Poor wording. That "is true" because you said it was, above. not because V is linearly independent. And you don't "let" the coeficients be 0, it is because V is linearly independent theymusteach be 0.

What a complicated way of showing a simple thing! The last equation, c3=0,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.tellsyou that c3= 0. Putting that into the second equation, c2+ c3= 0, gives immediately that c2= 0. Putting both of those into the first equation, c1+ c2+ c3= 0, gives c1= 0.

All you have written is, however, completely correct.

Again, it is true because you****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:said"c2u1+ c2u2+ c3u3= d". It because V spans D that the next equations are true.

Wouldn't it be a good idea to sayc1 + c2 + c3 = x

c2 + c3 = y

c3 = zfirstthat d= xv1+ yv2+ cv3? That is, that x, y, and z are the coefficients of v1, v2, and v3, respectively when you write d as a linear combination of v1, v2, and v3. Just throwing x, y, and z up there without any statement as to what they are is a bit shocking!

Personally, I think it would be simpler to show how to write v1, v2, and v3 as linear combinations of the u1, u2, and u3.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!

For example, if au1+ bu2+ cu3= v1, then a(v1)+ b(v1 + v2)+ c(v1 + v2 + v3)= (a+b+c)v1+ (b+c)v2+ cv3= v1. Then (a+b+c-1)v1+ (b+c)v2+ cv3= 0. Since {v1, v2, v3} are independent, we must have a+b+c- 1= 0, b+c= 0, c= 0 which immediately give c=0, b= 0, a-1= 0 so a= 1. u1= v1 (which was obvious from the start).

If if au1+ bu2+ cu3= v2, then a(v1)+ b(v1 + v2)+ c(v1 + v2 + v3)= (a+b+c)v1+ (b+c)v2+ cv3= v2. Then (a+b+c)v1+ (b+c-1)v2+ cv3= 0. Since {v1, v2, v3} are independent, we must have a+b+c= 0, b+c- 1= 0, and c= 0. Putting c= 0 into b+c- 1= 0, we get b= 1 and then a+b+c= 0 becomes a+1+0= 0 so a= -1. v2= -u1+ u2.

Do the same for v3.

Since any vector can be written in terms of v1, v2, and v3, and they can be written in terms of u1, u2, and u3, if follows immediately that any vector can be written in terms of u1, u2, and u3.

However, do you have the theorem:

" A basis for vector space V satisifies three properties:

1) They are independent.

2) They span V.

3) The number of vectors in the basis is equal to the dimension of V.

and anytwoof those is sufficient to prove the third."

Since {v1, v2, v3} is given as a basis for D, it follows that the dimension of D is 3. Since {u1, u2, u3} is a set of3vectors that are independent, it follows that "(2) They span D" and are a basis for D.

Your proof is good, I just want to help clean it up a little.