Question

Let \( A = I_n + xx^T \), where \( I_n \) is the \( n \times n \) identity matrix and \( x \in \mathbb{R}^n, x^T x = 1 \). 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

Hint:

For matrices of the form \( A = I_n + xx^T \), the rank is \( n \), and the matrix is always invertible as long as \( x \neq 0 \).

Answer 1

The Correct Option is A, B

We are given that \( A = I_n + xx^T \), where \( I_n \) is the identity matrix, and \( x \) is a vector in \( \mathbb{R}^n \) such that \( x^T x = 1 \).

Option (A): The rank of \( A \) is \( n \). Since \( I_n \) has rank \( n \) and \( xx^T \) is a rank-1 matrix, the rank of \( A = I_n + xx^T \) will be \( n \) as long as \( x \neq 0 \). Therefore, Option (A) is correct.
Option (B): Since \( A \) is a full-rank matrix (rank \( n \)), it is invertible. Thus, Option (B) is correct.
Option (C): Since \( A \) is a full-rank matrix, it does not have 0 as an eigenvalue. Therefore, Option (C) is incorrect.
Option (D): \( A^{-1} \) is invertible, but we do not have enough information to claim that \( A^{-1} \) has a negative eigenvalue. Therefore, Option (D) is incorrect.

Thus, the correct answers are Option (A) and Option (B).