Is there a contradiction here?

[yftach]

Let v₁,...,v_k,u,w be vectors in linear space V.
Given that the equation: x₁v₁+....+x_kv_k=u has a single solution, and the equation: x₁v₁+....+x_kv_k=w has no solution.
Find the dimension of: Sp{v₁,....,v_k,w}

So what I don't understand here is if we look at the narrow coefficient matrix, from the first given equation we can conclude that non of the rows reset and v₁,....,v_k are linear independent, but from the second given equation we can conclude that one or more rows resent and that v₁,....,v_k are linear dependent. Isn't it a contradiction?

 

[HallsofIvy]

I don't know what you mean by "non of the rows reset" or by "one or more rows resent". I assume "non" was supposed to be "none" and "resent" was supposed to be "reset" again but I don't know what "reset" was supposed to be.

However the point is that if the vectors v1, v2, ..., vk span all of V then the equation a1v1+ a2v2+...+ akvk= v has a solution for all v. If they do not- either the vectors v1, v2, ..., vk are not independent or the vectors are independent but k is less than the dimension of V- then they span some subspace of V. a1v1+ a2v2+ ...+ akvk= u has a solution if u is in that subspace and does not if u is not in the subspace.

 

[yftach]

I don't know what you mean by "non of the rows reset" or by "one or more rows resent". I assume "non" was supposed to be "none" and "resent" was supposed to be "reset" again but I don't know what "reset" was supposed to be.

However the point is that if the vectors v1, v2, ..., vk span all of V then the equation a1v1+ a2v2+...+ akvk= v has a solution for all v. If they do not- either the vectors v1, v2, ..., vk are not independent or the vectors are independent but k is less than the dimension of V- then they span some subspace of V. a1v1+ a2v2+ ...+ akvk= u has a solution if u is in that subspace and does not if u is not in the subspace.

What I meant by reset is that a row in the matrix becomes all 0.