# Linear independence again

1. Nov 1, 2006

Let V be a vector space over a field F, $$v_{1}, \cdots, v_{n} \in V$$ and $$\alpha_{1}, \cdots, \alpha_{n} \in F$$. Further on, let the set $$\left\{v_{1}, \cdots, v_{n}\right\}$$ be linearly independent, and b be a vector defined with $$b=\sum_{i=1}^n \alpha_{i}v_{i}$$. One has to find necessary and sufficient conditions on the scalars $$\alpha_{1}, \cdots, \alpha_{n}$$ such that the set S=$$\left\{b, v_{2}, \cdots, v_{n}\right\}$$ is linearly independent, too.

Well, I just used a simple proposition which states that a set is dependent if there exists at least one vector from that set which can be shown as a linear combination of the rest of the vectors from the same set. So, obviously, for $$\alpha_{1} = 0$$, the set S is dependent, which makes $$\alpha_{1} \neq 0$$ a necessary condition for S to be independent. Further on, a sufficient condition would be $$\alpha_{1} = \cdots = \alpha_{n} = 0$$, which leaves us with the set S\{b}. This set must be linearly independent, since it is a subset of {v1, ..., vn}, which we know is linearly independent.

I may be boring, but I'm just checking if my reasoning is allright..

Edit. I just realized, for $$\alpha_{1} = \cdots = \alpha_{n} = 0$$ we have b = 0, which makes the set dependent! So $$\alpha_{1} \neq 0$$ is a necessary condition, but what's the sufficient condition? Is it that at least one of the scalars $$\alpha_{2}, \cdots, \alpha_{n}$$must not be equal zero?

Last edited: Nov 1, 2006
2. Nov 1, 2006

### Office_Shredder

Staff Emeritus
The sufficient condition certainly includes this $$\alpha_{1} \neq 0$$

Basically, if a condition is necessary and sufficient, it means that the statement is true iff the condition is true. So search for a set of conditions that are true iff the set b, v 2-n is linearly independent.

Note that if all the alphas except $$\alpha_1$$ are zero, then b is just a multiple of v1, and the new set is certainly linearly independent

3. Nov 1, 2006

Yes, I realized that.. So, it seems $$\alpha_{1} \neq 0$$ is both necessary and sufficient condition, since the rest of the scalars can be any elements of F, all zero, all non zero, or combined.