GATE 2026 EE Q.55: solution and answer

GATE 2026 · EEQ.55 · 2 marks · MSQ

Consider an $n \times n$ orthogonal matrix $A$ with real entries and each column having unit Euclidean norm.

Which of the following statements is/are correct?

(A)
The value of the determinant of $A$ is either $+ 1$ or $- 1$
(B)
The eigenvalues of $A$ have modulus 1
(C)
$∥ A x ∥ = ∥ x ∥$ , for all $x \in R^{n}$ , where $∥ x ∥$ denotes the Euclidean norm of $x$ , and $(A x)^{T} (A y) \neq = x^{T} y$ , for all distinct $x, y \in R^{n}$
(D)
$∥ A x ∥ = ∥ x ∥$ , for all $x \in R^{n}$ , where $∥ x ∥$ denotes the Euclidean norm of $x$ , and $(A x)^{T} (A y) = x^{T} y$ , for all $x, y \in R^{n}$

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Answer and solution

ANSWER

The correct options are (A), (B) and (D). For a real orthogonal matrix, $det A = \pm 1$ , every eigenvalue has modulus 1, and $A$ preserves every length and every inner product. Option (C) is false because the equality $(A x)^{T} (A y) = x^{T} y$ holds for every pair of vectors, so it cannot fail for every distinct pair.

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

$A$ is a real orthogonal matrix, so $A^{T} A = I$ . Because $A$ is square, this also gives $A A^{T} = I$ . The condition that every column has unit Euclidean norm is already part of this: it is the diagonal of $A^{T} A = I$ . Every part below follows from $A^{T} A = I$ .
(A) Determinant. Take determinants of $A^{T} A = I$ and use $det (A^{T}) = det (A)$ : $det (A^{T}) det (A) = (det A)^{2} = det I = 1 \Rightarrow det A = \pm 1.$ So (A) is correct.
(B) Eigenvalues. Let $A v = λ v$ with $v \neq = 0$ . The eigenvector $v$ can be complex, so write $v^{*}$ for its conjugate transpose. $A$ is real, so $A^{*} = A^{T}$ . Then $∥ A v ∥^{2} = (A v)^{*} (A v) = v^{*} A^{T} A v = v^{*} v = ∥ v ∥^{2},$ and also $∥ A v ∥^{2} = ∥ λ v ∥^{2} = ∣ λ ∣^{2} ∥ v ∥^{2} .$ Since $∥ v ∥ \neq = 0$ , $∣ λ ∣^{2} = 1$ , so $∣ λ ∣ = 1$ . For example, the rotation matrix $[cos θ sin θ - sin θ cos θ]$ has eigenvalues $e^{\pm j θ}$ . They are complex, and each has modulus 1. So (B) is correct.
Inner products are preserved. For any real vectors $x, y$ : $(A x)^{T} (A y) = x^{T} A^{T} A y = x^{T} y .$ Put $y = x$ to get $∥ A x ∥^{2} = ∥ x ∥^{2}$ , so $∥ A x ∥ = ∥ x ∥$ .
(D) Both parts of (D) are exactly what the previous step proved: $∥ A x ∥ = ∥ x ∥$ for every $x$ , and $(A x)^{T} (A y) = x^{T} y$ for every pair. So (D) is correct.
(C) The first part of (C) is true. The second part says $(A x)^{T} (A y) \neq = x^{T} y$ for every pair of distinct vectors, but the previous step shows the two sides are equal for every pair. One concrete pair: $n = 2$ , $A = I$ , $x = (1, 0)^{T}$ and $y = (0, 1)^{T}$ are distinct, and both sides equal $0$ . Any orthogonal $A$ has such pairs, for instance $x = 0$ with any $y \neq = 0$ , where both sides are $0$ . The two parts of (C) are joined by "and", so (C) is false.
The correct options are A, B and D.

KEY IDEA

Every property here follows from $A^{T} A = I$ . Taking determinants gives $(det A)^{2} = 1$ . Writing $x^{T} A^{T} A y = x^{T} y$ shows that inner products are preserved, and so lengths are too. Applying the length result to a possibly complex eigenvector, using the conjugate transpose, gives $∣ λ ∣ = 1$ .

WATCH OUT FOR

One mistake is to read (B) as saying the eigenvalues are $\pm 1$ and reject it because rotation matrices have complex eigenvalues. The option only claims modulus 1, and $e^{\pm j θ}$ has modulus 1. Another is to accept (C) because its first half ( $∥ A x ∥ = ∥ x ∥$ ) is true, without noticing that its second half contradicts inner-product preservation. A third is to prove $∣ λ ∣ = 1$ with $v^{T}$ in place of $v^{*}$ . That argument fails when the eigenvector is complex.

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GATE 2026 EE Q.55: solution and answer

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