Determine if $\beta = \{b_1, b_2, b_3\}$ is linearly independent. Let $\beta = \{b_1, b_2, b_3\}$.
Suppose that all you know is that:


*

*$b_2$ is not a multiple of $b_3$.

*$b_1$ is not a linear combination of $b_2, b_3$.  


Can you determine if $\beta$ is linearly independent?  
From the second assumption, I think we can tell that $b_1$ isn't a multiple of $b_2$ nor $b_3$, Right?   
So, All in all, we get that no vector is a multiple of the other.
With that in mind, can we figure that $\beta$ is linearly independent? 
 A: Three vectors can be linearly independent without each being strictly a multiple of one other. You are correct that the vectors are linearly independent.
Assuming $b_1, b_2, b_3$ are non-zero:
We start with the set $\{b_2, b_3\}$ where we know that $b_2$ and $b_3$ must be linearly independent, since one is not a multiple of the other. We know that vectors are linearly independent if any one of them cannot be written as  a linear combination of the others. In the case of two vectors, this essentially means that they are linearly independent if one is not equal to a multiple of the other.
We have $b_1$ and are given that $b_1$ is not a linear combination of $b_2 $ and $b_3$. That tells us that the addition of a vector that is linearly independent of two linearly independent vectors gives us a set of three linearly independent vectors $\;\{b_1, b_2, b_3\}$.
A: Consider $b_1 = (1,0)$, $b_2 = (0,1)$ and $b_3 = (0,0)$ in $\mathbb{R}^2$. Are they linearly independent? Do they satisfy the hypotheses?
The common trap here is that two nonzero vectors are linearly independent iff the first is not a multiple of the second. When one of them is zero though...
