# Linear Independence of different combinations of vectors

In a previously asked question I missed a key ingredient, which will be mentioned in this question, so please do not tag this "duplicate".

Let V be a finite dimensional vector space with dimension suitably large ( say > 7 ). Let $\alpha_{1}, \alpha_{2}, \alpha_{3}, \beta_{1}, \beta_{2}, \beta_{3}, \gamma \in V$ such that all of them are distinct. Now if,

(i) $\alpha_{1}, \alpha_{2}, \alpha_{3}$ are linearly independent

(ii) $\beta_{1}, \beta_{2}, \beta_{3}$ are linearly independent.

(iii) $\alpha_{1},\beta_{1}$ are linearly independent.

(iv) $\alpha_{2},\beta_{2}$ are linearly independent.

(v) $\alpha_{3},\beta_{3}$ are linearly independent.

(vi) $\beta_{1}, \beta_{2}, \beta_{3}, \gamma$ are linearly independent.

Then prove or disprove (provide a counter example) whether $\alpha_{1},\alpha_{2}, \alpha_{3}, \gamma$ are linearly independent.

My attempt : I have thought in the Euclidean space (i.e. $\Bbb R^n$), and intuitively it seems true to me but haven't been able to come up with anything rigorous. Moreover V in question can be any arbitrary (suitably) finite dimensional vector space. So I couldn't come up with anything.... Thanks in advance for help....

## 1 Answer

Let $\alpha_1, \alpha_2, \alpha_3, \beta_1, \beta_2, \beta_3$ be a linear independent list in $V$. Choose $\gamma = \alpha_1 + \alpha_2 + \alpha_3$. All of your points are satisfied, however $\alpha_1, \alpha_2, \alpha_3, \gamma$ is obviously not linearly independent.