Orthogonal, Projection and Special Matrices Previous Year Questions (PYQs) for GATE DA: 3+ Solved Questions with Step-by-Step Solutions

Solve 3+ Orthogonal, Projection and Special Matrices previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Orthogonal, Projection and Special Matrices

Chapter Roadmap

Welcome to the geometry of linear transformations. This chapter bridges the gap between abstract matrix algebra and visual geometric intuition.

Your Learning Journey

1. Projection Matrices and Quadratic Forms

Classifying surfaces, Rayleigh quotient, centering matrix. High Weightage

2. Orthogonal and Involutory Matrices

Length-preserving transformations, rotations, own inverses. Moderate Weightage

By the end of this chapter, you will:

Instantly classify the shape of any quadratic form using eigenvalues.
Recognize projection matrices and deduce their eigenvalues without calculation.
Solve constrained optimization problems in seconds.

The Geometry of Quadratic Forms

A quadratic form is a scalar-valued polynomial where every term has a degree of exactly two.

For a column vector $x \in R^{n}$ and an $n \times n$ matrix $A$ , the quadratic form is written as:

Q (x) = x^{T} A x

Geometric Intuition

If you set $Q (x) = c$ for some constant $c$ , you get a geometric surface:

In 2D: Conic sections (ellipses, hyperbolas, parabolas).
In 3D: Quadric surfaces (ellipsoids, hyperboloids).

The matrix $A$ acts as the "DNA" of this surface, dictating its orientation, stretching, and curvature.

Orthogonal, Projection and Special Matrices: Solved Questions with Step-by-Step Explanations (3 Problems)

Question 1 · Linear Algebra MSQ

Let

M = (I_{n} - \frac{1}{n} 1 1^{T})

be a matrix, where

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

and

I_{n}

is the identity matrix of order

n

.

Which of the following options is/are correct?

A.
$M^{T} = M$
B.
$M^{2} = I_{n}$
C.
$T r a ce (M) = n$
D.
$M$ is a projection matrix

Correct Answer:

["A","D"]

Step-by-Step Solution

Key idea: This question tests the algebraic properties of the centering matrix $M = I_{n} - \frac{1}{n} 1 1^{T}$ .

Step 1: Analyze Option A ( $M^{T} = M$ ). The transpose of a sum is the sum of transposes, and $(1 1^{T})^{T} = (1^{T})^{T} 1^{T} = 1 1^{T}$ . Therefore, $M^{T} = (I_{n} - \frac{1}{n} 1 1^{T})^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Option A is correct.

Step 2: Analyze Option B ( $M^{2} = I_{n}$ ). Compute $M^{2} = (I_{n} - \frac{1}{n} 1 1^{T}) (I_{n} - \frac{1}{n} 1 1^{T}) = I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} 1 (1^{T} 1) 1^{T}$ . Notice that $1^{T} 1$ is the dot product of the all-ones vector with itself, which equals $n$ . Substituting this gives $I_{n} - \frac{2}{n} 1 1^{T} + \frac{1}{n ^{2}} (n) 1 1^{T} = I_{n} - \frac{1}{n} 1 1^{T} = M$ . Since $M^{2} = M$ (and $M \neq = I_{n}$ for $n > 1$ ), Option B is incorrect.

Step 3: Analyze Option C ( $T r a ce (M) = n$ ). Using the linearity of trace, $T r a ce (M) = T r a ce (I_{n}) - \frac{1}{n} T r a ce (1 1^{T})$ . We know $T r a ce (I_{n}) = n$ and $T r a ce (1 1^{T}) = 1^{T} 1 = n$ . Thus, $T r a ce (M) = n - \frac{1}{n} (n) = n - 1$ . Option C is incorrect.

Step 4: Analyze Option D ( $M$ is a projection matrix). A matrix is an orthogonal projection matrix if and only if it is symmetric ( $M^{T} = M$ ) and idempotent ( $M^{2} = M$ ). We proved both in Steps 1 and 2. Therefore, Option D is correct.

Answer: ["A", "D"]

Question 2 · Linear Algebra NAT

Let

A = (I_{n} - \frac{1}{n} 1 1^{T})

be a matrix, where

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

and

I_{n}

is the identity matrix of order

n

.

The value of max over

S

x^{T} A x

, where

S = {x \in R^{n} ∣ x^{T} x = 1}

, is __________ . (Answer in integer)

Correct Answer:

Step-by-Step Solution

Key idea: This is a Rayleigh quotient maximization problem involving a rank-1 update to the identity matrix (the centering matrix).

Step 1: Recognize that for any real symmetric matrix $A$ , the maximum value of $x^{T} A x$ subject to the constraint $x^{T} x = 1$ is exactly the largest eigenvalue of $A$ ( $λ_{m a x}$ ).

Step 2: Identify the matrix $A = I_{n} - \frac{1}{n} 1 1^{T}$ . This is a symmetric matrix.

Step 3: Find the eigenvalues of the rank-1 matrix $1 1^{T}$ . The trace of $1 1^{T}$ is $n$ (since it has $n$ ones on the diagonal). Because it is a rank-1 matrix, its only non-zero eigenvalue is $n$ (with multiplicity 1), and the remaining $n - 1$ eigenvalues are $0$ .

Step 4: Scale by $\frac{1}{n}$ . The eigenvalues of $\frac{1}{n} 1 1^{T}$ are $1$ (multiplicity 1) and $0$ (multiplicity $n - 1$ ).

Step 5: Apply the shift by $I_{n}$ . The eigenvalues of $A = I_{n} - \frac{1}{n} 1 1^{T}$ are $1 - 1 = 0$ (multiplicity 1) and $1 - 0 = 1$ (multiplicity $n - 1$ ).

Step 6: The maximum eigenvalue of $A$ is $1$ . Therefore, the maximum value of $x^{T} A x$ on the unit sphere is $1$ .

Answer: 1

Question 3 · Linear Algebra MSQ

A.
$A$ must be orthogonal
B.
$A = I$ , where $I$ denotes the identity matrix, is the only solution
C.
The eigenvalues of $A$ are either +1 or −1
D.
$A$ has full rank

Correct Answer:

["C","D"]

Step-by-Step Solution

Key idea: This question tests the fundamental properties of involutory matrices, defined by the condition $A^{2} = I$ .

Step 1: Analyze Option A (" $A$ must be orthogonal"). An orthogonal matrix requires $A^{T} A = I$ . While $A^{2} = I$ implies $A = A^{- 1}$ , it does not guarantee $A^{T} = A^{- 1}$ . Counterexample: Let $A = (10 1 - 1)$ . Then $A^{2} = (1001) = I$ , but $A^{T} A = (11 0 - 1) (10 1 - 1) = (1112) \neq = I$ . Thus, Option A is false.

Step 2: Analyze Option B (" $A = I$ is the only solution"). The counterexample above, or simply $A = - I$ , proves this is false.

Step 3: Analyze Option C ("The eigenvalues of $A$ are either +1 or −1"). Let $λ$ be an eigenvalue of $A$ with eigenvector $x \neq = 0$ . Then $A x = λ x$ . Multiplying by $A$ gives $A^{2} x = λ (A x) = λ^{2} x$ . Since $A^{2} = I$ , we have $I x = λ^{2} x$ , so $(λ^{2} - 1) x = 0$ . Since $x \neq = 0$ , $λ^{2} = 1$ , which means $λ = 1$ or $λ = - 1$ . Thus, Option C is true.

Step 4: Analyze Option D (" $A$ has full rank"). Taking the determinant of both sides of $A^{2} = I$ yields $det (A^{2}) = det (I) ⟹ (det A)^{2} = 1 ⟹ det A = \pm 1$ . Since $det A \neq = 0$ , the matrix $A$ is invertible and therefore has full rank. Thus, Option D is true.

Answer: ["C", "D"]

Orthogonal, Projection and Special Matrices Previous Year Questions (PYQs) for GATE DA: 3+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Orthogonal, Projection and Special Matrices

Chapter Roadmap

Your Learning Journey

1. Projection Matrices and Quadratic Forms

2. Orthogonal and Involutory Matrices

By the end of this chapter, you will:

The Geometry of Quadratic Forms

The Geometry of Quadratic Forms

Geometric Intuition

Orthogonal, Projection and Special Matrices: 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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