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 a 4 x 4 matrix A by

A =

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

What can be said about the rank of the matrix A?

I've written A as a sum of two other matrices. Would that help in any way? If not, I need a kickstart. Please provide a definitive hint.

  • $\begingroup$ meta.matheducators.stackexchange.com/questions/93/… . If you don't write your mathematics with a decent format many people won't even bother trying to decypher what is written there. $\endgroup$ – DonAntonio Apr 19 '14 at 11:32
  • $\begingroup$ I'm not sure how to write something in matrix form here. @DonAntonio, can you help in this regard? $\endgroup$ – A.Chakraborty Apr 19 '14 at 12:01
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    $\begingroup$ Sorry @A.Chakraborty, there is not explanation about matrices: use /begin{pmatrix}a&b\\c&d\end{pmatrix} between two pairs of dollar signs to write a $\;2\times 2\;$ matrix with first row $\;a,b\;$ and second row $\;c,d\;$ . Note that "\\" means jumping one line below, and by using it repeatedly you can write matrices with as many rows and columns as you wish. $\endgroup$ – DonAntonio Apr 19 '14 at 12:05
  • $\begingroup$ @DonAntonio, I've tried using the specified commands, but it does not appear to be solved as I would have wanted it to. Please help. $\endgroup$ – A.Chakraborty Apr 19 '14 at 12:24
  • $\begingroup$ @A. , use " x_1 " to do subindices, and "x_1^4" to do $\;x_1^4\;$ ...and all the times between a pair of dollar signs! And instead one dollar sign at each side, use TWO dollar signs together at each side to get a new line for that part which will also be centered... $\endgroup$ – DonAntonio Apr 19 '14 at 12:27

Your matrix is nothing but $$\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix} \begin{bmatrix} x_1& x_2& x_3& x_4\end{bmatrix} + \begin{bmatrix} y_1\\ y_2\\ y_3\\ y_4\end{bmatrix} \begin{bmatrix} y_1& y_2& y_3& y_4\end{bmatrix}$$ Hence, the rank of the matrix is $2$, in general. If $y_k = ax_k$ for all $k \in \{1,2,3,4\}$, where $a$ is some constant, then the rank is $1$.

  • $\begingroup$ This is specifically the part about which I said "I've written A as a product of two other matrices". But in general, R(A+B) $<=$ R(A) + R(B). So it can either be 1 or 2, since 0 is ruled out as a possibility here(acc to the given conditions). This is what I assumed apriori. $\endgroup$ – A.Chakraborty Apr 19 '14 at 12:47
  • $\begingroup$ @A.Chakraborty How is it a product of two matrices? And I have also mentioned when the rank is $1$ and when it is $2$. $\endgroup$ – user141421 Apr 19 '14 at 12:48
  • $\begingroup$ Extremely sorry, it's my bad. I actually meant the sum of two matrices. However, I had trouble as to why you were saying that it is in general 2, when the relation clearly states both the possibilities. I mean, you have put them together correctly, considering the linear dependence factor, but you cannot claim primarily that the rank will in general be 2. Can you? $\endgroup$ – A.Chakraborty Apr 19 '14 at 12:50
  • $\begingroup$ @A.Chakraborty When I mean in general, it means almost surely it is $2$, i.e., if you pick $4$ random numbers as $x_i$ and another $4$ random numbers as $y_i$, then the rank will be $2$. The only case in which the rank will be $1$ is if $\begin{bmatrix} y_1\\ y_2\\ y_3\\ y_4\end{bmatrix} = \begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}$ or if all the $x_i$'s are $0$ or if all the $y_i$'s are $0$. If all the $x_i$'s and $y_i$'s are $0$, then the rank is $0$. $\endgroup$ – user141421 Apr 19 '14 at 13:41
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    $\begingroup$ @A.Chakraborty I missed a factor $a$ infront of the second vector, i.e., $$\begin{bmatrix} y_1\\ y_2\\ y_3\\ y_4\end{bmatrix} = a\begin{bmatrix} x_1\\ x_2\\ x_3\\ x_4\end{bmatrix}$$ $\endgroup$ – user141421 Apr 19 '14 at 14:02

Actually,u can express the matrix as product of two matrices. one is 4*2 another one is 2*4,as u must notice the given matrix can be written as a product of two matrices viz. A and A(trns). And R(AA(trns))<= min(R(A),R(At)). A can hv atmost RANK=2.. so the possibilities of R(AAt) are 1 and 2. Moreover,R(AAt)=R(A).

  • $\begingroup$ Thanks for informing, Sir @user144300, but I believe you wanted to say 'sum' and not 'product' of two matrices... $\endgroup$ – A.Chakraborty Apr 20 '14 at 21:54
  • $\begingroup$ nope Sir @A.Chakraborty..i meant product.. $\endgroup$ – user144300 May 3 '14 at 12:46
  • $\begingroup$ Yeah I got it later actually, you must've used the augmented matrix. $\endgroup$ – A.Chakraborty May 3 '14 at 18:28

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