GATE 2025 DA Q.28: solution and answer

GATE 2025 · DAQ.28 · 1 mark · MSQ

Let $A = I_{n} + x x^{⊤}$ , where $I_{n}$ is the $n \times n$ identity matrix and $x \in R^{n}$ , $x^{⊤} x = 1$ . Which of the following options is/are correct?

(A)
Rank of $A$ is $n$
(B)
$A$ is invertible
(C)
$0$ is an eigenvalue of $A$
(D)
$A^{- 1}$ has a negative eigenvalue

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Answer and solution

ANSWER

The correct options are (A) and (B).

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

Since $x^{⊤} x = 1$ , we have $(x x^{⊤}) x = x (x^{⊤} x) = x$ . So $x$ is an eigenvector of $x x^{⊤}$ with eigenvalue $1$ .
If $x^{⊤} y = 0$ , then $(x x^{⊤}) y = x (x^{⊤} y) = 0$ . These vectors $y$ form an $(n - 1)$ -dimensional subspace, so $x x^{⊤}$ has eigenvalue $0$ with multiplicity $n - 1$ . The eigenvalues of $x x^{⊤}$ are therefore $1$ (once) and $0$ ( $n - 1$ times).
$A = I_{n} + x x^{⊤}$ has the same eigenvectors, and each eigenvalue is shifted by $1$ . The eigenvalues of $A$ are $2$ (once) and $1$ ( $n - 1$ times).
$det A = 2 \cdot 1^{n - 1} = 2 \neq = 0$ , so $A$ is invertible. Option (B) is correct.
A matrix with nonzero determinant has full rank, so $rank (A) = n$ . Option (A) is correct.
Every eigenvalue of $A$ is $1$ or $2$ , so $0$ is not an eigenvalue. Option (C) is wrong.
The eigenvalues of $A^{- 1}$ are the reciprocals $\frac{1}{2}$ and $1$ . All are positive, so $A^{- 1}$ has no negative eigenvalue. Option (D) is wrong.

KEY IDEA

A rank-one matrix $x x^{⊤}$ with $x^{⊤} x = 1$ has eigenvalues $1$ and $0$ . Adding $I_{n}$ shifts these to $2$ and $1$ , so $A$ is positive definite, invertible and of full rank.

WATCH OUT FOR

Forgetting that $x x^{⊤}$ is singular, with $n - 1$ zero eigenvalues, and then assuming $0$ is also an eigenvalue of $A$ . The identity shifts every eigenvalue by $1$ . Another slip is to take the eigenvalues of $A^{- 1}$ as negative.

From the GATE 2025 DA paper. Every question of this paper · Download the full paper · All 6 Eigenvalues questions, with notes

GATE 2025 DA Q.28: solution and answer

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