0
$\begingroup$

Let $p_0,p_1,p_2,p_3$ the vertices of a tetrahedron in $R^3$ and $f$ the affine transformation such that $$f(p_0)=p_1 \hspace{3ex} f(p_1)=p_2 \hspace{3ex} f(p_2)=p_3 \hspace{3ex} f(p_3)=p_0$$

Find the matrix of $f$ with $p_0 $ as origin and $B=\{\overline{p_0p_1},\hspace{2ex}\overline{p_1p_2},\hspace{2ex}\overline{p_2p_3}\}$ as basis.

My work

$\vec{f}(\overline{p_0p_1})=\overline{f(p_0)f(p_1)}=\overline{p_1p_2}$

$\vec{f}(\overline{p_1p_2})=\overline{f(p_1)f(p_2)}=\overline{p_2p_3}$

$\vec{f}(\overline{p_2p_3})=\overline{f(p_3)f(p_0)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$

Then $$M_{B,B}(\vec{f})=\left(\begin{array}{ccc} 0&0&-1\\ 1&0&-1\\ 0&1&-1\\ \end{array}\right) $$

So the matrix is $$\left(\begin{array}{cccc} 1&0&0&0\\ ?&0&0&-1\\ ?&1&0&-1\\ ?&0&1&-1\\ \end{array}\right) $$

Question: Is my work correct? How can I find the $f(p_0) $coordinates?

$\endgroup$
1
  • $\begingroup$ Each column is the image of a basis vector. You’re given that $f(p_0)=p_1$ and $\overline{p_0p_1}=p_1-p_0$, so... $\endgroup$
    – amd
    May 30, 2019 at 0:38

1 Answer 1

1
$\begingroup$

You wrote:

$\vec{f}(\overline{p_2p_3})=\overline{f(p_3)f(p_0)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$

but this is mistaken; I think it should be

$\vec{f}(\overline{p_2p_3})=\overline{f(p_\color{red}2)f(p_\color{red}3)}=\overline{p_3p_0}=-\overline{p_0p_3}=-(\overline{p_0p_1}+\overline{p_1p_2}+\overline{p_2p_3})$


You ask "how do I find the $f(p_0)$ coordinates?"

I'm not sure exactly how your text/teacher is expressing affine xforms as matrices, so I don't know how to fill in the first column, but you do know that $f(p_0) = p_1$, so you can write $$ f(p_0) = p_1 = p_0 + (p_1 - p_0) = p_0 + b_1 $$ where $b_1$ is the first element of your basis. That makes me suspect that your matrix should be something like $$ \left(\begin{array}{cccc} 1&0&0&0\\ \color{red}1&0&0&-1\\ \color{red}0&1&0&-1\\ \color{red}0&0&1&-1\\ \end{array}\right) $$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .