GATE 2025 DA Q.50: solution and answer

GATE 2025 · DAQ.50 · 2 marks · MSQ

Let $x_{1}, x_{2}, x_{3}, x_{4}, x_{5}$ be a system of orthonormal vectors in $R^{10}$ . Consider the matrix $A = x_{1} x_{1}^{⊤} + \dots + x_{5} x_{5}^{⊤}$ . Which of the following statements is/are correct?

(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

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Answer and solution

ANSWER

Options (A) and (B) are correct: $A = X X^{⊤}$ is an orthogonal projection of rank 5 in $R^{10}$ , with eigenvalues and singular values equal to five 1's and five 0's.

Official answer: A, BSource: Official answer key, GATE 2025 DA (IIT Roorkee)

Let $X = [x_{1} x_{2} x_{3} x_{4} x_{5}]$ , a $10 \times 5$ matrix. Then $A = \sum_{i = 1}^{5} x_{i} x_{i}^{⊤} = X X^{⊤}$ . Because the $x_{i}$ are orthonormal, $X^{⊤} X = I_{5}$ .
Hence $A^{⊤} = A$ and $A^{2} = X (X^{⊤} X) X^{⊤} = X X^{⊤} = A$ . So $A$ is a symmetric idempotent matrix, that is, the orthogonal projection onto $span {x_{1}, \dots, x_{5}}$ .
For $i \leq 5$ , $A x_{i} = \sum_{j} x_{j} (x_{j}^{⊤} x_{i}) = x_{i}$ , so $1$ is an eigenvalue with multiplicity at least $5$ . Any vector $v$ orthogonal to all $x_{i}$ satisfies $A v = 0$ , and this orthogonal complement has dimension $5$ . So the eigenvalues are exactly five $1$ 's and five $0$ 's.
$A$ is symmetric positive semidefinite, so its singular values are its eigenvalues: five $1$ 's and five $0$ 's. Statement (A) is true, and statement (B) is true because every singular value is $0$ or $1$ .
$det A = 1^{5} \cdot 0^{5} = 0$ , so (C) is false.
$rank A = 5 < 10$ , so $A$ is singular and not invertible. Statement (D) is false.
Correct options: (A) and (B).

KEY IDEA

A sum of $x_{i} x_{i}^{⊤}$ over orthonormal vectors is the orthogonal projection onto their span. It is symmetric and idempotent, so its eigenvalues are 0 or 1. Being positive semidefinite, its singular values equal its eigenvalues.

WATCH OUT FOR

Assuming the matrix is invertible or has determinant 1 because the vectors are orthonormal. That holds only when $A$ is built from a full set of 10 orthonormal vectors. With only 5 vectors in $R^{10}$ the rank is 5, so $A$ is singular and $det A = 0$ .

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GATE 2025 DA Q.50: solution and answer

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