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- Apr 14, 2013

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Let $1\leq n\in \mathbb{N}$ and $(p_0,\ldots , p_n)$ an affine basis (that means that the vectors $p_1-p_0, p_2-p_0,\ldots ,p_n-p_0$ build a basis of $\mathbb{R}^n$.

(a) Give a geometric description of affine bases of $\mathbb{R}^n$ for $1\leq n\leq 3$.

(b) For all $v\in \mathbb{R}^n$ show that $(p_0+v,\ldots , p_n+v)$ is an affine basis of $\mathbb{R}^n$.

(c) Let $a$ be an invertible matrix. Then show that $(ap_0,\ldots , ap_n)$ is an affine basis of $\mathbb{R}^n$.

(d) For each isometry $\beta\in \text{Isom}(\mathbb{R}^n)$ show that $(\beta (p_0),\ldots , \beta( p_n))$ an affine basis of $\mathbb{R}^n$.

(e) Let $p_0=0$ and $\beta\in \text{Isom}(\mathbb{R}^n)$, with $\beta (p_i)=p_i$ for all $0\leq i\leq n$. Then show that $\beta=\text{id}_{\mathbb{R}^n}$.

(f) Let $\beta\in \text{Isom}(\mathbb{R}^n)$, with $\beta (p_i)=p_i$ for all $0\leq i\leq n$. Then show that $\beta=\text{id}_{\mathbb{R}^n}$.

For (b) I have done the following :

We have that the vectors \begin{align*}&(p_1+v)-(p_0-v)=p_1+v-p_0-v=p_1-p_0 \\ &(p_2+v)-(p_0-v)=p_2+v-p_0-v=p_2-p_0 \\ &\ldots \\ &(p_n+v)-(p_0-v)=p_n+v-p_0-v=p_n-p_0 \end{align*}

build a basis of $\mathbb{R}^n$, since $(p_0, p_1, \ldots ,p_n)$ is an affine basis of $\mathbb{R}^n$.

Therefore $(p_0+v, p_1+v, \ldots ,p_n+v)$ is an affine basis of $\mathbb{R}^n$.

For (c) I have done the following :

We have that \begin{align*}&ap_1-ap_0=a(p_1-p_0) \\ &ap_2-ap_0=a(p_2-p_0) \\ &\ldots \\ &ap_n-ap_0=a(p_n-p_0) \end{align*}

We have that \begin{align*}\lambda_1[a(p_1-p_0)]+\lambda_2[a(p_2-p_0)]+\cdots +\lambda_n[a(p_n-p_0)]=0 &\Rightarrow a\left (\lambda_1(p_1-p_0)+\lambda_2(p_2-p_0)+\cdots +\lambda_n(p_n-p_0)\right )=0 \ \\ & \overset{a\in \text{GL}_n(\mathbb{R})}{\Rightarrow } \ a^{-1}a\left (\lambda_1(p_1-p_0)+\lambda_2(p_2-p_0)+\cdots +\lambda_n(p_n-p_0)\right )=a^{-1}\cdot 0 \\ & \Rightarrow \lambda_1(p_1-p_0)+\lambda_2(p_2-p_0)+\cdots +\lambda_n(p_n-p_0)=0\end{align*}

Since $(p_0, p_1, \ldots ,p_n)$ is an affine basis of $\mathbb{R}^n$ the $p_1-p_0, \ p_2-p_0, \ \ldots , \ p_n-p_0$ build a basis of $\mathbb{R}^n$ and so they are linearly independent, and so $\lambda_1=\lambda_2=\ldots =\lambda_n=0$.

So the $n$ linearly independent vectors $a(p_1-p_0), \ a(p_2-p_0), \ \ldots , \ a(p_n-p_0)$ in the $n$-dimensional space are a basis. So $(p_0+v, p_1+v, \ldots ,p_n+v)$ is anaffine basis of $\mathbb{R}^n$.

Is that correct so far?

Could you give me a hint for he remaining parts? For example for the isometries the distances are preserved but canwe say that the differences of $\beta(p_i)-\beta(p_0)$ are equal to $p_i-p_0$ ?