x=[x1x2…..xn]T is an n-tuple nonzero vector. The n×n matrix V=xxT    
  • a)
    has rank zero  
  • b)
    has rank l
  • c)
    is orthogonal  
  • d)
    has rank n 
Correct answer is option 'B'. Can you explain this answer?

Mechanical Engineering Question

Mitali Gupta answered Jun 13, 2018
The answer is a.
Given, x = [x1, x2, ....xn]T
Since, x is an n-tuple non-zero vector, that is, x is a non-singular matrix of order n, so it rank should be n.  
i.e, I (x) = n
The vector (xT) also have rank (n) because transpose of any matrix does not altered its rank. 
Then, matrix [v = x T'] must have the rank n, i. e., I(v) = n, because resultant of the multiplication of two same rank matrices also has the same rank as the rank of multiplicative matrix.

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