Let A=In+xx, where In is the n×n identity matrix and xn, xx=1. Which of the following options is/are correct?

A.

Rank of AA is nn

B.

AA is invertible

C.

0 is an eigenvalue of AA

D.

A1A^{-1} has a negative eigenvalue

Solution:

 

🔍 Step 1: Understanding A=In+xxA = I_n + xx^\top

  • xxxx^\top is a rank-1 symmetric matrix.

  • Since xx=1x^\top x = 1, it’s a unit vector, so xxxx^\top is a positive semi-definite matrix.

  • InI_n is identity matrix ⇒ positive definite.

So,

  • A=In+xxA = I_n + xx^\top is positive definite.

  • All eigenvalues of AA are positive.


Option-wise Analysis:

(A) Rank of AA is nnCorrect

  • Rank(In)=n\text{Rank}(I_n) = n

  • Rank(xx)=1\text{Rank}(xx^\top) = 1, but adding it to InI_n does not reduce rank.

  • So, Rank(A)=n\text{Rank}(A) = n


(B) AA is invertible Correct

  • Since all eigenvalues of AA are positive, it’s positive definiteinvertible.


(C) 0 is an eigenvalue of AAIncorrect

  • A positive definite matrix cannot have a zero eigenvalue. All eigenvalues are > 0.


(D) A1A^{-1} has a negative eigenvalue Incorrect

  • The inverse of a positive definite matrix is also positive definite.

  • So all eigenvalues of A1A^{-1} are positive, not negative.