There's an arrangement of vectors in linear space V over the field K. Is it possible that the field $lin(A)$ is infinitely dimensional?

Question

Let's take $A = \{α_1, α_2, ...\}$ as an arrangement of vectors in linear space V over the field K. Suppose that dim $lin \{α_i, α_{i+1}, α_{i+2}\} = 2$ for each $i \geq 1$. Is it possible that the field $lin(A)$ is infinitely dimensional?

Vectors $A = \{α_1, α_2, ...\}$ can be seen as a matrix of $2$ linearly independent columns for every $3$ in a row. This can be visualised as: $$A=\pmatrix{ind&ind&dep&ind&ind&dep&...\\ ind&ind&dep&ind&ind&dep&... \\ ind&ind&dep&ind&ind&dep&...\\ ind&ind&dep&ind&ind&dep&... \\ ...&...&...&...&...&... &...}$$

Then it possible that the field $lin(A)$ is infinitely dimensional.

Can somebody tell me if I think correctly?

Answer 1

It depends on whether many of the $\alpha_i$ can be zero. If infinitely many can be zero, you can have at least one in every three zero and the condition will hold. If infinitely many are not zero you can have an infinite dimensional space. If zero vectors are excluded you cannot have an infinite dimensional space. We are told that we can write $\alpha_3$ as a comibination of $\alpha_1$ and $\alpha_2$. Then we can write $\alpha_4$ as a combination of $\alpha_2$ and $\alpha_3$. What does that tell you? Continue

Answer 2

Let $V$ be the vector space of real sequences $(a_n)_{n=0}^{\infty}$. Let $e_j$ denote the sequence which has $1$ in the $j$-th place and $0$ elsewhere. Let $f_j=e_j$ when $j\not= 0\mod 3$, and let $f_{3k}=0$.

Then the vectors $f_0, f_1, f_2, \dots$ satisfy the required condition and span an infinite dimensional space.