GATE DA Projections and Orthogonality Previous Year Questions

GATE DA 2025 | Question 25

MSQ · 1 mark

Which of the following statements is/are correct?

\(\mathbb{R}^{n}\) has a unique set of orthonormal basis vectors \(\mathbb{R}^{n}\) does not have a unique set of orthonormal basis vectors Linearly independent vectors in \(\mathbb{R}^{n}\) are orthonormal Orthonormal vectors \(\mathbb{R}^{n}\) are linearly independent

Explanation

Answer: (B), (D)

A and B: orthonormal bases are not unique. Rotating the standard basis gives another one; in \(\mathbb{R}^{2}\), \(\frac{1}{\sqrt{2}}(1, 1)\) and \(\frac{1}{\sqrt{2}}(1, -1)\) also form an orthonormal basis. Even changing the sign of one vector gives a new basis. So A is false and B is true.
C is false: \((1, 0)\) and \((1, 1)\) are linearly independent, but they are not orthogonal and \((1, 1)\) is not a unit vector.
D is true: if \(\sum_{i} c_{i}v_{i} = 0\) for orthonormal \(v_{i}\), taking the dot product with \(v_{j}\) leaves \(c_{j} = 0\), so every coefficient is zero.

Answer: B, D

GATE DA 2025 | Question 50

MSQ · 2 marks

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

Singular values of \(A\) are also its eigenvalues Singular values of \(A\) are either 0 or 1 Determinant of \(A\) is 1 \(A\) is invertible

Explanation

Answer: (A), (B)

Because the \(x_{i}\) are orthonormal, \(Ax_{j} = \sum_{i} x_{i}(x_{i}^{\top}x_{j}) = x_{j}\) for \(j = 1, \ldots, 5\), while \(Av = 0\) for every \(v\) orthogonal to all five. So \(A\) is the orthogonal projection onto their 5-dimensional span: it is symmetric, \(A^{2} = A\), and its eigenvalues are 1 (five times) and 0 (five times).

A holds: the singular values of a symmetric matrix are the absolute values of its eigenvalues, and here the eigenvalues 0 and 1 are already non-negative.
B holds: so the singular values are 0 or 1.
C and D fail: 0 is an eigenvalue, so \(\det A = 0\) and \(A\) is not invertible.

Answer: A, B

GATE DA 2025 | Question 52

MSQ · 2 marks

An \(n \times n\) matrix \(A\) with real entries satisfies the property: \(\|Ax\|^{2} = \|x\|^{2}\), for all \(x \in \mathbb{R}^{n}\), where \(\| \cdot \|\) denotes the Euclidean norm. Which of the following statements is/are ALWAYS correct?

\(A\) must be orthogonal \(A = I\), where \(I\) denotes the identity matrix, is the only solution The eigenvalues of \(A\) are either \(+1\) or \(-1\) \(A\) has full rank

Explanation

Answer: (A), (D)

Write the condition as \(x^{\top}(A^{\top}A - I)x = 0\) for every \(x\). The matrix \(A^{\top}A - I\) is symmetric, and a symmetric matrix whose quadratic form is zero for every \(x\) is the zero matrix. So \(A^{\top}A = I\), and \(A\) is orthogonal: A holds.

D holds: \(A^{\top}A = I\) means \(A^{-1} = A^{\top}\) exists, so \(A\) has full rank.
B fails: \(-I\), or any rotation, also preserves lengths.
C fails: the rotation by 90\(^{\circ}\), \(\begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}\), is orthogonal, but its eigenvalues are \(\pm i\). An orthogonal matrix only guarantees eigenvalues of modulus 1.

Answer: A, D

GATE DA 2024 | Question 49

MSQ · 2 marks

Let \(\mathbb{R}\) be the set of real numbers, \(U\) be a subspace of \(\mathbb{R}^{3}\) and \(\boldsymbol{M} \in \mathbb{R}^{3 \times 3}\) be the matrix corresponding to the projection on to the subspace \(U\).

Which of the following statements is/are TRUE?

If \(U\) is a 1-dimensional subspace of \(\mathbb{R}^{3}\), then the null space of \(\boldsymbol{M}\) is a 1-dimensional subspace. If \(U\) is a 2-dimensional subspace of \(\mathbb{R}^{3}\), then the null space of \(\boldsymbol{M}\) is a 1-dimensional subspace. \(\boldsymbol{M}^{2} = \boldsymbol{M}\) \(\boldsymbol{M}^{3} = \boldsymbol{M}\)

Explanation

Answer: (B), (C), (D)

A projection onto \(U\) leaves every vector of \(U\) unchanged and sends the complementary directions to \(\boldsymbol{0}\). So its column space is \(U\), and by rank–nullity its null space has dimension \(3 - \dim U\).

A fails: a line \(U\) leaves a null space of dimension \(3 - 1 = 2\), a plane.
B holds: a plane \(U\) leaves a null space of dimension \(3 - 2 = 1\), a line.
C holds: \(\boldsymbol{M}\boldsymbol{x}\) already lies in \(U\), so projecting it again changes nothing, which gives \(\boldsymbol{M}^{2} = \boldsymbol{M}\).
D holds: \(\boldsymbol{M}^{3} = \boldsymbol{M}^{2}\boldsymbol{M} = \boldsymbol{M}\boldsymbol{M} = \boldsymbol{M}\).

Answer: B, C, D