What can we say about output of Gram–Schmidt process Given $\{x_1, \dots, x_{n-1}\}$ linearly independent vectors and $x_n \in \operatorname{span}\{x_1, \dots, x_{n-1}\}$ and let $\{\hat{x_1}, \dots, \hat{x_{n-1}}, \hat{{x_n}}\}$ be the output of the Gram–Schmidt process. 

I have to answer the question: what can we say about $\hat{x_n}$. Of course $\hat{x_n}$ is orthogonal to $\hat{x_1}, \dots \hat{x_{n-1}}$ and $\hat{x_n} \in \operatorname{span}\{x_1, \dots, x_{n-1}\}$ but can we say something more non-trivial?
Since we know that $\DeclareMathOperator{\span}{span}\span(x_i, i=1..n-1) = \span(x_i, i=1..n)$, Gram-Schmidt will produce $\hat x_i$ such that
$$\span(\hat x_i, i=1..n-1) = \span(\hat x_i, i=1..n) \ni \hat x_n$$
and $\langle \hat x_i, \hat x_j \rangle = \delta_{ij}$ (Kronecker delta). Thus we know that there is some representation
$$\hat x_n = \sum_{i=1}^{n-1} \alpha_i \hat x_i$$
and since
$$0 = \langle \hat x_i, \hat x_n \rangle = \sum_{j=1}^{n-1} \alpha_j \langle \hat x_i, \hat x_j \rangle = \alpha_i$$
for all $i=1..n-1$, we have
$$\hat x_n = \sum_{i=1}^{n-1} 0 \cdot \hat x_i = 0$$
Since we know that $\DeclareMathOperator{\span}{span}\span(x_i, i=1..n-1) = \span(x_i, i=1..n)$, Gram-Schmidt will produce $\hat x_i$ such that
$$\span(\hat x_i, i=1..n-1) = \span(\hat x_i, i=1..n) \ni \hat x_n$$
and $\langle \hat x_i, \hat x_j \rangle = \delta_{ij}$ (Kronecker delta). Thus we know that there is some representation
$$\hat x_n = \sum_{i=1}^{n-1} \alpha_i \hat x_i$$
and since
$$0 = \langle \hat x_i, \hat x_n \rangle = \sum_{j=1}^{n-1} \alpha_j \langle \hat x_i, \hat x_j \rangle = \alpha_i$$
for all $i=1..n-1$, we have
$$\hat x_n = \sum_{i=1}^{n-1} 0 \cdot \hat x_i = 0$$