Singular Values and Gram Matrices Previous Year Questions (PYQs) for GATE DA: 2+ Solved Questions with Step-by-Step Solutions

Solve 2+ Singular Values and Gram Matrices previous year questions for GATE DA with answers and detailed solutions. Free sample questions below.

Chapter Roadmap: Singular Values and Gram Matrices

Chapter Roadmap

Singular Values and Gram Matrices

1. Gram Matrices and Positive Definiteness

Inner products, geometry of vector sets, and conditions for positive definiteness.

2. Singular Values and Spectral Properties

SVD, pseudo-inverses, and spectral properties of special matrices.

What You Will Master

Construct and analyze Gram matrices to determine vector independence.
Rigorously test matrices for positive definiteness using Sylvester's criterion.
Decompose any matrix into its singular values and understand its geometric action.

Intuition: Measuring Geometry with Gram Matrices

Intuition: Measuring Geometry

Imagine you have a set of vectors in space. How do you capture all their lengths and the angles between them in a single, compact object?

The Gram matrix does exactly this.

It is a matrix composed entirely of inner products (dot products) between the vectors. Instead of looking at the vectors individually, the Gram matrix encodes the entire geometric relationship of the set. It tells you how much each vector "overlaps" with every other vector.

Singular Values and Gram Matrices: Solved Questions with Step-by-Step Explanations (2 Problems)

Question 1 · Linear Algebra MCQ

Let ${x_{1}, x_{2}, ..., x_{n}}$ be a set of linearly independent vectors in $R^{n}$ . Let the $(i, j)$ -th element of matrix $A \in R^{n \times n}$ be given by $A_{ij} = x_{i}^{T} x_{j}$ , $1 \leq i, j \leq n$ . Which one of the following statements is correct?

A.
$A$ is invertible
B.
$0$ is a singular value of $A$
C.
Determinant of $A$ is $0$
D.
$z^{T} A z = 0$ for some non-zero $z \in R^{n}$

Correct Answer:

Step-by-Step Solution

Key idea: This is a Gram matrix and positive definiteness question. We need to recognize that $A$ is formed by inner products of linearly independent vectors.

Step 1: Express $A$ in matrix form. Let $X$ be the $n \times n$ matrix whose columns are the vectors $x_{1}, x_{2}, \dots, x_{n}$ .

Step 2: Relate $X$ to $A$ . The $(i, j)$ -th element of $X^{T} X$ is the dot product of the $i$ -th row of $X^{T}$ (which is $x_{i}^{T}$ ) and the $j$ -th column of $X$ (which is $x_{j}$ ). Thus $(X^{T} X)_{ij} = x_{i}^{T} x_{j} = A_{ij}$ . So $A = X^{T} X$ .

Step 3: Since ${x_{1}, \dots, x_{n}}$ are linearly independent vectors in $R^{n}$ , the matrix $X$ is full rank (rank $n$ ) and therefore invertible.

Step 4: For any non-zero vector $z \in R^{n}$ : $z^{T} A z = z^{T} X^{T} X z = (X z)^{T} (X z) = ∥ X z ∥^{2}$ . Since $X$ is invertible, $X z \neq = 0$ for $z \neq = 0$ . Thus $∥ X z ∥^{2} > 0$ , which means $A$ is strictly positive definite.

Step 5: Evaluate options. Since $A$ is positive definite, all eigenvalues are strictly positive. Thus $det (A) > 0$ (not 0), 0 is not a singular value, and $z^{T} A z > 0$ for all non-zero $z$ . A positive definite matrix is always invertible.

Answer: Option A is correct.

Question 2 · Linear Algebra MSQ

A.
Singular values of $A$ are also its eigenvalues
B.
Singular values of $A$ are either 0 or 1
C.
Determinant of $A$ is 1
D.
$A$ is invertible

Correct Answer:

["A","B"]

Step-by-Step Solution

Key idea: This is a spectral properties of special matrices question, specifically involving a projection matrix formed by the outer products of orthonormal vectors.

Step 1: Understand the matrix $A$ . The matrix is $A = \sum_{i = 1}^{5} x_{i} x_{i}^{T}$ , where $x_{1}, \dots, x_{5}$ are orthonormal vectors in $R^{10}$ .

Step 2: Check symmetry. $A^{T} = (\sum_{i = 1}^{5} x_{i} x_{i}^{T})^{T} = \sum_{i = 1}^{5} x_{i} x_{i}^{T} = A$ . So $A$ is symmetric.

Step 3: Check idempotence. $A^{2} = (\sum_{i = 1}^{5} x_{i} x_{i}^{T}) (\sum_{j = 1}^{5} x_{j} x_{j}^{T}) = \sum_{i, j} x_{i} (x_{i}^{T} x_{j}) x_{j}^{T}$ . Since the vectors are orthonormal, $x_{i}^{T} x_{j} = δ_{ij}$ . Thus $A^{2} = \sum_{i = 1}^{5} x_{i} x_{i}^{T} = A$ .

Step 4: Since $A$ is idempotent, its eigenvalues can only be 0 or 1. Since $A$ is symmetric, its singular values equal the absolute values of its eigenvalues, which are also 0 or 1. Therefore the singular values of $A$ are also its eigenvalues. Options A and B are true.

Step 5: The rank of $A$ equals its trace: $tr (A) = \sum_{i = 1}^{5} x_{i}^{T} x_{i} = 5$ . Since $A$ is $10 \times 10$ with rank 5, it is not full rank. So $det (A) = 0$ and $A$ is not invertible. Options C and D are false.

Answer: Options A and B are correct.

A question from this chapter

Question 1

Let {x_{1}, x_{2}, ..., x_{n}} be a set of linearly independent vectors in R^{n} . Let the (i, j) -th element of matrix A \in R^{n \times n} be given by A_{ij} = x_{i}^{T} x_{j}, 1 \leq i, j \leq n . Which one of the following statements is correct?

Question 2

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Singular Values and Gram Matrices Previous Year Questions (PYQs) for GATE DA: 2+ Solved Questions with Step-by-Step Solutions

Chapter Roadmap: Singular Values and Gram Matrices

Chapter Roadmap

1. Gram Matrices and Positive Definiteness

2. Singular Values and Spectral Properties

Intuition: Measuring Geometry with Gram Matrices

Intuition: Measuring Geometry

Singular Values and Gram Matrices: Solved Questions with Step-by-Step Explanations (2 Problems)

Step-by-Step Solution

Step-by-Step Solution

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