1
$\begingroup$

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.

$\endgroup$
7
  • $\begingroup$ 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
  • 1
    $\begingroup$ 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
2
$\begingroup$

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 $1$.

$\endgroup$
6
  • $\begingroup$ 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
  • 1
    $\begingroup$ @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
2
$\begingroup$

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).

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.