Let $\mathcal{F}$ be a $m-$dimensional vector space of real-valued functions $f: \mathcal{X} \mapsto \mathbb{R}$. Let $n=m+2$.
Take any set of $n$ points $(x_{1},t_{1}), \ldots, (x_{n},t_{n})$. Show that: \begin{align} \left\{\left(f(x_{1}) - t_{1},\ldots,f(x_{n})-t_{n} \right): f \in \mathcal{F}\right\} \end{align} is contained within an $n-1$ dimensional subspace of $\mathbb{R}^{n}$.
It makes sense to me that the vector $\left(f(x_{1}),\ldots,f(x_{n}) \right)'$ for $f$ varying over $\mathcal{F}$ spans a $n-2$ dimensional subspace of $\mathbb{R}^{n}$. What I do not understand is how subtracting a fixed vector $t \in \mathbb{R}^{n}$ would ever allow the vector to span a $n-1$ dimensional subspace as $f$ varies over $\mathcal{F}$. Can anyone offer a simple explanation/proof?
Further background: This appears in Lemma 2.6.15 in Van der Vaart and Wellner (1996) Weak Convergence and Empirical Processes, although absolutely no knowledge of empirical process theory should be needed to solve this.
We denote the given set by $$ S := \{ (f(x_1) - t_1, \dotsc, f(x_n) - t_n) \mid f \in \mathcal{F} \}. $$ The set $$ U := \{ (f(x_1), \dotsc, f(x_n)) \mid f \in \mathcal{F} \} $$ is a linear subspace of $\mathbb{R}^n$ which is at most $m$-dimensional, since it is the image of the linear map $$ \mathcal{F} \to \mathbb{R}^n, \quad f \mapsto ( f(x_1), \dotsc, f(x_n) ), $$ with $\mathcal{F}$ being $m$-dimensional. It follows that $S$ is contained in the linear subspace \begin{align*} U' &:= U + \langle (t_1, \dotsc, t_n) \rangle \\ &= \{ u + \lambda (t_1, \dotsc, t_n) \mid u \in U, \lambda \in \mathbb{R} \} \\ &= \{ ( f(x_1) - \lambda t_1, \dotsc, f(x_n) - \lambda t_n ) \mid f \in \mathcal{F}, \lambda \in \mathbb{R} \}, \end{align*} which is at most $(m+1)$-dimensional, with $m+1 = n-1$.