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?

Question

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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).

Answer

Rank of $ A $ is $ n $

Answer

$ A $ is invertible

Answer

0 is an eigenvalue of $ A $

Answer

$ A^{-1} $ has a negative eigenvalue

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?

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The Correct Option is A, B

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