
Subspaces and bases
S={(1,1,4), (3,2,3)}
T= {(x,y,3yx) : x,y e R}
(a) Show that S C T, and write down a a vector in IR^{3 }that does not belong to T
(b) Show that T is a subspace of IR^{3}.
(c) Show that S is a basis for T. Write down the dimension of T, and describe T geometrically.
HELP..... my head is completely messed with this one!!! :(

Re: Subspaces and bases
to show (a) we need to find an x and y such that:
1) (x,y,3yx) = (1,1,4)
2) (x,y,3yx) = (3,2,3)
it's pretty obvious that for (1) we have to take x = 1, y = 1, so all we have to do is verify that 3(1)  1 = 4.
similarly for (2), we must have x = 3, y = 2, so it boils down to verifying that 3(2)  3 = 3.
so to find a vector in R^{3}, let's pick x = 1, y = 1. all we have to do is pick a 3rd coordinate which is NOT 3(1)  1 = 2. so, for example, (1,1,1) will work just fine.
(b) we have to verify 3 things:
1) T is closed under vector addition.
2) T is closed under scalar multiplication.
3) the 0vector is in T.
suppose u = (x,y,3yx) and v = (x',y',3y'x') are in T. we need to show that u+v is in T, as well.
now u+v = (x+x',y+y',(3yx)+(3y'x')). so we need to show the 3rd coordinate of u+v, which is (3yx)+(3y'x'), is 3 times the second coordinate of u+v minus the first coordinate of u+v. that is, show:
(3yx)+(3y'x') = 3(y+y')(x+x')....can you do this?
next suppose c is ANY real number. we need to show that cu is in T. now cu = c(x,y,3yx) = (cx,cy,c(3yx)). so the question now is, does:
c(3yx) = 3(cy)cx ?
finally, is (0,0,0) in T? that is: does 3(0)  0 = 0?
(c) we have to do TWO things to show S is a basis for T:
1) we need to show S is a linearly independent set. this means that if:
a(1,1,4) + b(3,2,3) = (0,0,0) the only possible choice for a and b is a = b = 0.
now a(1,1,4) + b(3,2,3) = (a,a,4a) + (3b,2b,3b) = (a+3b,2ba,3b4a). if this equals (0,0,0), this gives us 3 equations in 2 unknowns:
a+3b = 0
2ba = 0
3b4a = 0
if we add equations 1 and 2 together, we get:
5b = 0, which means b = 0. substituting b = 0 in any of the original 3 equations quickly gives a = 0, so we see that indeed this is the only possibility. so S is linearly independent.
2) we need to show S SPANS T. that is, given ANY member u of T, we must be able to find a and b such that:
a(1,1,4) + b(3,2,3) = u.
recall a typical element u of T is u = (x,y,3yx). so if:
a(1,1,4) + b(3,2,3) = (x,y,3yx) then:
(a+3b,2ba,3b4a) = (x,y,3yx). in this equation, "x" and "y" are constants, and we are solving for "a" and "b". again we have 3 equations:
a+3b = x
2ba = y
3b4a = 3yx
again, if we add the first two equations we obtain: 5b = x+y, thus b = (x+y)/5. substituting this value for b in the first equation gives:
a + 3(x+y)/5 = x
a = [5x  3(x+y)]/5 = (2x3y)/5
now, to verify this is a consistent solution, we could check all three equations with these values, but it is simpler just to verify that:
[(2x3y)/5](1,1,4) + [(x+y)/5](3,2,3) = (1/5)(2x3y,3y2x,12y8x) + (1/5)(3x+3y,2x+2y,3x+3y) = (1/5)(5x,5y,15y5x) = (x,y,3yx), as desired. so S spans T.
the dimension of a subspace is the number of elements in ANY basis. so S is a linearly independent spanning set for T, it IS a basis. since S has 2 elements, the dimension of T is 2. geometrically, a linear subspace of dimension 2 of R^{3} is called a plane, in this case, the plane determined by the 3 points:
(1,1,4),(3,2,3) and (0,0,0).