GATE 2026 DA Q.52: solution and answer

GATE 2026 · DAQ.52 · 2 marks · 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)
$Trace (M) = n$
(D)
$M$ is a projection matrix

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Answer and solution

ANSWER

Options (A) and (D) are correct: $M$ is symmetric and idempotent, so it is a projection matrix, while $M^{2} = M \neq = I_{n}$ and $tr (M) = n - 1$ .

Official answer: A, DSource: Official answer key, GATE 2026 DA (IIT Guwahati)

Since $1^{T} 1 = n$ , the matrix $P = \frac{1}{n} 1 1^{T}$ projects onto the line spanned by $1$ , and $M = I_{n} - P$ .
(A) $P^{T} = \frac{1}{n} (1 1^{T})^{T} = \frac{1}{n} 1 1^{T} = P$ , so $M^{T} = I_{n} - P^{T} = I_{n} - P = M$ . True.
(B) $P^{2} = \frac{1}{n ^{2}} 1 (1^{T} 1) 1^{T} = \frac{n}{n ^{2}} 1 1^{T} = P$ , so $M^{2} = I_{n} - 2 P + P^{2} = I_{n} - P = M$ . Since $P \neq = 0$ , $M \neq = I_{n}$ , so $M^{2} = M \neq = I_{n}$ . False.
(C) $tr (P) = \frac{1}{n} \sum_{i = 1}^{n} 1 = 1$ , so $tr (M) = n - 1 \neq = n$ . False.
(D) $M$ is symmetric ( $M^{T} = M$ ) and idempotent ( $M^{2} = M$ ), which is exactly an orthogonal projection matrix. It projects onto the subspace orthogonal to $1$ , and its rank $n - 1$ equals its trace. True.
Correct options: A and D.

KEY IDEA

$\frac{1}{n} 1 1^{T}$ is the orthogonal projection onto $1$ , so $I_{n}$ minus it is the orthogonal projection onto the complement: symmetric, idempotent, with trace equal to its rank $n - 1$ .

WATCH OUT FOR

Forgetting that a projection matrix satisfies $M^{2} = M$ rather than $M^{2} = I_{n}$ , or taking $tr (M) = n$ by ignoring the subtracted rank-one part, whose trace is 1.

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GATE 2026 DA Q.52: solution and answer

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