Hi guys, having some issues with some basic introductory linear algebra questions, any help much appreciated! - 1) Find the dimension and a basis of the following vector spaces V over the given field K.

(i) V is the set of all vectors (a,b,y) in R^3 with a+2b-2y=0 with K=R.

My answer: Here I'm completely unsure, rearranging the equation for each variable we get the set of vectors (2y-2b,(2y-a)/2, (a+2b)/2). So does this have dimension 3? Clearly any vector can be constructed by (2y-2b)(1,0,0)+((2y-a)/2)(0,1,0)+((a+2b)/2)(0,0,1) and of course (1,0,0),(0,1,0),(0,0,1) are linearly independant. Is this correct?

(ii) V set of vectors (a,b,y) in R^3 with a-b=-y and a+2b=y with K=R.

My answer: Again unsure what to do with the supplied equations. Setting both equations equal to a, then combining them and solving for y gives y=(3/2)b so we have set of vectors (a,b,(3/2)b). Now here I get really confused. Again any vector can be constructed by a(1,0,0)+b(0,1,0)+(3/2)b(0,0,1) and this is the standard basis whose components we know to be linearly independant (or it wouldn't be a basis) which would imply that the vector space has dimension 3. However the any vector can also be constructed by a(1,0,0)+b(0,1,(3/2)) (which are clearly LI) which would imply a dimension of two. Here I'm really confused as the answers are contradictory. I believe the correct dimension to be 2 since the vector space on visualisation is I believe a plane, and so one would expect a dimension of 2 but why do I get two different answers? What am I doing wrong?

This can be generalised to me being unsure of how to approach a situation where I have a set of vectors (a,b,b). This can be spanned in two ways: (1) a(1,0,0)+b(0,1,0)+b(0,0,1) implying a dimension of 3 or (2) a(1,0,0)+b(0,1,1) implying a dimension of 2. In both cases the set of spanning vectors are linearly independant. What gives?

(iii) V=C^2, K=R.

My answer: dimension 4, basis (1,0), (0,i), (0,1), (0,i). Pretty basic textbook example here so no problems (i hope). If K=C then the space would have dimension 2 and a basis (1,0), (0,1), correct?

(iv) V is set of all polynomials over R with degree at most n, in which the sum of the coefficients is 0, K=R.

My answer: So we have a_0+a_1x+a_2x^2+...+a_nx^n with a_0+a_1+a_2+...+a_n=0. Not really sure how the condition of the coefficients restricts things. Normally the set of all polynomials over R with degree at most n would have dimension n+1 and (standard) basis 1,x,x^2,...,x^n. Can someone give me a hint on how to start this one?

(v) K=R and and V is the set of functions from R to R which are solutions of the differential equation d^2f/dx^2-5f=0.

My answer: Solving the (second order) differential equation via standard methods gives a general solution f(x)=c_1e^(root(5)x)+c_2e^(-root(5)x) so does this just have dimension 2 and basis e^(root(5)x), e^(-root(5)x)?

Thanks in advance for any help!