GATE 2022 EC Q.37: solution and answer

GATE 2022 · ECQ.37 · 2 marks · MCQ

Let $α, β$ be two non-zero real numbers and $v_{1}, v_{2}$ be two non-zero real vectors of size $3 \times 1$ . Suppose that $v_{1}$ and $v_{2}$ satisfy $v_{1}^{T} v_{2} = 0$ , $v_{1}^{T} v_{1} = 1$ , and $v_{2}^{T} v_{2} = 1$ . Let $A$ be the $3 \times 3$ matrix given by:

A = α v_{1} v_{1}^{T} + β v_{2} v_{2}^{T}

The eigenvalues of $A$ are ________.

(A)
$0, α, β$
(B)
$0, α + β, α - β$
(C)
$0, \frac{α + β}{2}, α β$
(D)
$0, 0, α^{2} + β^{2}$

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Answer and solution

ANSWER

The eigenvalues of $A$ are $0, α, β$ , which is option (A).

Official answer: ASource: Official answer key, GATE 2022 EC (IIT Kharagpur)

The conditions $v_{1}^{T} v_{1} = 1$ , $v_{2}^{T} v_{2} = 1$ and $v_{1}^{T} v_{2} = 0$ say that $v_{1}, v_{2}$ are orthonormal vectors in $R^{3}$ .
Apply $A$ to $v_{1}$ . Matrix multiplication is associative, and $v_{1}^{T} v_{1}$ , $v_{2}^{T} v_{1}$ are scalars: $A v_{1} = α v_{1} (v_{1}^{T} v_{1}) + β v_{2} (v_{2}^{T} v_{1}) = α v_{1} (1) + β v_{2} (0) = α v_{1}$ So $v_{1}$ is an eigenvector with eigenvalue $α$ .
In the same way, $A v_{2} = α v_{1} (v_{1}^{T} v_{2}) + β v_{2} (v_{2}^{T} v_{2}) = 0 + β v_{2} = β v_{2}$ So $v_{2}$ is an eigenvector with eigenvalue $β$ .
In $R^{3}$ , take a non-zero $v_{3}$ orthogonal to both $v_{1}$ and $v_{2}$ (for example $v_{3} = v_{1} \times v_{2}$ , which is non-zero because $v_{1}, v_{2}$ are orthonormal). Then $v_{1}^{T} v_{3} = v_{2}^{T} v_{3} = 0$ , so $A v_{3} = α v_{1} (v_{1}^{T} v_{3}) + β v_{2} (v_{2}^{T} v_{3}) = 0 = 0 \cdot v_{3}$ So $v_{3}$ is an eigenvector with eigenvalue $0$ .
The vectors $v_{1}, v_{2}, v_{3}$ are non-zero and mutually orthogonal, so they are linearly independent and form a basis of $R^{3}$ . A $3 \times 3$ matrix with three independent eigenvectors has exactly those three eigenvalues (counted with multiplicity), so the eigenvalues of $A$ are $α$ , $β$ and $0$ .
Check with the trace: $tr (A) = α tr (v_{1} v_{1}^{T}) + β tr (v_{2} v_{2}^{T}) = α (v_{1}^{T} v_{1}) + β (v_{2}^{T} v_{2}) = α + β$ , which equals $0 + α + β$ . Also $A$ is a sum of two rank-1 matrices, so $rank (A) \leq 2$ . That makes $det A = 0$ , so one eigenvalue must be $0$ .
The eigenvalues of $A$ are $0, α, β$ : option (A).

KEY IDEA

$v_{1}$ and $v_{2}$ are orthonormal, so $A = α v_{1} v_{1}^{T} + β v_{2} v_{2}^{T}$ is a spectral decomposition. The matrix $v_{i} v_{i}^{T}$ projects onto $v_{i}$ , so $A v_{1} = α v_{1}$ and $A v_{2} = β v_{2}$ . Any vector orthogonal to both is sent to $0$ , which gives the third eigenvalue $0$ . Checks: trace $= α + β$ , and rank $\leq 2$ .

WATCH OUT FOR

Adding or combining $α$ and $β$ (options B, C, D) as if the two rank-1 terms mix. Because $v_{1} ⊥ v_{2}$ , each term acts only on its own direction, so the eigenvalues stay separate as $α$ and $β$ . Some students also forget the third eigenvalue. A $3 \times 3$ matrix of rank at most 2 always has $0$ as an eigenvalue.

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GATE 2022 EC Q.37: solution and answer

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