x=[x1x2…..xn]^{T} is an n-tuple nonzero vector. The n×n matrix V=xx^{T}

- 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?

By
Syed Zubair Ahmed
·
2 weeks ago ·Mechanical Engineering

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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