When trying to find a basis, I got (-3, -11, 1, 0 ) & (3, 1, 0, 10)
When i saw an example using a different technique to me, (3, 1, 0, 10) & (3, 0, 1, 11) was the basis?
Is that possible, generally and in this example? Thanks
all bases have the same "size" (literally if V is finite-dimensional).
in this case, however, one of the bases must be incorrect: (3,0,1,11) is not in span{(-3,-11,1,0),(3,1,0,10)}.
if it were, we would have (3,0,1,11) = a(-3,-11,1,0) + b(3,1,0,10), giving us:
-3a + 3b =3
-11a + b = 0
a = 1
10b = 11 --> b = 11/10. but -11 + 11/10 is not 0, so we cannot find any such a and b.
the 3rd coordinate of (3,0,1,11) is 1. the 3rd coordinate of a(-3,-11,1,0) + b(3,1,0,10) = (-3a+3b,-11a+b,a,10b) is a.
hence, if the two are to be equal, a MUST EQUAL 1. the 4 equations
-3a+3b = 3
-11a+b = 0
a=1
10b=11
must all have a SIMULTANEOUS solution for (3,1,0,11) to be in the span.