Question

Is it true that if x, y, z are linearly independent vectors, then so also are x+y, y +z and z +x ?
Please, explain me

Answer 1

@ranga

Answer 2

If x,y,z are linearly independent vectors, then there are no constants a,b,c (except the trivial case where a=b=c=0) such that ax + by + cz =0

x+y is independent if ax+by is not zero for any a,b (except trivial case).

So, does ax + by + cz not equal to zero for any a,b,c imply that ax+by is not zero for any a,b ?

If this is the case, then independence in 3 dimensions implies independence in 2 dimensions.

Note that if there is not independence in 2 dimensions then, there can not be independence in 3 because then we would not have 3 independent vectors.

Answer 3

is not that x, y , z are linearly independent to each other. And the question for x+y, y +z, z + x belongs to other vector space?

Answer 4

let say x+y = u
y + z = v
z + x = t
u, v, t are linearly independent

Answer 5

Put in matrix form and you can see if they are independent. If any two rows or 2 two columns are dependent, then automatically you have a determinant of zero, which implies that the matrix is not invertible and thus the rows / columns are dependent. This also implies that there is a combinations of the the rows or columns that produce the zero vector.

I can't tell by looking at your system of equations if they are dependent.

But using linear algebra ideas, you can see that ax + by + cy is not equal to zero for all a,b,c, means that ax + by is also not equal to zero for all a,b.

Answer 6

x + y vector space is a subspace of larger x + y + z

Answer 7

yes

Answer 8

x + y must be independent. Assume by way of contradiction that the are not. Then this means that the x+y+z is not independent, which contradicts our assumption. Therefore, x + y is independent.

Answer 9

hey, "x+y must be independent " compare to what? you cannot stop there.

Answer 10

because trivially, x +y is dependent compare to either x or y

Answer 11

When we say that x + y is independent, we mean that all combinations of vectors of x combined with all combinations of vectors in y when added can never produce the zero vector.

We assumed that ax + by + cz was independent, therefore x + y + z is independent, therefore x + y is independent. With contradiction proof, we assumed conclusion to be false and showed that it could not be because then our assumption would be invalid (think of matrix argument with determinant equal to zero).

Answer 12

x and y each can be two (basis) vectors that span a space. The question is, are they independent?

Answer 13

yes, they are.

Answer 14

oh, I got you. hehehe.

Answer 15

geometrically, x +y is a diagonal line of x and y. If we have its unit, it spans the new space, right?
the same with y+z and z + x
that's what I think

Answer 16

think of the 3 bisectors of the R^3, they spans a space which doesn't relate to x,y, z, right?

Answer 17

@ybarrap I see what you mean with the logic above

Answer 18

Nope, because the question is ask whether x+y , y+z and z +x are linearly independent or not, I understand that they are linearly independent to each other, not to x, y , z
To x, y ,z. trivially, they are NOT. how can x +y be independent to x and y? x +y is a combination of x and y with the coefficient 1 for both, right?

Answer 19

Ok. I finally see your point!

To check if x+y , y+z and z +x are linearly independent you need to check if
a(x + y) + b(y+z) + c(z +x) is not zero for any a,b c. Assuming that x,y and z are independent.

Did you do this?

Answer 20

a(x + y) + b(y+z) + c(z +x) = x(a + c) + y(b + a) + z(b+c)

But we know that mx + nx + py is not zero for ANY m,n, p therefore, those 3 spaces ARE independent. Note that a,b and c are constants.

Does this sound right to you?

Answer 21

yes, digest well, hehehe. thanks friend

Answer 22

NP