Your matrix is $8\times 16$ and therefore the nullspace is at least 8 dimensional but there is no guarantee that the overall rank is not less than 8. Did you check the overall rank? How did you calculate the $x_i$'s and $y_i$'s?

i checked the rank of 8 x 16 matrix, it is 8 and hence the null space is of rank 8.

i checked the rank of 8 x 16 matrix, it is 8 and hence the null space is of rank 8. i took $\mathbf{w}_1$ and and reversed the kronecker operation. That is i took the four elements of $\mathbf{w}_1$ and divided the first two elements with the third and fourth elements, element wise to check if both division results in a constant. I thinks kronecker produce is not uniquely reversible but to a scaling factor.

So the $w_i$ can be a linear combination of the 8 linearly independent basis vectors that span the nullspace. Hence whenever you pick up a vector and chop it up into $2\times 1$ sub components you can make up arbitrary solutions. Make sure you don't have $y_i \otimes x_i\$.

yep, you got it.

yes, i checked again $ w_i = x_i \otimes y_i$ is not satisfied.

OK. I think we are misunderstanding each other. If I give you a 8 by 1 vector and ask you to obtain two vectors such that $w_i = x_i\otimes y_i$ and both has nonzero components there are infinitely many solutions.

sorry, i dont understand how to find these solutions. could you tell me how to fint it. or some reference for such methods

sure. Let's start from the last point

yes

now we have a nullspace spanned by 8 vectors

then what we are trying to find next is :

Is there a vector let's say $[1 2]^T$

such that

(a*[1 2] b*[1 2] c*[1 2] d*[1 2] )^T

is a solution to the transpose of the null(A)

am i making any sense? I think that was a bad construction

sorry

i see the example. i am just trying to understand

OK. No problem. Are you using MATLAB by any chance?

yes.

i am using matlab

in your example [1 2]^T would be x and [a b c d]^T would be y

right?

exactly

now you compute the null space of A by null(A) right?

yes

a linear combination of those should give us the required solution

i mean the columns of null(A) let's call it N

ok

just a sec

ok

Thanks, sorry for the delay

So what we have is kron(eye(4),[1;2]) * [a; b; c; d]

no problem

just a min.

which is equal to N times some combination (N = null(A))

OK

there are w_1, w_2, w_3 and w_4

each of which is a 4 x 1 vector

did you already consider this?

ah yes the dimension doesn't fit sorry lets make it [1 2 3 4]' then

of course sorry

ok. no problem.

So, the matrix N' should solve (a*[1 2 3 4] b*[1 2 3 4] c*[1 2 3 4] d*[1 2 3 4] )^T

which is again impossible, hmm I am really confusing myself here.

just give me a few minutes

there is a war in the office right now

i also need few minutes to think

uuhh

ah yes

done, I have packed them up in the cupboard

anyway here is the code for you to come up with a particular solution

linsolve(N,kron(rand(4,1),rand(4,1)))

i understand your solution

so this is solving the N*r = x_i \otimes y_i

r is the required linear combination of your null space basis vectors

sorry, i caught up in a very bad situation. I will try to clean up this and post a clean answer for this later on if that is OK with you.

yes sure.

Thanks!

you're welcome.

good luck