If (a,b,c) is a Basis of R3 , Does (a+b,b+c,c+a) also a Basis of R3

[ThankYou]

Homework Statement

Well the same as the subject..
(a,b,c) is a Basis of $$R^{3}$$
does (a+b , b+c , c+a) Basis to $$R^{3}$$


I have another question ..
is (a-b , b-c , c-a) Basis to $$R^{3}$$
This is know is not true because if I use e1, e2 , e3 I got a error line.

Homework Equations

start of liner algebra one.

The Attempt at a Solution

I assume that this is true..
I think something like that.. Because (a,b,c)_ base of $$R^{3}$$ its means that I can get to a step matrix (Row echelon form)
and the same with (b,c,a) and then I can just add them and get to any soultion that I want... But maybe I'm wrong?
Thank you.


P.S
Can I write base instead of Basis ?

 

[Hurkyl]

Attacking questions like this is one of the reasons you learn about things like "row spaces", "Gaussian elimination", and "coordinates relative to a basis".

 

[HallsofIvy]

I would be inclined to just use the fact that three vectors form a basis for a three dimensional space if and only if they are independent- and use the definition of "independent".
a+ b, b+ c, and a+ c form a basis for R³ if and only if the only values of x, y, and z that make x(a+ b)+ y(b+ c)+ z(a+ c)= 0 are x= y= z= 0. Re write that as expressions in x, y, and z times "a", "b", and "c" separately and use the fact that a, b, and c are independent.

 

[hunt_mat]

Presumably by (a,b,c) is a basis for R^3, you mean that the vectors (1,0,0), (0,1,0) and (0,0,1) form a basis for R^3, it seems that the question is asking you if (1,0,1), (1,1,0) and (0,1,1) form a basis for R^3. The question is then can you find three non-zero real numbers such that:
$$a_{1}(1,0,1)+a_{2}(1,1,0)+a_{3}(0,1,1)=0$$
If the answers is yes then the vectors don't form as basis, if the only solution is a_{i}=0, then they do form a basis.

 

[gomunkul51]

It can be shown, through elemental row procedures (on a 3x3 basis matrix) that the second basis: (a+b,b+c,c+a), could be reduces to the first one: (a,b,c).
thus it is a basis of R3.

I've checked, it can be done.

Try it for yourself :)

*P.S: the point of n vectors forming a basis in R^n space is that they aren't linearly dependent !

 

[Mark44]

Presumably by (a,b,c) is a basis for R^3, you mean that the vectors (1,0,0), (0,1,0) and (0,0,1) form a basis for R^3

No, you can't assume that a, b, and c are any specific vectors - only that they form a basis for R³.

, it seems that the question is asking you if (1,0,1), (1,1,0) and (0,1,1) form a basis for R^3. The question is then can you find three non-zero real numbers such that:
$$a_{1}(1,0,1)+a_{2}(1,1,0)+a_{3}(0,1,1)=0$$
If the answers is yes then the vectors don't form as basis, if the only solution is a_{i}=0, then they do form a basis.

To the OP, to answer your question - you can have one basis or two bases. That's how I remember it working.

 

[ThankYou]

Thank you all , Its a yes no question, But I were interested n how to prove it..

I would be inclined to just use the fact that three vectors form a basis for a three dimensional space if and only if they are independent- and use the definition of "independent".
a+ b, b+ c, and a+ c form a basis for R³ if and only if the only values of x, y, and z that make x(a+ b)+ y(b+ c)+ z(a+ c)= 0 are x= y= z= 0. Re write that as expressions in x, y, and z times "a", "b", and "c" separately and use the fact that a, b, and c are independent.

Well I've tried this ... even before I posted the question, But I could not get to any solution.
Now I've tried again and made it
a(x+z)+b(z+y)+c(y+z)=0
then , x+z = z+y = y+z = 0
Thank you.

It can be shown, through elemental row procedures (on a 3x3 basis matrix) that the second basis: (a+b,b+c,c+a), could be reduces to the first one: (a,b,c).
thus it is a basis of R3.

I've checked, it can be done.

Try it for yourself :)

*P.S: the point of n vectors forming a basis in R^n space is that they aren't linearly dependent !

Thank you, I've also tried this before I've posted the question.. But for some reason I didn't made it..
After you told me you did it I looked at it again and it work..
Thank you
(R3-R2) -> (R3+R1) -> (R1-0.5R3) -> (R2-R1) Done...