No. y = (a+2b)/2. There is no need to write the other elements of the vector, because that doesn't give you anything. Instead what you have is (a,b,[a+2b]/2), where a and b are real numbers. It's not too hard to show that the basis of this space is (1,0,1/2), (0,1,1).So does this have dimension 3?
You defined a to be one of b-y or y-2b. As you have defined it, a is not a "free variable".Now here I get really confused.
You have y=b-a and y=a+2b. This is only possible if b=-2a. Then y = -3a. Then the set of vectors is in fact those of the form (a,-2a,-3a), where a is a real number. The dimension should be obvious here.What am I doing wrong?
dimension 4, basis (1,0), (0,i), (0,1), (i,0).If you know all but one of the coefficients, then you know ALL of the coefficients. Suppose we are in p^2, with the condition on the coefficients. Then a vector is ax^2+ bx+c. But a+b+c= 0 implies that c = -a-b. Then (in vector form) what you really have is (a,0,-a) + (0,b,-b). It should be clear that the basis vectors are (1,0,-1) and (0,1,-1) - with dimension 2.Not really sure how the condition of the coefficients restricts things.