Vector Spaces, Rank, Nullity and Orthogonality Short Notes for GATE CS
Vector Spaces, Rank, Nullity and Orthogonality short notes for GATE CS: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved pract
vector spaces rank nullity and orthogonality short notes
Summary: The Homogeneous System Checklist
Summary Checklist
The Homogeneous Execution Plan
1
The Baseline
Ax=0 always has the trivial solution x=0.
2
Square Matrices (m=n)
Nontrivial solutions exist ⟺det(A)=0⟺rank(A)<n.
3
Rectangular Matrices (m=n)
If variables exceed equations (n>m), nontrivial solutions are guaranteed. Otherwise, check if rank(A)<n.
4
The Null Space
N(A) is a subspace of Rn. It is closed under addition and scalar multiplication.
5
The Basis Vector Shortcut
If a standard basis vector ek∈N(A), the entire k-th column of A is exactly zero.
Summary: The Rank-Nullity Checklist
Summary Checklist
The Rank-Nullity Execution Plan
1
Identify dimensions
Matrix is m×n. Remember: rank + nullity = n.
2
If you have the matrix
Row reduce → count pivots (rank) and non-pivots (nullity).
3
If null space is described
Count basis vectors → that is nullity.
4
Apply the theorem
rank = n - nullity (or vice versa).
5
Verify the bound
Check: rank ≤min(m,n).
Quick Revision: Orthogonality Limits
Quick Revision: Orthogonality Limits
Summary Checklist
1Definition:u⊥v⟺u⋅v=0.
2Independence: Any set of non-zero pairwise orthogonal vectors is Linearly Independent.
3Maximum Limit: In Rn, max number of non-zero mutually orthogonal vectors is n.
4Exam Pattern: Identify "sum of products" as the dot product → Check dimensions → Check "non-zero" constraint → Answer is the dimension n.
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Question 1
Level 1: Warm-up
What is the maximum number of non-zero, mutually orthogonal vectors that can exist in R4?
Question 2
Level 1: Warm-up
What is the maximum number of non-zero, mutually orthogonal vectors that can exist in R7?
Question 3
Level 1: Warm-up
Let A be an m×n matrix. According to the Rank-Nullity Theorem, which of the following equations correctly relates the rank and nullity of A?
Question 4
Level 1: Warm-up
Two non-zero vectors u and v in Rn are orthogonal if and only if:
Question 5
Level 1: Warm-up
Let S be a set of non-zero vectors in Rn that are mutually orthogonal. Which of the following properties is ALWAYS guaranteed for S?
Question 6
Level 1: Warm-up
Let A be a 4×6 matrix. If the rank of A is 3, what is the nullity of A?
Question 7
Level 1: Warm-up
For a square n×n matrix A, which of the following conditions guarantees that the homogeneous system Ax=0 has a non-trivial solution?
Question 8
Level 1: Warm-up
Consider the matrix A=[132k]. For what value of k does the homogeneous system Ax=0 have a non-trivial solution?
Question 9
Level 1: Warm-up
Let A be a 3×5 matrix. Which of the following CANNOT be the rank of A?
Question 10
Level 1: Warm-up
Which of the following pairs of vectors in R3 is orthogonal?
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Vector Spaces, Rank, Nullity and Orthogonality Short Notes for GATE CS
Vector Spaces, Rank, Nullity and Orthogonality short notes for GATE CS: 3 study cards covering concepts, formulas, shortcuts and exam traps, plus solved practice questions.
Summary: The Homogeneous System Checklist
Summary Checklist
The Homogeneous Execution Plan
1
The Baseline
Ax=0 always has the trivial solution x=0.
2
Square Matrices (m=n)
Nontrivial solutions exist ⟺det(A)=0⟺rank(A)<n.
3
Rectangular Matrices (m=n)
If variables exceed equations (n>m), nontrivial solutions are guaranteed. Otherwise, check if rank(A)<n.
4
The Null Space
N(A) is a subspace of Rn. It is closed under addition and scalar multiplication.
5
The Basis Vector Shortcut
If a standard basis vector ek∈N(A), the entire k-th column of A is exactly zero.
Summary: The Rank-Nullity Checklist
Summary Checklist
The Rank-Nullity Execution Plan
1
Identify dimensions
Matrix is m×n. Remember: rank + nullity = n.
2
If you have the matrix
Row reduce → count pivots (rank) and non-pivots (nullity).
3
If null space is described
Count basis vectors → that is nullity.
4
Apply the theorem
rank = n - nullity (or vice versa).
5
Verify the bound
Check: rank ≤min(m,n).
Quick Revision: Orthogonality Limits
Quick Revision: Orthogonality Limits
Summary Checklist
1Definition:u⊥v⟺u⋅v=0.
2Independence: Any set of non-zero pairwise orthogonal vectors is Linearly Independent.
3Maximum Limit: In Rn, max number of non-zero mutually orthogonal vectors is n.
4Exam Pattern: Identify "sum of products" as the dot product → Check dimensions → Check "non-zero" constraint → Answer is the dimension n.
Vector Spaces, Rank, Nullity and Orthogonality: Solved Questions with Step-by-Step Explanations (10 Problems)
Question 1 · Engineering MathematicsMCQ
What is the maximum number of non-zero, mutually orthogonal vectors that can exist in R4?
A.
2
B.
3
C.
4
D.
8
Correct Answer:
C
Step-by-Step Solution
Key idea: This is a "maximum orthogonal set" question, recognisable because it asks for the maximum cardinality of a set of pairwise orthogonal non-zero vectors in a specific dimension.
Step 1: Recall the key theorem. Any set of non-zero, mutually orthogonal vectors in Rn is linearly independent.
Step 2: Apply the dimension bound. In Rn, you cannot have more than n linearly independent vectors. Therefore, the maximum number of non-zero mutually orthogonal vectors is exactly n.
Step 3: Substitute the given dimension. For R4, the maximum number is 4.
Answer: 4
Question 2 · Engineering MathematicsMCQ
What is the maximum number of non-zero, mutually orthogonal vectors that can exist in R7?
A.
5
B.
6
C.
7
D.
8
Correct Answer:
C
Step-by-Step Solution
Key idea: This is a "maximum orthogonal set size" question, recognisable because it asks for the maximum cardinality of a set of pairwise orthogonal non-zero vectors in a specific dimensional space.
Step 1: Recall the link between orthogonality and independence. Non-zero mutually orthogonal vectors are linearly independent.
Step 2: Recall the dimension bound. In Rn, you cannot have more than n linearly independent vectors.
Step 3: Apply to the given space. For R7, the maximum number of linearly independent vectors is 7. Therefore, the maximum number of non-zero mutually orthogonal vectors is exactly 7.
Answer: 7
Question 3 · Engineering MathematicsMCQ
Let A be an m×n matrix. According to the Rank-Nullity Theorem, which of the following equations correctly relates the rank and nullity of A?
A.
extrank(A)+extnullity(A)=m
B.
extrank(A)+extnullity(A)=n
C.
extrank(A)−extnullity(A)=n
D.
extrank(A)imesextnullity(A)=mimesn
Correct Answer:
B
Step-by-Step Solution
Key idea: This is a direct recall question about the statement of the Rank-Nullity Theorem.
Step 1: Recall the Rank-Nullity Theorem. It states that for an m×n matrix A, the sum of the rank and the nullity equals the number of columns.
Step 2: Identify the number of columns. For an m×n matrix, the number of columns is n.
Step 3: Write the equation: rank(A)+nullity(A)=n.
Answer: B
Question 4 · Engineering MathematicsMCQ
Two non-zero vectors u and v in Rn are orthogonal if and only if:
A.
u · v = 1
B.
u · v = 0
C.
||u|| = ||v||
D.
u × v = 0
Correct Answer:
B
Step-by-Step Solution
Key idea: This is a "definition of orthogonality" question, recognisable because it asks for the algebraic condition that defines when two vectors are perpendicular.
Step 1: Recall the geometric meaning of orthogonality. Two vectors are orthogonal if the angle between them is 90∘.
Step 2: Recall the algebraic definition. The dot product u⋅v=∥u∥∥v∥cos(θ). If θ=90∘, then cos(90∘)=0, so u⋅v=0.
Step 3: Evaluate the options. The condition u⋅v=0 is the exact definition of orthogonality.
Answer: u · v = 0
Question 5 · Engineering MathematicsMCQ
Let S be a set of non-zero vectors in Rn that are mutually orthogonal. Which of the following properties is ALWAYS guaranteed for S?
A.
S spans the entire space Rn.
B.
Every vector in S is a unit vector.
C.
S is a linearly independent set.
D.
S contains exactly n vectors.
Correct Answer:
C
Step-by-Step Solution
Key idea: This is an "orthogonal sets and linear independence" question, recognisable because it describes a set of mutually orthogonal non-zero vectors and asks for a guaranteed structural property.
Step 1: Recall the fundamental theorem. Any set of non-zero, pairwise orthogonal vectors is automatically linearly independent.
Step 2: Evaluate the other options to see why they are not guaranteed.
Spanning Rn: Not guaranteed. S could just be a single vector, or a subset of a basis.
Unit vectors: Not guaranteed. Orthogonal vectors can have any non-zero magnitude.
Exactly n vectors: Not guaranteed. The set could have fewer than n vectors (e.g., just 2 orthogonal vectors in R5).
Step 3: Conclude. Linear independence is the only property strictly guaranteed by the given conditions.
Answer: S is a linearly independent set.
Question 6 · Engineering MathematicsMCQ
Let A be a 4×6 matrix. If the rank of A is 3, what is the nullity of A?
A.
1
B.
2
C.
3
D.
4
Correct Answer:
C
Step-by-Step Solution
Key idea: This is a "rank-nullity computation" question, recognisable because it gives the dimensions of the matrix and the rank, and asks for the nullity.
Step 1: Recall the Rank-Nullity Theorem. For any m×n matrix A, the sum of the rank and the nullity equals the number of columns n: rank(A)+nullity(A)=n.
Step 2: Identify the number of columns. The matrix A is 4×6, so it has n=6 columns.
Step 3: Solve for the nullity. We are given rank(A)=3. Substituting into the theorem: 3+nullity(A)=6⟹nullity(A)=3.
Answer: 3
Question 7 · Engineering MathematicsMCQ
For a square n×n matrix A, which of the following conditions guarantees that the homogeneous system Ax=0 has a non-trivial solution?
A.
$\det(A)
eq 0$
B.
det(A)=0
C.
extrank(A)=n
D.
The rows of A are linearly independent.
Correct Answer:
B
Step-by-Step Solution
Key idea: This is a direct recall question about the determinant condition for non-trivial solutions in square homogeneous systems.
Step 1: Recall the property of homogeneous systems. The system Ax=0 always has the trivial solution x=0.
Step 2: For a square matrix A, a non-trivial solution exists if and only if the columns of A are linearly dependent.
Step 3: The columns of A are linearly dependent if and only if the determinant of A is zero.
Therefore, the condition that guarantees a non-trivial solution is det(A)=0.
Answer: B
Question 8 · Engineering MathematicsMCQ
Consider the matrix A=[132k]. For what value of k does the homogeneous system Ax=0 have a non-trivial solution?
A.
5
B.
6
C.
7
D.
8
Correct Answer:
B
Step-by-Step Solution
Key idea: This is a "determinant condition for non-trivial solutions" question, recognisable because it asks for a parameter that makes a square homogeneous system have non-trivial solutions.
Step 1: Recall the condition for non-trivial solutions. For a square matrix A, the system Ax=0 has a non-trivial solution if and only if det(A)=0.
Step 2: Compute the determinant of A. det(A)=(1)(k)−(2)(3)=k−6.
Step 3: Set the determinant to zero and solve for k. k−6=0⟹k=6.
Answer: 6
Question 9 · Engineering MathematicsMCQ
Let A be a 3×5 matrix. Which of the following CANNOT be the rank of A?
A.
1
B.
2
C.
3
D.
4
Correct Answer:
D
Step-by-Step Solution
Key idea: This is a "rank boundary" question, recognisable because it asks for an impossible value for the rank of a matrix with given dimensions.
Step 1: Recall the bounds on the rank of a matrix. For any m×n matrix A, the rank is bounded by the minimum of its dimensions: 0≤rank(A)≤min(m,n).
Step 2: Apply the bounds to the given matrix. The matrix A is 3×5, so m=3 and n=5. The maximum possible rank is min(3,5)=3.
Step 3: Evaluate the options. The rank can be 0, 1, 2, or 3. It cannot be 4, because that would exceed the number of rows.
Answer: 4
Question 10 · Engineering MathematicsMCQ
Which of the following pairs of vectors in R3 is orthogonal?
A.
[1,2,3] and [2,4,6]
B.
[1,0,−1] and [1,2,1]
C.
[1,1,1] and [−1,−1,−1]
D.
[0,1,0] and [0,2,0]
Correct Answer:
B
Step-by-Step Solution
Key idea: This is an "orthogonality check via dot product" question, recognisable because it asks to identify a perpendicular pair of vectors from a list of options.
Step 1: Recall the condition for orthogonality. Two vectors u and v are orthogonal if and only if their dot product u⋅v=0.
Step 2: Compute the dot product for each option.
Option A: (1)(2)+(2)(4)+(3)(6)=2+8+18=28=0.
Option B: (1)(1)+(0)(2)+(−1)(1)=1+0−1=0.
Option C: (1)(−1)+(1)(−1)+(1)(−1)=−1−1−1=−3=0.
Option D: (0)(0)+(1)(2)+(0)(0)=0+2+0=2=0.
Step 3: Conclude. Only the pair in Option B has a dot product of zero.