Linear Algebra Exercises: Transformations and
Subspaces - Prof. Mumo (2026/2027)
Core Linear Algebra Concepts | Key Domains: Linear Transformations & Their Matrices, Kernel (Null
Space) & Image (Column Space), Basis & Dimension, Eigenvalues & Eigenvectors, Orthogonality &
Projections, and Vector Space Axioms | Expert-Aligned Structure | Multiple-Choice Exercise Format
Introduction
This structured Linear Algebra exercise set for 2026/2027 provides 55 multiple-choice questions
with correct answers and rationales. It focuses on the abstract and geometric understanding of
linear transformations, subspaces, and their properties, which are fundamental to advanced
mathematics, engineering, and data science.
Exercise Structure:
● Conceptual & Computational Exercise Bank: (55 MULTIPLE-CHOICE QUESTIONS)
Answer Format
All correct answers, vector space determinations, and computational results must appear in bold
and cyan blue, accompanied by concise rationales explaining the application of a definition or
theorem (e.g., "The set is a subspace because it is closed under addition and scalar multiplication"),
the steps of a calculation (e.g., finding a basis for the null space by row reduction), the properties of
a linear transformation (e.g., one-to-one if kernel is trivial), the geometric interpretation (e.g.,
eigenvectors indicate directions unchanged by the transformation), and why the alternative
multiple-choice options violate linear algebra principles or contain algebraic errors.
Conceptual & Computational Exercise Bank (55
Multiple-Choice Questions)
1. Which of the following sets is a subspace of ℝ³?
, A. {(x, y, z) | x + y + z = 1}
B. {(x, y, z) | x ≥ 0}
C. {(x, y, z) | x + y + z = 0}
D. {(x, y, z) | xyz = 0}
C. {(x, y, z) | x + y + z = 0}
Rationale: A subspace must contain the zero vector, be closed under addition, and closed under scalar
multiplication. Option C contains (0,0,0), and if u and v satisfy x+y+z=0, so does u+v and cu. Option A
fails (0,0,0) ∉ set; B is not closed under scalar multiplication (e.g., -1·(1,0,0) = (-1,0,0) ∉ set); D is not
closed under addition (e.g., (1,0,0)+(0,1,0)=(1,1,0) ∈ set, but (1,0,0)+(0,0,1)=(1,0,1) ∉ set since 1·0·1=0
is true—wait, actually (1,0,1) has product 0, so it is in D. However, (1,1,0) and (1,0,1) are in D, but their
sum (2,1,1) has product 2≠0, so not in D. Thus D is not closed under addition.
2. Let T: ℝ² → ℝ² be defined by T(x, y) = (2x, 3y). What is the matrix of T with respect to the
standard basis?
A. [[2, 0], [0, 3]]
B. [[2, 3], [0, 0]]
C. [[0, 2], [3, 0]]
D. [[1, 2], [3, 1]]
A. [[2, 0], [0, 3]]
Rationale: The standard matrix is formed by T(e₁) and T(e₂) as columns. T(1,0) = (2,0); T(0,1) = (0,3).
So the matrix is diagonal with entries 2 and 3. Other options do not map basis vectors correctly.
3. What is the dimension of the null space of the matrix A = [[1, 2, 3], [2, 4, 6]]?
A. 0
B. 1
C. 2
D. 3
C. 2
, Rationale: Row reduce A: R₂ ← R₂ - 2R₁ gives [[1,2,3],[0,0,0]]. Rank = 1. By Rank-Nullity Theorem,
dim(Nul A) = n - rank = 3 - 1 = 2. Options A, B, D contradict this fundamental theorem.
4. A linear transformation T: ℝⁿ → ℝᵐ is one-to-one if and only if:
A. The columns of its standard matrix span ℝᵐ
B. The kernel of T contains only the zero vector
C. n = m
D. T is onto
B. The kernel of T contains only the zero vector
Rationale: T is one-to-one ⇔ ker(T) = {0}. Option A describes onto (surjective); C is unnecessary (e.g., T:
ℝ²→ℝ³ can be one-to-one); D is the opposite property. The kernel condition is the precise
characterization.
5. Which vector is an eigenvector of A = [[2, 0], [0, 3]]?
A. (1, 1)
B. (1, 0)
C. (0, 1)
D. Both B and C
D. Both B and C
Rationale: A(1,0) = (2,0) = 2·(1,0); A(0,1) = (0,3) = 3·(0,1). So both are eigenvectors (with eigenvalues 2
and 3). (1,1) maps to (2,3), which is not a scalar multiple of (1,1), so not an eigenvector.
6. The set S = {(1,0,1), (0,1,1), (1,1,0)} in ℝ³ is:
A. Linearly dependent
B. A basis for ℝ³
C. Orthogonal
D. Contains the zero vector
B. A basis for ℝ³
Subspaces - Prof. Mumo (2026/2027)
Core Linear Algebra Concepts | Key Domains: Linear Transformations & Their Matrices, Kernel (Null
Space) & Image (Column Space), Basis & Dimension, Eigenvalues & Eigenvectors, Orthogonality &
Projections, and Vector Space Axioms | Expert-Aligned Structure | Multiple-Choice Exercise Format
Introduction
This structured Linear Algebra exercise set for 2026/2027 provides 55 multiple-choice questions
with correct answers and rationales. It focuses on the abstract and geometric understanding of
linear transformations, subspaces, and their properties, which are fundamental to advanced
mathematics, engineering, and data science.
Exercise Structure:
● Conceptual & Computational Exercise Bank: (55 MULTIPLE-CHOICE QUESTIONS)
Answer Format
All correct answers, vector space determinations, and computational results must appear in bold
and cyan blue, accompanied by concise rationales explaining the application of a definition or
theorem (e.g., "The set is a subspace because it is closed under addition and scalar multiplication"),
the steps of a calculation (e.g., finding a basis for the null space by row reduction), the properties of
a linear transformation (e.g., one-to-one if kernel is trivial), the geometric interpretation (e.g.,
eigenvectors indicate directions unchanged by the transformation), and why the alternative
multiple-choice options violate linear algebra principles or contain algebraic errors.
Conceptual & Computational Exercise Bank (55
Multiple-Choice Questions)
1. Which of the following sets is a subspace of ℝ³?
, A. {(x, y, z) | x + y + z = 1}
B. {(x, y, z) | x ≥ 0}
C. {(x, y, z) | x + y + z = 0}
D. {(x, y, z) | xyz = 0}
C. {(x, y, z) | x + y + z = 0}
Rationale: A subspace must contain the zero vector, be closed under addition, and closed under scalar
multiplication. Option C contains (0,0,0), and if u and v satisfy x+y+z=0, so does u+v and cu. Option A
fails (0,0,0) ∉ set; B is not closed under scalar multiplication (e.g., -1·(1,0,0) = (-1,0,0) ∉ set); D is not
closed under addition (e.g., (1,0,0)+(0,1,0)=(1,1,0) ∈ set, but (1,0,0)+(0,0,1)=(1,0,1) ∉ set since 1·0·1=0
is true—wait, actually (1,0,1) has product 0, so it is in D. However, (1,1,0) and (1,0,1) are in D, but their
sum (2,1,1) has product 2≠0, so not in D. Thus D is not closed under addition.
2. Let T: ℝ² → ℝ² be defined by T(x, y) = (2x, 3y). What is the matrix of T with respect to the
standard basis?
A. [[2, 0], [0, 3]]
B. [[2, 3], [0, 0]]
C. [[0, 2], [3, 0]]
D. [[1, 2], [3, 1]]
A. [[2, 0], [0, 3]]
Rationale: The standard matrix is formed by T(e₁) and T(e₂) as columns. T(1,0) = (2,0); T(0,1) = (0,3).
So the matrix is diagonal with entries 2 and 3. Other options do not map basis vectors correctly.
3. What is the dimension of the null space of the matrix A = [[1, 2, 3], [2, 4, 6]]?
A. 0
B. 1
C. 2
D. 3
C. 2
, Rationale: Row reduce A: R₂ ← R₂ - 2R₁ gives [[1,2,3],[0,0,0]]. Rank = 1. By Rank-Nullity Theorem,
dim(Nul A) = n - rank = 3 - 1 = 2. Options A, B, D contradict this fundamental theorem.
4. A linear transformation T: ℝⁿ → ℝᵐ is one-to-one if and only if:
A. The columns of its standard matrix span ℝᵐ
B. The kernel of T contains only the zero vector
C. n = m
D. T is onto
B. The kernel of T contains only the zero vector
Rationale: T is one-to-one ⇔ ker(T) = {0}. Option A describes onto (surjective); C is unnecessary (e.g., T:
ℝ²→ℝ³ can be one-to-one); D is the opposite property. The kernel condition is the precise
characterization.
5. Which vector is an eigenvector of A = [[2, 0], [0, 3]]?
A. (1, 1)
B. (1, 0)
C. (0, 1)
D. Both B and C
D. Both B and C
Rationale: A(1,0) = (2,0) = 2·(1,0); A(0,1) = (0,3) = 3·(0,1). So both are eigenvectors (with eigenvalues 2
and 3). (1,1) maps to (2,3), which is not a scalar multiple of (1,1), so not an eigenvector.
6. The set S = {(1,0,1), (0,1,1), (1,1,0)} in ℝ³ is:
A. Linearly dependent
B. A basis for ℝ³
C. Orthogonal
D. Contains the zero vector
B. A basis for ℝ³
