y). ✓

Quiz

Q1: A function T: V → W is a linear transformation if:

A) T(u + v) = T(u) + T(v) and T(cv) = cT(v) for all u, v and scalar c B) T is continuous C) T(0) = 0 only D) T is invertible

Correct: A)

• If you chose A: These two properties — additivity and homogeneity — define a linear transformation. Correct!
• If you chose B: Continuity is not part of the definition (though linear maps on finite-dimensional spaces are automatically continuous).
• If you chose C: T(0) = 0 is a consequence of linearity, not the definition.
• If you chose D: Invertibility is a special property, not the definition.

Q2: The kernel (null space) of a linear transformation T is:

A) The set of all outputs T(v) B) {v ∈ V : T(v) = 0} C) The dimension of the domain D) The set of all scalars c

Correct: B)

• If you chose B: ker(T) = {v : T(v) = 0} — the set of vectors mapped to the zero vector. Correct!
• If you chose A: That's the image (range) of T.
• If you chose C: That's dim(V), not the kernel.
• If you chose D: Scalars live in the field, not in ker(T).

Q3: A linear transformation T is one-to-one (injective) if and only if:

A) T is onto (surjective) B) ker(T) = {0} C) T(v) ≠ 0 for all v D) dim(V) = dim(W)

Correct: B)

• If you chose B: A linear map is injective exactly when its kernel contains only the zero vector. Correct!
• If you chose A: These are different properties — a map can be injective without being surjective.
• If you chose C: T(v) ≠ 0 for v ≠ 0 is equivalent to ker(T) = {0}, but stated imprecisely.
• If you chose D: Dimensions being equal is necessary for invertibility, not just injectivity.

Q4: The Rank-Nullity Theorem states that for T: V → W:

A) rank(T) = nullity(T) B) rank(T) + nullity(T) = dim(V) C) rank(T) · nullity(T) = dim(V) D) rank(T) − nullity(T) = dim(W)

Correct: B)

• If you chose B: dim(Im T) + dim(ker T) = dim(V) — the rank plus the nullity equals the dimension of the domain. Correct!
• If you chose A: They're generally not equal (e.g., projection onto a line has rank 1, nullity n−1).
• If you chose C: It's a sum, not a product.
• If you chose D: The right side is dim(V), not dim(W).

Q5: The composition of two linear transformations is:

A) Always commutative B) Always a linear transformation C) Never defined D) Always invertible

Correct: B)

• If you chose B: If S and T are linear, then S ∘ T is also linear (satisfies additivity and homogeneity). Correct!
• If you chose A: Composition is generally not commutative (ST ≠ TS).
• If you chose C: Composition is defined when codomains and domains match.
• If you chose D: Composition can be non-invertible.

Q6: If T: R³ → R² is linear, what can rank(T) be?

A) Only 3 B) Only 2 C) 0, 1, or 2 D) 0, 1, 2, or 3

Correct: C)

• If you chose C: By Rank-Nullity: rank(T) + nullity(T) = 3. Since rank(T) ≤ dim(W) = 2, rank can be 0, 1, or 2. Correct!
• If you chose A: Rank can't exceed the dimension of the codomain (2).
• If you chose B: Rank could be 0 or 1 as well.
• If you chose D: Rank cannot be 3 in a map to R².

Practice Problems

(Answers are below. Try each problem before checking.)

Problem 1: Determine whether T : R^2 → R^3, T(x, y) = (x², x + y, 0) is linear.

Problem 2: Find ker(T) and range(T) for T : R^3 → R^2, T(x, y, z) = (2x − y, y + 3z). Is T injective? Surjective?

Problem 3: Let T : P₂ → P₃, T(p(x)) = x·p(x). Find the matrix of T relative to the standard bases {1, x, x²} of P₂ and {1, x, x², x³} of P₃.

Problem 4: T : R^2 → R^2 rotates vectors by 90° counterclockwise. Find [T]E (standard basis) and [T]{B←B} where B = {(1, 1), (−1, 1)}.

Problem 5: If S : R^3 → R^2 and T : R^2 → R^4 are linear transformations, can T ∘ S be an isomorphism? Explain.

Problem 6: Find the matrix of T : R^2 → R^3, T(x, y) = (x, x+y, x−y) relative to B = {(1, 1), (1, −1)} on the domain and C = {e₁, e₂, e₃} on the codomain.

Problem 7: Is T : M_{2×2} → R, T(A) = tr(A), a linear transformation? If so, what are ker(T) and range(T)?

$[T] = [0  0  0]
      [1  0  0]
      [0  1  0]
      [0  0  1]
$

$[T]_E = [0  -1]
        [1   0]
$

$[T]_{B←B} = [0  -1]
            [1   0]
$

$[T]_{C←B} = [1  1]
            [2  0]
            [0  2]
$

Summary

1. A linear transformation preserves the vector space structure: T(u+v) = T(u)+T(v) and T(cv) = cT(v); this single property underlies all of linear algebra
2. The kernel (what maps to zero) and range (what can be reached) are subspaces; the rank-nullity theorem dim(V) = dim(ker) + dim(range) is a fundamental structural law
3. T is injective ⇔ ker(T) = {0}; T is surjective ⇔ range(T) = W; an isomorphism is both, and two spaces are isomorphic exactly when they have the same dimension
4. Every linear transformation can be represented by a matrix once bases are chosen: [T]_{C←B} has columns [T(b_j)]_C — different bases give different matrices for the same transformation
5. Composition of linear transformations corresponds to matrix multiplication: [S ∘ T] = [S][T] — this is the deep reason behind the definition of matrix multiplication

Pitfalls

1. Forgetting to check both additivity AND homogeneity when verifying linearity. A function can be additive but not homogeneous (over R), or homogeneous but not additive. Always verify T(u + v) = T(u) + T(v) AND T(cv) = cT(v) — or use the combined form T(cu + v) = cT(u) + T(v).

2. Confusing the kernel and range with the null space and column space. For a matrix transformation T(x) = Ax, ker(T) = N(A) and range(T) = C(A). But for general linear transformations (differentiation, integration, trace), the same concepts apply — you must find them from the definition, not from a matrix.

3. Assuming injectivity and surjectivity are equivalent. They are independent properties. A transformation from R^2 to R^3 can be injective but never surjective (range dimension ≤ 2 < 3). A transformation from R^3 to R^2 can be surjective but never injective (kernel dimension ≥ 1). They coincide only when domain and codomain have the same dimension.

4. Constructing the matrix [T]_{C←B} with columns in the wrong order. Column j is [T(b_j)]_C — the coordinates of T applied to the j-th basis vector of B. A common error is using [T]_C as columns or putting C-basis vectors before B-basis vectors in the construction.

5. Misapplying the rank-nullity theorem. dim(V) = dim(ker(T)) + dim(range(T)) relates dimensions, but it doesn't tell you what the kernel and range ARE. Two different transformations can have the same kernel dimension (nullity) and range dimension (rank) while being entirely different maps.

Next Steps

Move on to 08-04 — Matrices as Linear Transformations to explore matrix-vector multiplication in depth, the matrix of a linear transformation, similarity, trace, and the rank-nullity theorem in its full matrix form.