Proving {f1,f2,f3} is a Basis for V Over F

[boneill3]

Homework Statement

Let {e1,e2,e3} be a basis for the vgector space V over the field F.
Put f1 = -e1, f2 = e1+e2 and f3 =e1 + e3

Prove that {f1,f2,f3} is also a basis for V

Homework Equations

The Attempt at a Solution

I made e1,e2,e3 be the unit bases.

1
0
e1= 0


0
0
e2= 1



0
0
e3= 1


which makes {f1,f2,f3}


-1
0
f1= 0


1
1
f2= 0


1
0
f3= 1



so {f1,f2,f3} =

-1 1 1 0
a1 * 0 + a2 * 1 + a3 * 0 = 0
0 0 1 0

What i did was set this matrix equal to zero and solve.

The solution was only the trivial solution.
As this proves that {f1,f2,f3} is linear independent is that enough to show that it also is a basis ?

regards

 

[Mark44]

You're given an arbitrary vector space V over some field K, so you can't assume that e₁, e₂, and e₃ are the standard basis vectors for R³. Your set {f₁, f₂, f₃} is a basis for V iff this set is linearly independent and spans V.

For the first part, show that the equation c₁f₁ + c₂f₂ + c₃f₃ = 0 has exactly one solution. For the second part, show that any vector v in V can be written as a linear combination of the vectors f₁, f₂, and f₃.

 

[gabbagabbahey]

What is your definition for basis? And why are you assuming that $e_i$ are unit vectors?...They can have any length and still form a basis.

 

[boneill3]

For the first part show that c1f1 + c2f2 + c3f3 = 0

is the equation to prove

c1(-e1) + c2(e1+e2) + c3(e1 + e3) = 0

regards

 

[Mark44]

For the first part show that c1f1 + c2f2 + c3f3 = 0

is the equation to prove

c1(-e1) + c2(e1+e2) + c3(e1 + e3) = 0

regards

No. You need to prove that c₁(-e₁) + c₂(e₁ + e₂) + c₃(e₁ + e₃) = 0 has exactly one solution for the constants c₁, c₂, and c₃.

Rewrite this equation so that it is ___e₁ + ___e₂ + ___e₃ = 0 (you need to fill in the blanks). Use what you know about e₁, e₂, and e₃ being a basis for V.

 

[boneill3]

So I have:

f1 = -e1
f2 = e1+e2
f3 = e1+e3

I need to try and eliminate e2 and e3

so


f1 = -e1
f2+f3 = 2e1+e2+e3
f3+f2 = 2e1+e3+e2


Then

f1 = -e1
f2+f3 = 2e1+e2+e3
-1*(f3+f2) = -2e1 -e3 -e2



f1 = -e1
(f2+f3)+(-1*f3+f2) = 0

f1 = -e1
(f2+f3)+(-1*f3+f2)+f1= -e1

(f2+f3)+(-1*f3+f2)+f1= f1


(f2+f3)+(-1*f3+f2)+ (f1-f1)= 0

showing that f1,f2,f3 are independant

 

[Mark44]

showing that f1,f2,f3 are independant

And how exactly does your work show this? Look at what I outlined for you in post 5.

 

[boneill3]

HI Guys
In my lecture notes it says.

a set S = {U1,U2...Un} of vectors is a basis of V if every v an element of V can be written uniquely as a linear combination of the basis vectors.

I am told that:

Let {e1,e2,e3} be a basis for the vector space V over the field F.

Can I assume that they are lineary independent and therefore only equal to zero if all coefficients are zero?


so using "a set S = {U1,U2...Un} of vectors is a basis of V if every v an element of V can be written uniiquely as a linear combination of the basis vectors."


don't I just have to show that each vectore {f1,f2,f3} can be written as an linear combination of the basis vectors.

eg. f2 = e1+e3 = 1*e1 + 0*e2 + 1*e3


You said:
Rewrite this equation so that it is ___e1 + ___e2 + ___e3 = 0 (you need to fill in the blanks). Use what you know about e1, e2, and e3 being a basis for V.


As e1, e2, and e3 is a basis they are Linearly independant and span the the vector Space. So can only equal zero if all coefficients are zero

 

[VeeEight]

Let {e1,e2,e3} be a basis for the vector space V over the field F.

Can I assume that they are linearly independent and therefore only equal to zero if all coefficients are zero?

The vectors that form a basis are linearly independent.

don't I just have to show that each vector {f1,f2,f3} can be written as an linear combination of the basis vectors.

You need to show that any vector v in V can be written as a linear combination of {f1, f2, f3} - this shows that the set generates V

 

[boneill3]

So say you got the vector e₁

It can be generated by

e₁ = -1*f₁ + 0*f₂ + 0*f₃

And as The vectors e1,e2,e3 are independent so are f1 , f2 , f3

regards