# Thread: basis of a vector space proof

1. ## basis of a vector space proof

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 all d in 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?

2. Originally Posted by sprinks13
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:
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 they must each be 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.
What a complicated way of showing a simple thing! The last equation, c3=0, tells you 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.

****span <--- problem

so i pretty much do the same thing, except:
c1u1 + c2u2 + c3u3 = d
for all d in 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:
Again, it is true because you said "c2u1+ c2u2+ c3u3= d". It because V spans D that the next equations are true.

c1 + c2 + c3 = x
c2 + c3 = y
c3 = z
Wouldn't it be a good idea to say first that 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!

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?

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.

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 any two of 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 of 3 vectors 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.

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### let b={v1,v2,v3,v4} be a basis of a vector space v. find the matrix with respect to b of the linear operator

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