matrix

Question

Let $x_1, x_2, x_3, x_4, y_1, y_2, y_3$ and $y_4$ be fixed real numbers, not all of them equal to zero. Define 4 $\times$ matrix A by

$A = \begin{pmatrix} x_1^2+y_1^2 & x_1x_2+y_1y_2 & x_1x_3+y_1y_3 & x_1x_4+y_1y_4 \\ x_2x_1+y_2y_1 & x_2^2+y_2^2 & x_2x_3+y_2y_3 & x_2x_4+y_2y_4 \\ x_3x_1+y_3y_1 & x_3x_2+y_3y_2 & x_3^2+y_3^2 & x_3x_4+y_3y_4 \\ x_4x_1+y_4y_1 & x_4x_2+y_4y_2 & x_4x_3+y_4y_3 & x_4^2+y_4^2 \end{pmatrix}$

Then rank(A) equals

• A. 1 or 2
• B. 0
• C. 4
• D. 2 or 3

Answer 1

Option A should be the correct answer.

The matrix $A$ can be written as a product of two matrices, namely matrix $B$ and $B^{T}$, as follows:

$A = \begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\\ x_{3} & y_{3}\\ x_{4} & y_{4} \end{bmatrix} \times \begin{bmatrix} x_{1} & x_{2} & x_{3} & x_{4}\\ y_{1} & y_{2} & y_{3} & y_{4} \end{bmatrix}$

where $B = \begin{bmatrix} x_{1} & y_{1} \\ x_{2} & y_{2}\\ x_{3} & y_{3}\\ x_{4} & y_{4} \end{bmatrix}$.

Rank of matrix $B$ is the dimension of the vector space spanned by the columns of $B$.

$B$ has two columns.

If both of the columns vectors are independent then their linear combinations can represent every vector on the plane in which both of them are contained in and hence the span of the column vectors of $B$ would be a two dimensional plane in the four dimensional space and hence the rank of $B$ would be 2.

If both the column vectors of $B$ are linearly dependent then their linear combinations will only span the one dimensional line(on which both of the columns vectors lie) in the four dimesional space.Thus in this case rank would be 1.

It is given that not all x's & y's can be zero which implies $B$ can not be a zero matrix so rank of B can not be zero.

So rank of $B$ could be 1 or 2.

FACT: A matrix and its transpose have the same rank. (for proof see http://linear.ups.edu/html/section-PD.html)

So $B$ and $B^{T}$ will have the same rank. 

Now, $A = BB^{T}$ 

FACT: Rank($A$) <= min{Rank($B$), Rank($B^{T}$)}  ( for proof see 1.4 b) of http://ocw.mit.edu/courses/electrical-engineering-and-computer-science/6-241j-dynamic-systems-and-control-spring-2011/assignments/MIT6_241JS11_assn02_sol.pdf )

So Rank($A$) <= 1 or 2 

but all x's & y's can not be zero so $A$ is not a zero matrix for sure and hence rank of $A$ can not be zero.

So Rank of $A$ could be 1 or 2.

Answer 2

since its given that not every element i.e x1, x2, x3, x4, y1, y2, y3, y4 be equal to zero.

then lets consider x1=x2=x3=y1=y2=y3=0 and y3,y4 to be real numbers.

then the given matrix would become as $\begin{pmatrix} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 1 \\ \end{pmatrix}$ and (then converting it into row echelon form) which is $\begin{bmatrix} 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ \end{bmatrix}$ (operation R1<->R4)and hence the martix is now in row echelon form thus the rank is coming out to be 1 = no of non-zero rows and non-zeros columns  thus option A is correct choice as it says rank = 1 OR 2.