Chapter 1. Once $$ U=\{u_1,\ldots,u_n\} $$ is a basis for $V$. Then the associated Gram-Schmidt process yields a new basis $U^\prime=\{u_1^\prime,\ldots,u_n^\prime\}$ where $$ u_i^\prime=\sum_{j=1}^ni_ju_j $$ and where the first $n-1$ of the $u_i^\prime$ vectors are independent. However, since the $u_i$ basis is a standard basis, the vectors $u_i^\prime$ will have integer coordinates in this case. In other words, with respect to the standard basis, one can only obtain a basis with integer coordinates (and thus a sublattice of ${\mathbb Z}^n$) by using the Gram-Schmidt process on the standard basis. If one allows more general $U$, then one can obtain vectors with all rational coordinates. Example. Let $U=\{1,-1,2,-2,3,-3\}$ and use the Gram-Schmidt process to obtain a new basis $U^\prime=\{u_1^\prime,\ldots,u_6^\prime\}$. Then \begin{align*} u_1^\prime=&2u_1-u_2+2u_3-3u_4+u_5-2u_6\\ u_2^\prime=&u_2-u_1+u_3-2u_4+3u_5-2u_6\\ u_3^\prime=&u_3-u_1+2u_2-u_4+3u_5-2u_6\\ u_4^\prime=&u_4-2u_1+u_5-u_2+3u_3-2u_6\\ u_5^\prime=&u_5-u_1+2u_2-3u_3+u_4-2u_6\\ u_6^\prime=&u_6-u_1+2u_2-3u_3+2u_4-u_5-2u_6 \end{align*} In the example given in the text, \begin{align*} &4u_1-3u_2+2u_3-3u_4+u_5-u_6,\\ &2u_1+2u_2+3u_3+3u_4+u_5+u_6,\\ &u_1+u_2+2u_3+2u_4+3u_5+2u_6,\\ &4u_1+2u_2+u_3+u_4+2u_5+u_6,\\ &2u_1+3u_2+u_3+2u_4+u_5+u_6. \end{align*} A: No, this is wrong. We can take $e_1,e_2,e_3$ such that $U=Span(e_1,e_2,e_3)$ is a basis, $2e_1+3e_2$ is not in this basis. Then the vectors cannot be integer valued for any $a,b\in\mathbb{Z}$, and a different proof is just by the definition of a basis. A: If you start with $U = \{v_1,v_2,v_3,v_4\}$, you can’t guarantee that $2e_1+3e_2\in\operatorname{Span}(U)$. For example, you can have $$ U = \{2e_1,3e_1-e_2,4e_1-e_3,e_4\} $$ or $$ U = \{4e_1,5e_1-2e_2,3e_1-e_2,e_4\} $$ or any other number of options in between. One interesting property that a lattice always has is that $e_1+e_2\in\operatorname{Span}(U)$. It turns out that this property is what makes this so much easier, and the reason why the statement is correct. Without knowing why, I can probably guess that you know the most obvious statement about a vector being in the span of a set of linearly independent vectors. That is to say, if $v_1,v_2,\ldots,v_n$ are linearly independent, then $k_1v_1+k_2v_2+\cdots+k_nv_n=0\iff k_1=\cdots=k_n=0$. This follows immediately from the linear independence of the $v_i$ and the fact that all the $k_i$ must be zero. (For the sake of concreteness, I’ll assume $k_i\in\mathbb{Z}$ and $v_i\in\mathbb{R}$, but there’s nothing that’s important about this.) Theorem: If $U$ is a set of $n$ linearly independent vectors in $\mathbb{R}^n$, then $U\cap\mathbb{Z}^n = \{\vec 0\}$. The only thing that this might not be obvious for is that $U\cap\mathbb{Z}^n$ is empty. After that, you should know that $\operatorname{Span}(U\cap\mathbb{Z}^n)=\operatorname{Span}(U)$. A vector is in $\operatorname{Span}(U)$ iff it is in $\operatorname{Span}(U\cap\mathbb{Z}^n)$ because $\mathbb{R}^n$ is $\mathbb{Z}$-spaned, but $\mathbb{Z}^n\subseteq\operatorname{Span}(U)$ by definition. Since $\mathbb{Z}^n$ is the set of vectors that are the sum of $n$ integers, any vector in $U$ that’s not equal to $\vec 0$ cannot have all its components integers, and the proof is complete.