Vector Spaces, Subspaces and Bases Notes for GATE DA
Vector Spaces, Subspaces and Bases notes for GATE DA: 16 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
vector spaces subspaces and bases notes
Chapter Roadmap: Vector Spaces, Subspaces and Bases
Chapter Journey: Vector Spaces & Bases
1
Vector Spaces & Subspaces
Defining the playground. What makes a set a space?
2
Linear Independence
Removing redundancy. Do we really need all these vectors?
3
Bases & Orthonormal Bases
The minimal building blocks. Constructing perfect coordinate systems.
4
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={v1,v2,…,vn} 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.
∀v∈V,∃c1,…,cn such that v=c1v1+⋯+cnvn
Why "Orthonormal" is the Gold Standard
An orthonormal basis is a basis where:
Orthogonal: All vectors are perpendicular to each other (vi⋅vj=0 for i=j).
Normal: Each vector has length 1 (∥vi∥=1).
The Superpower
If {u1,…,un} is an orthonormal basis, finding the coefficients ci for any vector v becomes trivial:
ci=v⋅ui
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.
Subspace Checklist: The Three Rules
Subspace Checklist: The Three Rules
A subset W of a vector space V is a subspace if and only if it passes these three checks:
1
Zero Vector
0∈W. The space must contain the origin.
2
Closed under Addition
If u,v∈W, then u+v∈W.
3
Closed under Scalar Multiplication
If u∈W and c is a scalar, then cu∈W.
Common Examples in Rn
Lines through origin Yes, subspace
Planes through origin Yes, subspace
Lines NOT through origin Fails zero vector
First Quadrant (x≥0) Fails scaling
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Question 1
Level 1: Warm-up
Assertion (A): In the Gram-Schmidt process, the first step to find u1 is to normalize the initial vector v1.
Reason (R): Normalization ensures that u1 is orthogonal to all other vectors in the basis.
Question 2
Level 1: Warm-up
Assertion (A): In the Gram-Schmidt process, the vector w2 is constructed by subtracting the projection of v2 onto u1 from v2.
Reason (R): This subtraction step ensures that the resulting vector w2 has a unit length.
Question 3
Level 1: Warm-up
Assertion (A): In the Gram-Schmidt process, the final step to find u2 is to divide the orthogonalized vector w2 by its norm ∥w2∥.
Reason (R): This division step ensures that the resulting vector u2 is orthogonal to u1.
Question 4
Level 1: Warm-up
Assertion (A): In the Gram-Schmidt process, the projection of v2 onto u1 is calculated as (v2⋅u1)u1.
Reason (R): This projection vector is orthogonal to u1.
Question 5
Level 1: Warm-up
Let {u1,u2} be an orthonormal basis for R2, where u1=(0,1) and u2=(−1,0). If a vector v=(3,4) is expressed as v=c1u1+c2u2, what is the value of the coefficient c2?
Question 6
Level 1: Warm-up
What is the minimum number of vectors in R4 that guarantees the set is linearly dependent, regardless of the specific vectors chosen?
Question 7
Level 1: Warm-up
Let {u1,u2} be an orthonormal basis for R2 with u1=(1,0) and u2=(0,−1). If a vector v=(4,−5) is expressed as v=c1u1+c2u2, what is the value of c1−c2?
Question 8
Level 1: Warm-up
If a matrix A is formed by placing 4 vectors from R6 as its columns, what is the minimum rank A must have for these vectors to be linearly independent?
Question 9
Level 1: Warm-up
Let {u1,u2} be an orthonormal basis for R2, where u1=(53,54) and u2=(−54,53). If a vector v=(10,5) is expressed as v=c1u1+c2u2, what is the value of c1+c2?
Question 10
Level 1: Warm-up
If a matrix A is formed by placing 3 vectors from R5 as its columns, what is the minimum rank A must have for these vectors to be linearly independent?
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Vector Spaces, Subspaces and Bases Notes for GATE DA
Vector Spaces, Subspaces and Bases notes for GATE DA: 16 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Chapter Roadmap: Vector Spaces, Subspaces and Bases
Chapter Journey: Vector Spaces & Bases
1
Vector Spaces & Subspaces
Defining the playground. What makes a set a space?
2
Linear Independence
Removing redundancy. Do we really need all these vectors?
3
Bases & Orthonormal Bases
The minimal building blocks. Constructing perfect coordinate systems.
4
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={v1,v2,…,vn} 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.
∀v∈V,∃c1,…,cn such that v=c1v1+⋯+cnvn
Why "Orthonormal" is the Gold Standard
An orthonormal basis is a basis where:
Orthogonal: All vectors are perpendicular to each other (vi⋅vj=0 for i=j).
Normal: Each vector has length 1 (∥vi∥=1).
The Superpower
If {u1,…,un} is an orthonormal basis, finding the coefficients ci for any vector v becomes trivial:
ci=v⋅ui
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.
Subspace Checklist: The Three Rules
Subspace Checklist: The Three Rules
A subset W of a vector space V is a subspace if and only if it passes these three checks:
1
Zero Vector
0∈W. The space must contain the origin.
2
Closed under Addition
If u,v∈W, then u+v∈W.
3
Closed under Scalar Multiplication
If u∈W and c is a scalar, then cu∈W.
Common Examples in Rn
Lines through origin Yes, subspace
Planes through origin Yes, subspace
Lines NOT through origin Fails zero vector
First Quadrant (x≥0) Fails scaling
Testing Linear Independence: The Rank Method
Testing Linear Independence: The Rank Method
Given vectors v1,…,vk in Rn:
Step 1: Form matrix A=[v1∣v2∣⋯∣vk].
Step 2: Compute the Rank of A.
Decision Rule
If Rank(A)=k: The vectors are Linearly Independent.
If Rank(A)<k: The vectors are Linearly Dependent.
Special Case (Square Matrix)
If k=n (square matrix), compute det(A):
det(A)=0⟹ Independent.
det(A)=0⟹ Dependent.
Vector Spaces, Subspaces and Bases: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Linear AlgebraMCQ
Assertion (A): In the Gram-Schmidt process, the first step to find u1 is to normalize the initial vector v1.
Reason (R): Normalization ensures that u1 is orthogonal to all other vectors in the basis.
A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.
Correct Answer:
C
Step-by-Step Solution
Key idea: Distinguish between normalization (unit length) and orthogonalization (perpendicularity).
Step 1: Evaluate Assertion (A). The Gram-Schmidt process starts by taking v1 and dividing by its norm to get u1. This is normalization. So, A is true.
Step 2: Evaluate Reason (R). Normalization only ensures the vector has a length of 1 (∥u1∥=1). It does not make the vector orthogonal to anything. Orthogonality is achieved in subsequent steps by subtracting projections. So, R is false.
Step 3: Match with options. A is true, R is false.
Answer: A is true, but R is false.
Question 2 · Linear AlgebraMCQ
Assertion (A): In the Gram-Schmidt process, the vector w2 is constructed by subtracting the projection of v2 onto u1 from v2.
Reason (R): This subtraction step ensures that the resulting vector w2 has a unit length.
A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.
Correct Answer:
C
Step-by-Step Solution
Key idea: Distinguish between orthogonalization (subtraction) and normalization (division by norm).
Step 1: Evaluate Assertion (A). The formula for w2 is indeed w2=v2−(v2⋅u1)u1. This is correct.
Step 2: Evaluate Reason (R). The subtraction step removes the component of v2 that is parallel to u1, making w2 orthogonal to u1. It does not ensure unit length. Normalization (u2=w2/∥w2∥) is required for unit length.
Step 3: Conclude that A is true, but R is false.
Answer: A is true, but R is false.
Question 3 · Linear AlgebraMCQ
Assertion (A): In the Gram-Schmidt process, the final step to find u2 is to divide the orthogonalized vector w2 by its norm ∥w2∥.
Reason (R): This division step ensures that the resulting vector u2 is orthogonal to u1.
A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.
Correct Answer:
C
Step-by-Step Solution
Key idea: Distinguish between orthogonalization (subtraction) and normalization (division by norm).
Step 1: Evaluate Assertion (A). The formula for u2 is indeed u2=w2/∥w2∥. This is correct.
Step 2: Evaluate Reason (R). The division step scales w2 to have a length of 1. It does not ensure orthogonality. Orthogonality was already achieved in the previous step when w2 was constructed by subtracting the projection of v2 onto u1.
Step 3: Conclude that A is true, but R is false.
Answer: A is true, but R is false.
Question 4 · Linear AlgebraMCQ
Assertion (A): In the Gram-Schmidt process, the projection of v2 onto u1 is calculated as (v2⋅u1)u1.
Reason (R): This projection vector is orthogonal to u1.
A.
Both A and R are true, and R is the correct explanation of A.
B.
Both A and R are true, but R is not the correct explanation of A.
C.
A is true, but R is false.
D.
A is false, but R is true.
Correct Answer:
C
Step-by-Step Solution
Key idea: Distinguish between the projection vector and the orthogonalized remainder vector.
Step 1: Evaluate Assertion (A). The formula for the projection of v2 onto the unit vector u1 is indeed (v2⋅u1)u1. This is correct.
Step 2: Evaluate Reason (R). The projection vector (v2⋅u1)u1 is a scalar multiple of u1, meaning it is parallel to u1, not orthogonal to it. Orthogonality is achieved by the remainder vector w2=v2−(v2⋅u1)u1.
Step 3: Conclude that A is true, but R is false.
Answer: A is true, but R is false.
Question 5 · Linear AlgebraMCQ
Let {u1,u2} be an orthonormal basis for R2, where u1=(0,1) and u2=(−1,0). If a vector v=(3,4) is expressed as v=c1u1+c2u2, what is the value of the coefficient c2?
A.
-3
B.
3
C.
4
D.
-4
Correct Answer:
A
Step-by-Step Solution
Key idea: In an orthonormal basis, coefficients are found using the dot product.
Step 1: Recall the formula for coordinates in an orthonormal basis: ci=v⋅ui.
Step 2: We need c2, so we compute the dot product of v and u2.
What is the minimum number of vectors in R4 that guarantees the set is linearly dependent, regardless of the specific vectors chosen?
A.
3
B.
4
C.
5
D.
6
Correct Answer:
C
Step-by-Step Solution
Key idea: In Rn, any set of more than n vectors is linearly dependent.
Step 1: Identify the dimension of the space. Here, the space is R4, so n=4.
Step 2: Apply the theorem: Any set of k vectors in Rn is linearly dependent if k>n.
Step 3: We need the minimum k such that k>4. The smallest integer greater than 4 is 5.
Answer: 5
Question 7 · Linear AlgebraMCQ
Let {u1,u2} be an orthonormal basis for R2 with u1=(1,0) and u2=(0,−1). If a vector v=(4,−5) is expressed as v=c1u1+c2u2, what is the value of c1−c2?
A.
-1
B.
9
C.
-9
D.
1
Correct Answer:
A
Step-by-Step Solution
Key idea: In an orthonormal basis, coordinates are found via dot products.
Step 4: Compute the requested value: c1−c2=4−5=−1.
Answer: -1
Question 8 · Linear AlgebraMCQ
If a matrix A is formed by placing 4 vectors from R6 as its columns, what is the minimum rank A must have for these vectors to be linearly independent?
A.
2
B.
4
C.
6
D.
10
Correct Answer:
B
Step-by-Step Solution
Key idea: The rank method states that vectors are independent if the rank equals the number of vectors.
Step 1: Identify the number of vectors. Here, there are k=4 vectors.
Step 2: Apply the decision rule from the rank method: The vectors are linearly independent if and only if Rank(A)=k.
Step 3: Therefore, the minimum (and exact) rank required is 4. The dimension of the space (n=6) is irrelevant as long as k≤n.
Answer: 4
Question 9 · Linear AlgebraMCQ
Let {u1,u2} be an orthonormal basis for R2, where u1=(53,54) and u2=(−54,53). If a vector v=(10,5) is expressed as v=c1u1+c2u2, what is the value of c1+c2?
A.
5
B.
21
C.
-5
D.
15
Correct Answer:
A
Step-by-Step Solution
Key idea: In an orthonormal basis, coordinates are found via dot products.
Step 4: Compute the requested sum: c1+c2=10+(−5)=5.
Answer: 5
Question 10 · Linear AlgebraMCQ
If a matrix A is formed by placing 3 vectors from R5 as its columns, what is the minimum rank A must have for these vectors to be linearly independent?
A.
2
B.
3
C.
5
D.
8
Correct Answer:
B
Step-by-Step Solution
Key idea: The rank method states that vectors are independent if the rank equals the number of vectors.
Step 1: Identify the number of vectors. Here, there are k=3 vectors.
Step 2: Apply the decision rule from the rank method: The vectors are linearly independent if and only if Rank(A)=k.
Step 3: Therefore, the minimum (and exact) rank required is 3. The dimension of the space (n=5) is irrelevant as long as k≤n.