Vector Spaces, Rank, Nullity and Orthogonality PYQs for GATE CS

Solve 4+ Vector Spaces, Rank, Nullity and Orthogonality previous year questions for GATE CS with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Vector Spaces, Rank, Nullity and Orthogonality

Chapter Roadmap

Vector Spaces, Rank, Nullity and Orthogonality

Master the geometry of linear equations, from solution spaces to orthogonal bases.

Topic 1 · High Weightage

Homogeneous Systems and Nontrivial Null Spaces

Understand when systems have infinite solutions and define the null space.

Topic 2

Rank-Nullity Computations

Compute dimensions of row, column, and null spaces using the fundamental theorem.

Topic 3

Orthogonality and Maximum Orthogonal Sets

Find orthogonal bases and understand the limits of orthogonal sets in n-dimensional space.

Homogeneous Systems and Nontrivial Null Spaces

Vector Spaces · Topic 1

Homogeneous Systems & Nontrivial Null Spaces

Move beyond just solving equations to understanding the geometry of solutions.

Determine exactly when a system has solutions other than zero.
Identify the null space and use its subspace properties.
Crack tricky exam questions using standard basis vectors.

Vector Spaces, Rank, Nullity and Orthogonality: Solved Questions with Step-by-Step Explanations (4 Problems)

Question 1 · Engineering Mathematics · 2026_Set1 MSQ

Let

n > 1

. Consider an

n \times n

matrix

M

with its elements from

R

. Let the vector

(0, 1, 0, 0, \dots, 0) \in R^{n}

be in the null space of

M

.

Which of the following options is/are always correct?

A.
Determinant of $M$ is 1
B.
Determinant of $M$ is 0
C.
Rank of $M$ is 1
D.
There are at least two non-zero vectors in the null space of $M$

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This is a "null space membership implies matrix properties" question, recognisable because a specific non-zero vector is given as belonging to the null space, and we must deduce which matrix properties are guaranteed.

Step 1: Interpret the given information.

The vector $v = (0, 1, 0, 0, \dots, 0)^{T} \in R^{n}$ is in the null space of $M$ . This means:

$M v = 0$

Since $v \neq = 0$ , the homogeneous system $M x = 0$ has a non-trivial solution.

Step 2: Evaluate Option A (det $M = 1$ ).

A non-trivial null space means $M$ is singular. For a singular matrix, $det (M) = 0 \neq = 1$ . Option A is FALSE.

Step 3: Evaluate Option B (det $M = 0$ ).

Since $M v = 0$ with $v \neq = 0$ , the columns of $M$ are linearly dependent (specifically, the second column must be the zero vector, since $M v$ picks out the second column). A matrix with linearly dependent columns has determinant zero. Option B is TRUE.

Step 4: Evaluate Option C (rank $M = 1$ ).

We know $det (M) = 0$ , so rank $(M) < n$ . But the rank could be any value from $0$ to $n - 1$ . For example:

If $M = 0$ (zero matrix), rank = 0.
If $M = diag (1, 0, 1, 1, \dots, 1)$ , rank = $n - 1$ .

Neither is necessarily 1. Option C is FALSE.

Step 5: Evaluate Option D (at least two non-zero vectors in null space).

The null space is a vector subspace. If $v$ is in the null space, then every scalar multiple $c v$ is also in the null space. Since $v \neq = 0$ , the vectors $v$ and $2 v$ are both non-zero and distinct. In fact, there are infinitely many non-zero vectors in the null space. Option D is TRUE.

Answer: B, D

Question 2 · Engineering Mathematics · 2024_Set1 MSQ

Let $A$ be any $n \times m$ matrix, where $m > n$ . Which of the following statements is/are TRUE about the system of linear equations $A x = 0$ ?

A.
There exist at least $m - n$ linearly independent solutions to this system
B.
There exist $m - n$ linearly independent vectors such that every solution is a linear combination of these vectors
C.
There exists a non-zero solution in which at least $m - n$ variables are 0
D.
There exists a solution in which at least $n$ variables are non-zero

Correct Answer:

["A","C"]

Step-by-Step Solution

Key idea: This is a "wide matrix null space" question, recognisable because the matrix $A$ is $n \times m$ with $m > n$ (more columns than rows). The key tool is the Rank-Nullity Theorem combined with a structural result about sparse solutions.

Step 1: Set up the Rank-Nullity framework.

Let $r = rank (A)$ . Since $A$ has $n$ rows, $r \leq n$ . By the Rank-Nullity Theorem applied to the $m$ columns:

$nullity (A) = m - r \geq m - n > 0$

So the null space has dimension at least $m - n$ .

Step 2: Evaluate Option A.

The null space has dimension $m - r \geq m - n$ . Any basis of the null space contains $m - r$ linearly independent vectors. Since $m - r \geq m - n$ , there exist at least $m - n$ linearly independent solutions. Option A is TRUE.

Step 3: Evaluate Option B.

This option claims that $m - n$ vectors span the entire null space. That would require the null space dimension to be exactly $m - n$ , i.e., $r = n$ . But the rank could be less than $n$ . For example, if $A$ is the zero matrix, $r = 0$ and nullity $= m$ , which is strictly greater than $m - n$ . In that case $m - n$ vectors cannot span an $m$ -dimensional space. Option B is FALSE.

Step 4: Evaluate Option C.

A standard result in linear algebra states: for $A x = 0$ with $A$ being $n \times m$ and $m > n$ , there exists a non-trivial solution with at most $n$ non-zero entries. This means at least $m - n$ entries are zero. To see why, pick any $n + 1$ columns of $A$ ; they must be linearly dependent, giving a non-trivial combination involving at most $n + 1$ columns. A more careful argument using basic/non-basic variables in RREF shows we can find a solution with at most $n$ non-zero variables. Option C is TRUE.

Step 5: Evaluate Option D.

Consider $A = [I_{n} 0_{n \times (m - n)}]$ . The null space is spanned by $e_{n + 1}, \dots, e_{m}$ . Every non-zero solution has at most $m - n$ non-zero entries. If $m - n < n$ (e.g., $n = 5, m = 6$ ), then no solution has $n$ or more non-zero variables. Option D is FALSE.

Answer: A, C

Question 3 · Engineering Mathematics · 2021_Set2 MCQ

For two

n

-dimensional real vectors

P

and

Q

, the operation

s (P, Q)

is defined as follows:

s (P, Q) = i = 1 \sum n (P [i] \cdot Q [i])

Let

L

be a set of 10-dimensional non-zero real vectors such that for every pair of distinct vectors

P, Q \in L

s (P, Q) = 0

. What is the maximum cardinality possible for the set

L

A.
9
B.
10
C.
11
D.
100

Correct Answer:

Step-by-Step Solution

Key idea: This is a "maximum orthogonal set" question, recognisable because it defines the dot product

s (P, Q)

and asks for the maximum number of pairwise orthogonal non-zero vectors in a given dimension. Step 1: Decode the notation. The operation

s (P, Q) = \sum_{i = 1}^{n} P [i] \cdot Q [i]

is simply the standard dot product

P \cdot Q

. The condition

s (P, Q) = 0

for all distinct

P, Q \in L

means the vectors in

L

are pairwise orthogonal. Step 2: Apply the key theorem. A fundamental result in linear algebra states: any set of pairwise orthogonal non-zero vectors in

R^{n}

is linearly independent. Proof sketch: Suppose

\sum_{i = 1}^{k} c_{i} v_{i} = 0

where the

v_{i}

are pairwise orthogonal and non-zero. Take the dot product with

v_{j}

c_{j} (v_{j} \cdot v_{j}) = 0

Since

v_{j} \neq = 0

, we have

v_{j} \cdot v_{j} > 0

, so

c_{j} = 0

. This holds for all

j

. Step 3: Determine the maximum. Since the vectors in

L

are linearly independent and live in

R^{10}

, we cannot have more than

10

of them (the dimension of the space). The standard basis

{e_{1}, e_{2}, \dots, e_{10}}

achieves exactly

10

pairwise orthogonal non-zero vectors. Step 4: Eliminate wrong options. - 9 is achievable but not the maximum. - 11 exceeds the dimension of

R^{10}

, so impossible. - 100 is far beyond the dimension. Answer: B

Question 4 · Engineering Mathematics · 2021_Set2 NAT

Suppose that $P$ is a $4 \times 5$ matrix such that every solution of the equation $P x = 0$ is a scalar multiple of $[25431]^{T}$ . The rank of $P$ is __________.

Correct Answer:

Step-by-Step Solution

Key idea: This is a "read nullity from a geometric description" question, recognisable because the problem describes the entire null space as "every solution is a scalar multiple of" a single vector. This directly tells us the nullity.

Step 1: Extract the nullity from the problem statement.

The problem says every solution of $P x = 0$ is a scalar multiple of $v = [2, 5, 4, 3, 1]^{T}$ . This means the null space is:

$Null (P) = span {v}$

Since $v$ is a single non-zero vector, the null space is one-dimensional. Therefore:

$nullity (P) = 1$

Step 2: Apply the Rank-Nullity Theorem.

$P$ is a $4 \times 5$ matrix, so it has $5$ columns. The theorem states:

$rank (P) + nullity (P) = number of columns = 5$

Step 3: Solve for the rank.

$rank (P) = 5 - 1 = 4$

Step 4: Sanity check.

The rank of a $4 \times 5$ matrix cannot exceed $min (4, 5) = 4$ . Our answer of $4$ is consistent with this bound.

Answer: 4

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Vector Spaces, Rank, Nullity and Orthogonality PYQs for GATE CS

Chapter Roadmap: Vector Spaces, Rank, Nullity and Orthogonality

Vector Spaces, Rank, Nullity and Orthogonality

Homogeneous Systems and Nontrivial Null Spaces

Homogeneous Systems & Nontrivial Null Spaces

Vector Spaces, Rank, Nullity and Orthogonality: Solved Questions with Step-by-Step Explanations (4 Problems)

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

Step-by-Step Solution

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