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Orthogonality, Projections and Linear Systems PYQs for GATE DA

Solve 2+ Orthogonality, Projections and Linear Systems previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

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Question 1

2024 PYQ

Level 3: Exam Standard

Which of the following statements is/are TRUE? Note: R denotes the set of real numbers.

Question 2

2024 PYQ

Level 3: Exam Standard

Let R be the set of real numbers, U be a subspace of R^{3} and M \in R^{3 \times 3} be the matrix corresponding to the projection on to the subspace U . Which of the following statements is/are TRUE?

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Orthogonality, Projections and Linear Systems PYQs for GATE DA

Solve 2+ Orthogonality, Projections and Linear Systems previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Orthogonality, Projections and Linear Systems

Chapter Journey

Step 1: Projection Matrices, Null Space and Idempotence (Current)

Geometric intuition of orthogonal projections. Algebraic properties: Idempotence and Symmetry. Column space, Null space, and Eigenvalues.

Step 2: Consistency and Solution Sets of Linear Systems

Conditions for a linear system to have a solution. Geometric interpretation of consistency. Structuring the complete solution set.

Goal: Master the transition from projecting vectors to solving and analyzing linear systems.

The Geometry of Projection Matrices

The Core Idea

A projection matrix $P$ maps any vector $b \in R^{n}$ to a vector $p$ in a target subspace $U$ , such that the error vector $e = b - p$ is orthogonal to $U$ .

Geometric Meaning

$p$ is the "shadow" or closest point in $U$ .
$e$ is the "error" or perpendicular drop.
$P$ isolates the component of $b$ that lives inside $U$ .

Orthogonality, Projections and Linear Systems: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Linear Algebra · 2024 MSQ

Which of the following statements is/are TRUE?
Note:

R

denotes the set of real numbers.

A. There exist $M \in R^{3 \times 3}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has a unique
solution and $M x = q$ has infinite solutions.
B. There exist $M \in R^{3 \times 3}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has no solutions
and $M x = q$ has infinite solutions.
C. There exist $M \in R^{2 \times 3}$ , $p \in R^{2}$ , and $q \in R^{2}$ such that $M x = p$ has a unique
solution and $M x = q$ has infinite solutions.
D. There exist $M \in R^{3 \times 2}$ , $p \in R^{3}$ , and $q \in R^{3}$ such that $M x = p$ has a unique
solution and $M x = q$ has no solutions.

Correct Answer:

["B","D"]

Step-by-Step Solution

Key idea: This question tests the consistency conditions of linear systems $A x = b$ based on the shape of matrix $A$ (square, wide, or tall) and its rank.

Step 1: Analyze Option A ( $M \in R^{3 \times 3}$ ).

For a square matrix $M$ , if $M x = p$ has a unique solution, then $det (M) \neq = 0$ ( $M$ is invertible).

If $M$ is invertible, $M x = q$ must also have a unique solution ( $x = M^{- 1} q$ ) for ANY $q$ .

It is impossible for an invertible matrix to have infinite solutions for any RHS.

Thus, Option A is FALSE.

Step 2: Analyze Option B ( $M \in R^{3 \times 3}$ ).

Can $M x = p$ have no solution and $M x = q$ have infinite solutions?

Yes. Let $M$ be singular (Rank < 3).

Example: $M = 100010000$ .

Let $p = 001$ . System is inconsistent (no solution) because $0 \neq = 1$ .
Let $q = 000$ . System is $x = 0, y = 0, 0 = 0$ . $z$ is free. Infinite solutions.

Thus, Option B is TRUE.

Step 3: Analyze Option C ( $M \in R^{2 \times 3}$ ).

Can $M x = p$ have a unique solution?

$M$ has 3 columns (variables) and 2 rows (equations).

Max Rank is 2.

Number of free variables = $3 - Rank (M) \geq 3 - 2 = 1$ .

If a solution exists, there is at least 1 free variable, implying infinite solutions.

A unique solution is impossible for a wide matrix ( $n > m$ ).

Thus, Option C is FALSE.

Step 4: Analyze Option D ( $M \in R^{3 \times 2}$ ).

Can $M x = p$ have a unique solution?

$M$ has 2 columns. Max Rank is 2.

If Rank( $M$ )=2, the columns are linearly independent.

If $p$ is in the column space, the solution is unique (0 free variables).

Can $M x = q$ have no solution?

Yes, if $q$ is not in the column space (which is a 2D plane in $R^{3}$ ).

So, Unique Solution AND No Solution are both possible for different RHS vectors.

Thus, Option D is TRUE.

Answer: Options B and D are TRUE.

Question 2 · Linear Algebra · 2024 MSQ

Let

R

be the set of real numbers,

U

be a subspace of

R^{3}

and

M \in R^{3 \times 3}

be the
matrix corresponding to the projection on to the subspace

U

.
Which of the following statements is/are TRUE?

A. If $U$ is a 1-dimensional subspace of $R^{3}$ , then the null space of $M$ is a
1-dimensional subspace.
B. If $U$ is a 2-dimensional subspace of $R^{3}$ , then the null space of $M$ is a
1-dimensional subspace.
C.
$M^{2} = M$
D.
$M^{3} = M$

Correct Answer:

["B","C","D"]

Step-by-Step Solution

Key idea: This question tests the fundamental algebraic and geometric properties of orthogonal projection matrices, specifically the Rank-Nullity Theorem and idempotence.

Step 1: Analyze the Null Space dimension (Options A and B).

Let $M$ be the projection matrix onto subspace $U \subseteq R^{3}$ .

The column space of $M$ , denoted $C (M)$ , is exactly the target subspace $U$ . Thus, $rank (M) = dim (U)$ .

By the Rank-Nullity Theorem for a $3 \times 3$ matrix:

$rank (M) + nullity (M) = 3$

$dim (Null (M)) = 3 - dim (U)$

Check Option A: If $dim (U) = 1$ , then $dim (Null (M)) = 3 - 1 = 2$ . Option A claims it is 1-dimensional. This is FALSE.

Check Option B: If $dim (U) = 2$ , then $dim (Null (M)) = 3 - 2 = 1$ . Option B claims it is 1-dimensional. This is TRUE.

Step 2: Analyze Idempotence (Option C).

A matrix $M$ is a projection matrix if and only if it is idempotent, meaning $M^{2} = M$ .

Geometrically, if you project a vector $v$ onto $U$ to get $p$ , projecting $p$ again yields $p$ itself because $p$ is already in $U$ .

Algebraically: $M (M v) = M v ⟹ M^{2} = M$ .

Thus, Option C is TRUE.

Step 3: Analyze Higher Powers (Option D).

Since $M^{2} = M$ , we can compute $M^{3}$ :

$M^{3} = M^{2} \cdot M = M \cdot M = M^{2} = M$

By induction, $M^{k} = M$ for all integers $k \geq 1$ .

Thus, $M^{3} = M$ is TRUE.

Answer: Options B, C, and D are TRUE.

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Honest progress

Unlock the whole course

Orthogonality, Projections and Linear Systems PYQs for GATE DA

Chapter Roadmap: Orthogonality, Projections and Linear Systems

Chapter Journey

The Geometry of Projection Matrices

The Core Idea

Orthogonality, Projections and Linear Systems: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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