Is the Second Set of Vectors {v1, v1 + v2, v1 + v2 + v3} Linearly Independent?

[simmonj7]

1. Homework Statement

Suppose that {v1, v2, v3} is a linearly independent subset of R^M. Show that the set {v1, v1 + v2, v1 + v2 + v3} is also linearly independent.

3. The Attempt at a Solution
So I know that {v1, v2, v3} is contained in R^M. And that since the set is linearly independent, the only solution to the vector equation a1v1 + a2v2 + a3v3 = 0 is a1 = 0, a2 = 0, and a3 = 0.

It is true to say that the vector {v1, v1 + v2, v1 + v2 + v3} is linearly independent because for the first vector to be linearly independent, the dimension (M) would have to be greater than the number of vectors (which in this case is three). We know from a theorem that if p (the number of vectors) is greater than m (the dimensional space) then the set is linearly dependent. In this case, for the second set to be linearly independent, M would have to be equal to or less then the number of vectors. However, since the number of vectors is still the same in both the first and second set, the set is linearly dependent...

Correct?

 

[radou]

Of course, the whole idea doesn't make sense unless M >= 3.

But this is pretty easy. What does α v1 + β (v1 + v2) + γ (v1 + v2 + v3) = 0 imply (if you use the fact that {v1, v2, v3} is linearly independent)?

 

[vela]

It is true to say that the vector {v1, v1 + v2, v1 + v2 + v3} is linearly independent because for the first vector to be linearly independent, the dimension (M) would have to be greater than the number of vectors (which in this case is three). We know from a theorem that if p (the number of vectors) is greater than m (the dimensional space) then the set is linearly dependent. In this case, for the second set to be linearly independent, M would have to be equal to or less then the number of vectors. However, since the number of vectors is still the same in both the first and second set, the set is linearly dependent...

Correct?

I think you mixed up independent and dependent a couple of times. Either that or you have a major misconception about linear independence. I'll assume you just accidentally used one word when you meant the other.

If I understand your argument, you're essentially saying: (1) there are 3 vectors that are linearly independent, so $M\ge 3$; (2) the second set of vectors contains 3 vectors; (3) since this number is less than or equal to M, the vectors must be linearly independent. The problem is the last step doesn't work. Take the vectors (1,0,0), (0,1,0), and (1,1,0) in R³. There are only 3 which is equal to the dimension of the vector space, but they're clearly not independent since (1,1,0)=(1,0,0)+(0,1,0).

The theorem you're referring to tells you if you have more vectors than the dimension of the space, then they are linearly dependent. It doesn't tell you anything about independence or dependence, however, if you have fewer vectors than the dimension.