Let $\beta = \{ e_1, e_2, …, e_n\}$ be a basis of the vector space $(\Bbb R, V, +)$. Is $ \alpha = \{ e_1 – e_2, e_2 – e_3, …, e_{n-1} – e_n, e_n – e_1\}$ a linearly independent set?
I think the answer is yes and here's my argument
We can always take the basis $\beta$ and reorder its elements to get $\beta' = \{e_2, e_3, …, e_n, e_1\}$. Given that both $\beta$ and $\beta'$ are linearly independent we get
$$k_1e_1 + \dots + k_n e_n = 0, \qquad k_1' e_2 + \dots + k_n' e_1 = 0 \Rightarrow (k_1-k_n')e_1 + \dots + (k_n-k_{n-1})e_n = 0$$
Given that $k_1 = k'_n$ and so on, $ \alpha = \{ e_1 – e_2, e_2 – e_3, …, e_{n-1} – e_n, e_n – e_1\}$ is linearly independent.
Do you agree? 🙂
Best Answer
No.
$$1 \cdot (e_1 - e_2) + 1 \cdot (e_2 - e_3) + \cdots + 1 \cdot (e_n - e_1) = \mathbf{0}.$$