# Vector space has dimension less than d

#### UOAMCBURGER

1. The problem statement, all variables and given/known data
Problem given to me for an assignment in a math course. Haven't learnt about roots of unity at all though. Finding this problem super tricky any help would be appreciated. Screenshot of problem below.

2. Relevant equations
Unsure of relevant equations

3. The attempt at a solution
so far i am just trying to understand what the question is asking me to do, i am showing that the vector space V, which is just the rational numbers over the field Q? has dimension strictly less than d, where d is >= 1?

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#### Math_QED

Homework Helper
The vector space $V$ is spanned by $d$ elements. Therefore, its dimension is at most $d$. The question asks you to show that the generators are not linearly independent, which means you can remove one of the generators of $V$ such that the remaining generators still span $V$. It will then follow that the soace has dimension $d-1$.

The question gives a hint how to show that the generators are not linearly independent.

Also, the space is not the rational numbers.

It are $\mathbb{Q}$-linear combinations of powers of roots of unity.

"Vector space has dimension less than d"

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