Let V = {v1, v2, v3} be a basis for a vector space D. Show that U = {u1, u2, u3} is a basis for V, where u1 = v1, u2 = v1 + v2, and u3 = v1 + v2 + v3.

attempt at solving:

linearly independent

c1u1 + c2u2 + c3u3 =0

iff c1=c2=c3=0

so sub in for u's and rearrange:

(c1 + c2 + c3)v1 + (c2 + c3)v2 + c3v3 =0

which is true, since V is linearly independant, so let:

c1 + c2 + c3 = 0

c2 + c3 = 0

c3 = 0

1 1 1 | 0

0 1 1 | 0

0 0 1 | 0

in RREF is just the identity matrix with a row of zeroes, so c1=c2=c3=0 and therefore linearly independent.

****span <--- problem

so i pretty much do the same thing, except:

c1u1 + c2u2 + c3u3 =d

for alldin D

so sub in for u's and rearrange:

(c1 + c2 + c3)v1 + (c2 + c3)v2 + c3v3 =d

which is true, since V spans D, so let:

c1 + c2 + c3 = x

c2 + c3 = y

c3 = z

except my matrix is now

1 1 1 | x

0 1 1 | y

0 0 1 | z

and if i get it into RREF, i still have an identity matrix, and all my c's are combinations of x,y,z, which are just scalars. So I want to say i'm done... but,

Have I proved that V spans D?

thanks in advance!