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S={(1,-1,-4), (3,2,3)}
T= {(x,y,3y-x) : x,y e R}
(a) Show that S C T, and write down a a vector in IR3 that does not belong to T
(b) Show that T is a subspace of IR3.
(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!!! :(
to show (a) we need to find an x and y such that:
1) (x,y,3y-x) = (1,-1,-4)
2) (x,y,3y-x) = (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 R3, 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 0-vector is in T.
suppose u = (x,y,3y-x) 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',(3y-x)+(3y'-x')). so we need to show the 3rd coordinate of u+v, which is (3y-x)+(3y'-x'), is 3 times the second coordinate of u+v minus the first coordinate of u+v. that is, show:
(3y-x)+(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,3y-x) = (cx,cy,c(3y-x)). so the question now is, does:
c(3y-x) = 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,2b-a,3b-4a). if this equals (0,0,0), this gives us 3 equations in 2 unknowns:
a+3b = 0
2b-a = 0
3b-4a = 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,3y-x). so if:
a(1,-1,-4) + b(3,2,3) = (x,y,3y-x) then:
(a+3b,2b-a,3b-4a) = (x,y,3y-x). in this equation, "x" and "y" are constants, and we are solving for "a" and "b". again we have 3 equations:
a+3b = x
2b-a = y
3b-4a = 3y-x
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 = (2x-3y)/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:
[(2x-3y)/5](1,-1,-4) + [(x+y)/5](3,2,3) = (1/5)(2x-3y,3y-2x,12y-8x) + (1/5)(3x+3y,2x+2y,3x+3y) = (1/5)(5x,5y,15y-5x) = (x,y,3y-x), 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 R3 is called a plane, in this case, the plane determined by the 3 points:
(1,-1,-4),(3,2,3) and (0,0,0).