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Affine basis

mathmari

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Apr 14, 2013
4,713
Hey!! :giggle:

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? :unsure:

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 $p_i-p_0$ ? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
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 $p_i-p_0$ ?
Hey mathmari !!

Every isometry in $\mathbb R^n$ with the usual metric can be written as an orthogonal transformation plus a translation.
We can use that for (d), (e), and (f), can't we? 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
Every isometry in $\mathbb R^n$ with the usual metric can be written as an orthogonal transformation plus a translation.
We can use that for (d), (e), and (f), can't we? 🤔
How do we use that? I got stuck right now. :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
How do we use that? I got stuck right now.
Use that $\beta(x)=ax+v$ where $a$ is an invertible matrix and use the same logic as in (b) and (c)? 🤔
We also have that $a^Ta=\text{id}_{\mathbb R^n}$.
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
Use that $\beta(x)=ax+v$ where $a$ is an invertible matrix and use the same logic as in (b) and (c)? 🤔
We also have that $a^Ta=\text{id}_{\mathbb R^n}$.
For (d) I have done the following :

We have that \begin{align*}&\beta (p_1)-\beta (p_0)=(ap_1+v)-(ap_0+v)=ap_1-ap_0=a(p_1-p_0) \\ &\beta (p_2)-\beta (p_0)=(ap_2+v)-(ap_0+v)=ap_2-ap_0=a(p_2-p_0) \\ &\ldots \\ &\beta (p_n)-\beta (p_0)=(ap_n+v)-(ap_0+v)=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 $\mathbb{R}^n$, it follows that $p_1-p_0, \ p_2-p_0, \ \ldots , \ p_n-p_0$ is a basis of $\mathbb{R}^n$, and so these are linearly independent, so $\lambda_1=\lambda_2=\ldots =\lambda_n=0$.
So it follows that the $n$ linearly independent vectors $a(p_1-p_0), \ a(p_2-p_0), \ \ldots , \ a(p_n-p_0)$, i.e. $\beta (p_1)-\beta (p_0), \ \beta (p_2)-\beta (p_0), \ \ldots , \ \beta (p_n)-\beta (p_0)$, in $n$-dimensional space form a basis.
Therefore $(\beta (p_0), \beta (p_1), \ldots ,\beta (p_n))$ is an affine basis of $\mathbb{R}^n$.



For (e) I have done the following :

We have that $\beta(p_i)=p_i \Rightarrow ap_i+v=p_i\Rightarrow (a-u)p_i=-v$.
How do we continue? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
We have that $\beta(p_i)=p_i \Rightarrow ap_i+v=p_i\Rightarrow (a-u)p_i=-v$.
How do we continue?
Suppose we substitute $p_0=0$? :unsure:
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
Suppose we substitute $p_0=0$? :unsure:
From $\beta(p_i)=p_i \Rightarrow ap_i+v=p_i\Rightarrow (a-u)p_i=-v$ for $i=0$ we have $p_0=0$ so we get $(a-u)\cdot 0=-v \Rightarrow v=0$. Therefore $ap_i=p_i$. Do we have to show that $a$ is the identity matrix? :unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,591
From $\beta(p_i)=p_i \Rightarrow ap_i+v=p_i\Rightarrow (a-u)p_i=-v$ for $i=0$ we have $p_0=0$ so we get $(a-u)\cdot 0=-v \Rightarrow v=0$. Therefore $ap_i=p_i$. Do we have to show that $a$ is the identity matrix?
Yep. (Nod)
 

mathmari

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Apr 14, 2013
4,713
Do we consider for that the system $(a-I)p_i=0$ ? Since the $p_i$'s cannot be zero because the form an affine basis, it follows that $a-I=0$, i.e. $a=I$.

Is that correct? :unsure:
 

Klaas van Aarsen

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Staff member
Mar 5, 2012
9,591
Do we consider for that the system $(a-I)p_i=0$ ? Since the $p_i$'s cannot be zero because the form an affine basis, it follows that $a-I=0$, i.e. $a=I$.

Is that correct?
It's not sufficient that the $p_i$ are non-zero. We actually need that the $p_i$ for $n=1,\ldots,n$ form a basis of $\mathbb R^n$.
It follows from there that $a$ must indeed be the identity matrix. 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
It's not sufficient that the $p_i$ are non-zero. We actually need that the $p_i$ for $n=1,\ldots,n$ form a basis of $\mathbb R^n$.
It follows from there that $a$ must indeed be the identity matrix. 🤔
We have that the $p_i$ for $n=1,\ldots,n$ form a basis of $\mathbb R^n$. But how does it follow that $a$ must indeed be the identity matrix? :unsure:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
9,591
We have that the $p_i$ for $n=1,\ldots,n$ form a basis of $\mathbb R^n$. But how does it follow that $a$ must indeed be the identity matrix? :unsure:
Suppose we put the $p_i$ as column vectors in a basis matrix $b$, which is invertible.
Then $ap_i=p_i\implies ab=b \implies abb^{-1}=bb^{-1}\implies a=I$. 🤔
 

mathmari

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MHB Site Helper
Apr 14, 2013
4,713
Suppose we put the $p_i$ as column vectors in a basis matrix $b$, which is invertible.
Then $ap_i=p_i\implies ab=b \implies abb^{-1}=bb^{-1}\implies a=I$. 🤔
Ahh I see!!

So at (f) we do that in a similar way:
We have that $\beta(p_i)=p_i \Rightarrow ap_i+v=p_i\Rightarrow (a-u)p_i=-v$. But how do we get that $v=0$ ? We don't have any additional condition that we could use as before that $p_0=0$.
Could you give me a hint? :unsure:
 

Klaas van Aarsen

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Mar 5, 2012
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What is $\beta(p_i-p_0)$? 🤔
 

mathmari

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Apr 14, 2013
4,713

Klaas van Aarsen

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Mar 5, 2012
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How about $\beta p_i - \beta p_0$?
And in particular for $i=0$? 🤔: