Vector Spaces, Subspaces and Bases PYQs for GATE DA

Solve 3+ Vector Spaces, Subspaces and Bases previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Vector Spaces, Subspaces and Bases

Chapter Journey: Vector Spaces & Bases

Vector Spaces & Subspaces

Defining the playground. What makes a set a space?

Linear Independence

Removing redundancy. Do we really need all these vectors?

Bases & Orthonormal Bases

The minimal building blocks. Constructing perfect coordinate systems.

Geometry of Subspaces

Visualizing norms, balls, and intersections in higher dimensions.

Why this matters for GATE DA: Understanding bases allows you to change perspectives, simplify matrices, and compress data. This topic is the foundation for almost all advanced linear algebra applications in data science.

Hero Concept: The Basis as a Minimal Coordinate System

What is a Basis?

A basis for a vector space $V$ is a sequence of vectors $B = {v_{1}, v_{2}, \dots, v_{n}}$ that satisfies two critical conditions:

Linear Independence: No vector in the set can be written as a combination of the others. There is no "waste" or redundancy.
Spanning: Every vector in $V$ can be written as a linear combination of vectors in $B$ .

\forall v \in V, \exists c_{1}, \dots, c_{n} such that v = c_{1} v_{1} + \dots + c_{n} v_{n}

Why "Orthonormal" is the Gold Standard

An orthonormal basis is a basis where:

Orthogonal: All vectors are perpendicular to each other ( $v_{i} \cdot v_{j} = 0$ for $i \neq = j$ ).
Normal: Each vector has length 1 ( $∥ v_{i} ∥ = 1$ ).

The Superpower

If ${u_{1}, \dots, u_{n}}$ is an orthonormal basis, finding the coefficients $c_{i}$ for any vector $v$ becomes trivial:

c_{i} = v \cdot u_{i}

You do not need to solve a system of linear equations. You just project $v$ onto each basis vector. This is the core idea behind Fourier series, PCA, and many signal processing techniques.

Vector Spaces, Subspaces and Bases: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Linear Algebra · 2026 MCQ

Consider a set

S_{1} = {x = (x_{1}, x_{2}, x_{3})^{T} \in R^{3} ∣ x^{T} x \leq 16}

. Let

S_{2}

be another set which is a subspace of

R^{3}

with dimension two.
Which of the following gives the area of

S_{1} \cap S_{2}

A.
$16 π$
B.
$4 π$
C.
$4 π^{2}$
D.
$16 π^{2}$

Correct Answer:

Step-by-Step Solution

Insight: The intersection of a solid n-dimensional ball with a k-dimensional subspace passing through the origin is a solid k-dimensional ball of the exact same radius.

Exam route: $x^{T} x \leq 16$ means $∥ x ∥_{2} \leq 4$ , so it's a solid 3D ball of radius $R = 4$ . $S_{2}$ is a 2D subspace (a plane through the origin). The intersection is a 2D solid disk of radius 4. Area = $π R^{2} = 16 π$ .

Learning route:

Identify $S_{1}$ : The condition $x^{T} x \leq 16$ is equivalent to $∥ x ∥_{2}^{2} \leq 16$ , which means $∥ x ∥_{2} \leq 4$ . This defines a solid 3-dimensional $L_{2}$ ball centered at the origin with radius $R = 4$ .
Identify $S_{2}$ : A subspace of $R^{3}$ with dimension 2 is a flat plane that passes exactly through the origin.
Find the intersection $S_{1} \cap S_{2}$ : Slicing a solid 3D sphere with a plane that passes through its exact center (the origin) yields a solid 2D disk (a "great disk") with the same radius $R = 4$ .
Calculate the area: The area of a 2D disk of radius $R$ is $π R^{2}$ . Substituting $R = 4$ , we get Area = $π (4)^{2} = 16 π$ .

Question 2 · Linear Algebra · 2025 MSQ

Which of the following statements is/are correct?

A.
$R^{n}$ has a unique set of orthonormal basis vectors
B.
$R^{n}$ does not have a unique set of orthonormal basis vectors
C.
Linearly independent vectors in $R^{n}$ are orthonormal
D.
Orthonormal vectors $R^{n}$ are linearly independent

Correct Answer:

["B","D"]

Step-by-Step Solution

Insight: Orthonormal bases are not unique (any rotation works), but orthonormality strictly guarantees linear independence.

Exam route: Evaluate each option. A is false because we can rotate the standard basis. B is true. C is false because independent vectors need not be orthogonal. D is true because orthonormal vectors are always independent.

Learning route:

An orthonormal basis for $R^{n}$ requires vectors to be mutually orthogonal and unit length. The standard basis is one, but any rotation of it (e.g., in $R^{2}$ , using $(cos θ, sin θ)$ and $(- sin θ, cos θ)$ ) is another. Thus, it is not unique (A is false, B is true).
For C, consider vectors $(1, 0)$ and $(1, 1)$ in $R^{2}$ . They are linearly independent, but their dot product is $1 \neq = 0$ , so they are not orthogonal. Thus, independent vectors are not necessarily orthonormal.
For D, let $v_{1}, \dots, v_{k}$ be orthonormal. Suppose $\sum c_{i} v_{i} = 0$ . Taking the dot product with $v_{j}$ gives $c_{j} (v_{j} \cdot v_{j}) = 0 ⟹ c_{j} = 0$ . Thus, they are linearly independent.

Question 3 · Linear Algebra · 2024 MSQ

Select all choices that are subspaces of

R^{3}

.
Note:

R

denotes the set of real numbers.

A.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : x = α 110 + β 100, α, β \in R}$
B.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : x = α^{2} 120 + β^{2} 101, α, β \in R}$
C.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : 5 x_{1} + 2 x_{3} = 0, 4 x_{1} - 2 x_{2} + 3 x_{3} = 0}$
D.
${x = [x_{1} \x_{2} \x_{3}] \in R^{3} : 5 x_{1} + 2 x_{3} + 4 = 0}$

Correct Answer:

["A","C"]

Step-by-Step Solution

Insight: Subspaces must be closed under all scalar multiplications (including negative) and contain the origin. Spans of vectors and solutions to homogeneous linear equations are always subspaces.

Exam route: Check each option against the subspace criteria. A is a span (subspace). B uses squared parameters (only non-negative scalars, fails closure). C is a homogeneous system (subspace). D is a non-homogeneous system (fails zero vector).

Learning route:

Option A: This is the span of two vectors, $span {v_{1}, v_{2}}$ . The span of any set of vectors is always a subspace because it is closed under addition and scalar multiplication by definition.
Option B: The coefficients are $α^{2}$ and $β^{2}$ . Since squares of real numbers are always non-negative ( $α^{2} \geq 0$ ), we can only form non-negative linear combinations. If we multiply a vector in this set by $- 1$ , we cannot guarantee it remains in the set. It fails closure under scalar multiplication.
Option C: This is the solution set to a system of homogeneous linear equations ( $A x = 0$ ). The null space of any matrix is always a subspace. It contains the zero vector and is closed under addition and scalar multiplication.
Option D: The equation is $5 x_{1} + 2 x_{3} = - 4$ . To be a subspace, it must contain the zero vector $(0, 0, 0)$ . Substituting gives $5 (0) + 2 (0) = 0 \neq = - 4$ . It fails the zero vector test.

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Vector Spaces, Subspaces and Bases PYQs for GATE DA

Chapter Roadmap: Vector Spaces, Subspaces and Bases

Chapter Journey: Vector Spaces & Bases

Hero Concept: The Basis as a Minimal Coordinate System

What is a Basis?

Why "Orthonormal" is the Gold Standard

Vector Spaces, Subspaces and Bases: Solved Questions with Step-by-Step Explanations (3 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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