rank of matrix

Question

If X=[x₁ x₂ x₃… x_n ]^T is an n-tuple non-zero vector. Then m×n matrix V=XX^T will be

1.   Is orthogonal
2.   Having a rank 1
3. .  Having a rank n-1
4.   Having a rank zero

Comments

Can't I do in this way??

 X=[x₁ x₂ x₃… x_n ]^T  is 1*n matrix 

 X^T   = n* 1 matrix

V=XX^T which is of size 1*1 and since it is non zero so i can say its rank is 1

   

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@akash

exactly the same mistake i did

 X=[x₁ x₂ x₃… x_n ]^T  which means X is n *1 matrix

and X^T is 1*n

X is itself said to be taken as transpose

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Aish,

I m not getting u!!

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$X=\begin{bmatrix} x_{1},x_{2},x_{3....x_{n}} \end{bmatrix}^{T}=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ .\\ .\\ x_{n}\\ \end{bmatrix}$
So X is n*1 matrix not 1*n.Read question carefully.

Answer 1

Given $X=\begin{bmatrix} x_{1},x_{2},x_{3....x_{n}} \end{bmatrix}^{T}=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\\ .\\ .\\ x_{n}\\ \end{bmatrix}$

Take n=2

$X=\begin{bmatrix} x_{1}\\ x_{2}\\ \end{bmatrix}$

$XX^{T}=\begin{bmatrix} x_{1}\\ x_{2}\\ \end{bmatrix}\begin{bmatrix} x_{1}\\ x_{2}\\ \end{bmatrix}^{T}$

$XX^{T}=\begin{bmatrix} x_{1}\\ x_{2}\\ \end{bmatrix}\begin{bmatrix} x_{1},x_{2} \end{bmatrix}$

$XX^{T}= \begin{bmatrix} x_{1}^{2} & x_{1}x_{2}\\ x_{1}x_{2}& x_{2}^{2 } \end{bmatrix}$

$x_{1} \times x_{2} \times \begin{bmatrix} x_{1} & x_{2}\\ x_{1} & x_{2} \end{bmatrix}$

$x_{1} \times x_{2} \times\begin{bmatrix} x_{1} & x_{2}\\ 0 & 0 \end{bmatrix}$$ \ \ \ \ R_{2} \rightarrow R_{2}-R{1}$

Rank =1

For orthogonal matrix

$XX^{T}=I$

Where I is an identity matrix.

we can eliminate option (1) and (4) directly because of $XX^{T} \neq I $ and Rank of the matrix is not 0.

Now take n=3

$X=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3}\end{bmatrix}$

$XX^{T}=\begin{bmatrix} x_{1}\\ x_{2}\\ x_{3} \end{bmatrix}\begin{bmatrix} x_{1},x_{2},x_{3} \end{bmatrix}$

$XX^{T}= \begin{bmatrix} x_{1}^{2} & x_{1}x_{2}&x_{1}x_{3}\\ x_{1}x_{2}& x_{2}^{2 }&x_{2}x_{3} \\ x_{3}x_{1}& x_{2}x_{3}&x_{3}^{2} \end{bmatrix}$

$x_{1} \times x_{2} \times x_{3} \times \begin{bmatrix} x_{1} & x_{2}&x_{3}\\ x_{1}& x_{2}&x_{3} \\ x_{1}& x_{2}&x_{3} \end{bmatrix}$

$x_{1} \times x_{2} \times x_{3} \times \begin{bmatrix} x_{1} & x_{2}&x_{3}\\ x_{1}& x_{2}&x_{3} \\ 0& 0&0 \end{bmatrix} \ \ \ \ R_{3}\rightarrow R_{3}-R{1}$

$x_{1} \times x_{2} \times x_{3} \times \begin{bmatrix} x_{1} & x_{2}&x_{3}\\ 0& 0&0 \\ 0& 0&0 \end{bmatrix} \ \ \ \ R_{2}\rightarrow R_{2}-R{1}$

Rank = 1

 We can eliminate option(3) because rank can't be $n-1$.

Hence, Option(2)Having rank 1 is the correct choice.