# Are there finitely many integral vectors $v \in V$ satisfying $||v|| \leq 1$?

Let $V$ be an inner product space over $\mathbb{Q}$ with a basis $\mathcal{E} = (e_1, ... , e_N)$ and call a vector $\sum_{i=1}^N \lambda_i e_i \in V$ an integral (w.r.t $\mathcal{E}$ ) if $\lambda_i \in \mathbb{Z}$ for all $i = 1, \ldots , n$.

My question is: are there finitely many integral vectors $v \in V$ satisfying $||v|| \leq 1$?

My attempt as been to obtain , via Gram-Schmidt, an orthonormal basis $\mathcal{F} = (f_1, \ldots, f_n)$ and an automorphism $\phi$ corrseponding to the change of basis map. Then we have a new norm, $||\cdot||_{\star}$, on $V$ given by $||v||_{\star} = ||\phi (v) ||$. And, by equivalence of norms on $\mathbb{R}$, there exists an $M \in \mathbb{N}$ such that for all $v\in V$ we have $||v||_{\star} \leq M ||v||$.

And there are finitely integral points w.r.t $\mathcal{F}$ satisfying $||v||_{\star} \leq M$ as $||\sum_{i=1}^N \mu_i f_i||_{\star} = \sum_{i=1}^N \mu_i^2$. However I am struggling to relate integral vectors w.r.t to $\mathcal{E}$ to integral vectorss w.r.t to $\mathcal{F}$.

I'm grateful for any help.

I assume that with "inner product" here you mean a $\mathbb{Q}$-valued positive definite quadratic form. Its Gram matrix is rational and therefore non-singular (otherwise it would have a null vector with rational coefficients). Then as a real quadratic form it is still positive definite: The only other option is that it is semi-definite but semi-definite forms are singular.
For each indek $k$ there exists a positive $d_k \in \mathbb{R}$ such that $$\lVert e_k + \sum_{m \neq k} \lambda_m e_m \rVert \geq d_k$$ for any choice of coefficients $\lambda_m$. (Just note that the hyperplane of all such points does not pass through the origin.) Therefore $$\lVert \sum_{k=1}^n \lambda_k e_k \rVert \geq \max \{\lvert \lambda_1\rvert d_1, \ldots, \lvert\lambda_n\rvert d_n \}.$$