The Points $p_1,…,p_k$ are Affinely Dependent if there exist scalars $α_1,…,α_k$, not all zero, such that
$α_1 p_1+…+α_k p_k=0$
$α_1+…+α_k=0.$
We know that $d+1$ vectors in $\mathbb{R}^d$ are always linearly dependent, use this how can I prove that $d+2$ vectors in $\mathbb{R}^d$ are always affinely dependent?
Let $p_1,\dots,p_{d+2}$ be $d+2$ points in $\mathbb R^d$. Consider the $d+1$ vectors $p_1-p_{d+2}, p_2-p_{d+2},\dots, p_{d+1}-p_{d+2}$ in $\mathbb R^d$. Then using the fact that these vectors must be linearly dependent, do you see how to construct scalars $\alpha_1,\dots,\alpha_{d+2}$ not all zero such that $$ \alpha_1p_1+\cdots+\alpha_{d+2}p_{d+2} = 0, \\ \alpha_1+\cdots+\alpha_{d+2} = 0? $$ Let me know if you would like extra help. In general, it is true that points $p_0,\dots,p_k$ in any affine space $A$ are affinely independent if and only if the vectors $p_1-p_0,p_2-p_0,\dots,p_k-p_0$ are linearly independent vectors in the tangent space $V$ of $A$.
To prove that $d+2$ vectors in $\mathbb{R}^d$ are always affinely dependent, we'll use the fact that $d+1$ vectors in $\mathbb{R}^d$ are always linearly dependent.
Linear dependence: A set of vectors $\{v_1, v_2, \dots, v_k\}$ is linearly dependent if there exist scalars $\beta_1, \beta_2, \dots, \beta_k$, not all zero, such that:
$$\beta_1 v_1 + \beta_2 v_2 + \dots + \beta_k v_k = 0.$$
Affine dependence: A set of points $\{p_1, p_2, \dots, p_k\}$ is affinely dependent if there exist scalars $\alpha_1, \alpha_2, \dots, \alpha_k$, not all zero, such that:
$$\alpha_1 p_1 + \alpha_2 p_2 + \dots + \alpha_k p_k = 0 \quad \text{and} \quad \alpha_1 + \alpha_2 + \dots + \alpha_k = 0.$$
Consider $d+2$ points $p_1, p_2, \dots, p_{d+2}$ in $\mathbb{R}^d$. To prove that these points are affinely dependent, we need to show that there exist scalars $\alpha_1, \alpha_2, \dots, \alpha_{d+2}$ , not all zero, such that:
$$\alpha_1 p_1 + \alpha_2 p_2 + \dots + \alpha_{d+2} p_{d+2} = 0 \quad \text{and} \quad \alpha_1 + \alpha_2 + \dots + \alpha_{d+2} = 0.$$
Fix one of the points, say $p_1$ , and consider the vectors $v_i = p_i - p_1$ for $i = 2, 3, \dots, d+2$ . These vectors $v_2, v_3, \dots, v_{d+2}$ are $d+1$ vectors in $\mathbb{R}^d$.
Since $d+1$ vectors in $\mathbb{R}^d$ are linearly dependent, there exist scalars $\beta_2, \beta_3, \dots, \beta_{d+2}$ , not all zero, such that:
$$\beta_2 v_2 + \beta_3 v_3 + \dots + \beta_{d+2} v_{d+2} = 0.$$
Substituting $v_i = p_i - p_1$ , we have:
$$\beta_2 (p_2 - p_1) + \beta_3 (p_3 - p_1) + \dots + \beta_{d+2} (p_{d+2} - p_1) = 0.$$
Expanding this, we get:
$$\beta_2 p_2 + \beta_3 p_3 + \dots + \beta_{d+2} p_{d+2} - (\beta_2 + \beta_3 + \dots + \beta_{d+2}) p_1 = 0.$$
Let $\alpha_1 = -(\beta_2 + \beta_3 + \dots + \beta_{d+2})$ and $\alpha_i = \beta_i$ for $i = 2, 3, \dots, d+2$ . Then the above equation becomes:
$$\alpha_1 p_1 + \alpha_2 p_2 + \dots + \alpha_{d+2} p_{d+2} = 0.$$
Also, by construction:
$$\alpha_1 + \alpha_2 + \dots + \alpha_{d+2} = -(\beta_2 + \beta_3 + \dots + \beta_{d+2}) + \beta_2 + \beta_3 + \dots + \beta_{d+2} = 0.$$
Since not all $\beta_i$ are zero, not all $\alpha_i$ are zero. Therefore, the points $p_1, p_2, \dots, p_{d+2}$ are affinely dependent.
Define the $d \times (d+2)$ matrix by columns $$ P = \pmatrix{p_1|\cdots |p_{d+2}}.$$ One trick is to add a row of ones, obtaining the following $(d+1)\times(d+2)$ matrix: $$\hat P = \pmatrix{1 \\ P}$$ (Here $1$ stands for the row vector $(1,\dots,1).$ Now, the system $$ \hat P \pmatrix{x_1 \\ \vdots \\ x_{d+2}} = 0.$$ has more unknowns than equations, so it has a nontrivial solution $\alpha = (\alpha_1|\cdots|\alpha_{d+2})^T.$ In general these larger matrices are often helpful in affine spaces to reduce problems to the linear case.
