AkilMAI said:
Homework Statement
For each of the following subsets U of the vector space V I have to decide whether or not U is a subspace of V . In each case when U is a subspace, I also must find a
basis for U and state dim U:
(i) V = R^4; U = {x = (x1; x2; x3; x4) : 3x1 - x2 -2x3 + x4 = 0}:
(ii) V = R^3; U = {x = (x1; x2; x3) : x1^2 = x2^2 + 4*x3^2}
(iii) V = P3; U = {p in P3 : p'(0) = p(1)}
I have some questions for each problem...
Homework Equations
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The Attempt at a Solution
I need to know that my solutions are ok or not and if so, how to find the basis and dim of U.
(i) we take x,y in U=> x+y=0 and a(a scalar),a*x=a*0=0 ,so it is closed under addition and scalar multiplication,therefore it is a subspace.If my approach is correct how can I find basis U and dim U?
What do you mean by "x+ y= 0" and "a*x= a*0= 0"? If x and y are arbitrary members of U, they they are NOT necessarily the 0 vector and neither x+ y nor a*x is the 0 vector. You need to use the definition of U: that if x= (x1, x2, x3, x4) then 3x1- x2- 2x3+x4= 0 and you have made no reference to that equation.
As for a basis, any vector in the set is of the form (x1, x2, x3, x4) but you can use the single equation, 3x1- x2- 2x3+x4= 0, to write one of those in terms of the other three. replace that one, in (x1, x2, x3, x4), by its expression as the other three, and then write it as a linear combination specific vectors with the x1, x2, etc. as coefficients.
(ii)Here I did something similar,I wrote x1^2 = x2^2 + 4*x3^2 as x1^2 - x2^2 - 4*x3^2=0
=> x,y in U where x+y =0+0=0 and a*x=a*0=0.Again If my approach is correct how cand I find basis U and dim U?
No, your approach is NOT correct. Saying that x= (x1, x2, x3) such that x1^2 = x2^2 + 4*x3^2 does NOT mean that x= 0!
(iii)take p and g two polyn from U=>(g+p)(1)=p(1)+g(1)=p'(0)+g'(0)=(g+p)(0) and (a*p)(1)=a*p(1)=a*p(0)=(a*p)(0).Same thing, If my approach is correct how cand I find basis U and dim U?
Again, all you have done in all three is
state the conditions for a subspace- that it be closed under addition and scalar multiplication. You have NOT shown that this is true for these particular subsets. In fact, that is NOT true for at least one of these sets. For ii and iii, see my suggestions for i.
I won't do your problems for you but I will give other examples:
Show that (x1, x2, x3), such that 2x1- x2+ 3x3= 0 is a subspace of R^3.
If x and y are vectors in this subspace, then we can write x= (x1, x2, x3) with 2x1- x2+ 3x3= 0 and we can write y= (y1, y2, y3) with 2y1- y2+ 3y3= 0. Are x+y= (x1+ y1, x2+ y2, x3+ y3) and ax= (ax1, ax2, ax3) also in that set? That is, do they satisfy that defining equation?
2(x1+ y1)- (x2+ y2)+ 3(x3+ y3)= 2x1+ 2y1- x2- y2+ 3x3+ 3y3= (2x1- x2+ 3x3)+ (2y1- y2+ 3y3)= 0+ 0= 0.
2(ax1)- (ax2)+ 3(ax3)= a(2x1- x2+ 3x3)= a(0)= 0 so, yes, they do. This is a subspace of R^3.
We can solve 2x1- x2+ 3x3= 0 for x2: x2= 2x1+ 3x3. That means we can write an arbitrary member of the subspace as x= (x1, x2, x3)= (x1, 2x1+3x3, x3)= (x1, 2x1, 0)+ (0, 3x3, x3)= x1(1, 2, 0)+ x3(0, 3, 1). Now can you see a basis and the dimension?
Another example. Let U be the set of all vetors (x1, x2) in R^2 such that x1^2- x2^2= 0. Is this a subspace? Let x= (x1, x2) and y= (y1, y2) be in this set. Then x+y= (x1+y1, x2+ y2). Is this in the same set? Does it satisfy (x1+ y1)^2- (x2+ y2)^2= 0?
(x1+ y1)^2- (x2+ y2)^2= (x1^2+ 2x1y1+ y1^2)- (x2^2+ 2x2y2+ y2^2)= (x1^2- x2^2)+ (y1^2- y2^2)+ (2x1y1- 2x2y2)= (0)+ (0)+ 2(x1y1- x2y2). That is equal to 0 if and only if x1y1= x2y2 but that is NOT always true. In order to show that it is NOT always true we only need to find a counter-example.
Look at x=(1, 1) and y= (1, -1) (so that x1y1= 1(1)= 1 and x2y2= (1)(-1)= -1 are NOT equal). Both are in the set because (1)^2- (1)^2= 0 and (1)^2- (-1)^2= 0. But x+ y= (1+ 1, 1- 1)= (1, 0). Now (1)^2- 0^2= 1, not 0, so these are two vectors in the subset whose sum is NOT in the subset. This subset is not closed under addition so the subset is NOT a subspace.
Since this is not a subspace, there is NO basis and NO dimension.