Find the formula for the projection P:C^3--->C^3 on W along U, where:
W=span{(1,i,0),(0,i,1)} and
U=span{(0,1,0)}.
Attempt at solution:
According to my textbook: A function T: V -> V is called the projection on W1 along W2 if, for x = x1 + x2 with x1 E W1 and x2 E W2 we have T(x)=x1, but this requires that V be the direct sum of W1 and W2. For the above question W is taken over the real field F=R so for W to be a subspace of C^3, C^3 must be taken over the field R but then C^3 has dimension 6 (example of basis: {(1,0,0),(0,1,0),(0,0,1),(i,0,0),(0,i,0),(0,0,i)}) . Since the vectors in span W and span U are independent they form a basis for W and U respectively and although the intersection of the 2 basis is the empty set the sum of the dimensions is 3 and not 6 so C^3 is not the direct sum of W and U. Or do we just assume that it is the direct sum then x1= a(1,i,0)+b(0,i,1) and x2=c(0,1,0) and x=x1+x2=(a,ai+bi+c,b) and T(x)=T(a,ai+bi+c,b)=(a,ai+bi,b) where a,b,c are scalars and since T=T^2 this is a projection.
This is the only definition the textbook gives for a projection so I'm really uncertain about how to approach the question and what to do. Any help would be greatly appreciated.Thanks in advance.
In fact, if , real numbers are sufficient to form a basis of the total space (e.g., for your exercise), but it's certainly not a necessity !
Just think about the -vector space . Isn't or a basis of this space ?
Here is (given the context) the same. Even more, some subspaces of (as a -vector space) have no basis formed by just real numbers. Like the -line (else, there's a such that and , which isn't possible).
And that the case (verify it) of your subspace here. You have to understand that offers more directions than , hence the fact that (exploiting a -direction that isn't a -direction) can not be expressed with a real basis.
So even though there are complex numbers in the basis of W it can still be a subspace of C^3 taken over C? Then since (1,i,0),(0,i,1),(0,1,0) are linearly independent and since C^3 is 3 dimensional taken over C (1,i,0),(0,i,1),(0,1,0) is a basis for C^3 and C^3 is the direct sum of U and W and the formula for the projection is T(a,ai+bi+c,b)=(a,ai+bi,b) where a,b,c are elements of C? But shouldn't the formula have the form T(z1,z2,z3)=... where z1 z2 and z3 are elements of C? Or is it enough that (a,ai+bi+c,b) represent all elements of C^3 because it is made up of the basis of C^3?
Oh I get it we can solve for A,B,C are elements of C in terms of z1=a+ib z2=c+id z3=e+if because we have the basis!
So we end up with A=a+ib B=e+if and C=c+id-ai+b-ei+f then from the initial definition that T(x)=x1 we have T(z1,z2,z3)=A(1,i,0)+B(0,i,1)=(a+ib,ia-b+ie-f,e+if)=(a+ib,i(a+e)-(b+f),e+if) I get it. Thank you so much!I really appreciate it.