# Vectors transformation

Give a necessary and sufficient condition ("if and only if") for when three vectors $a, b, c, \in \mathbb{R^2}$ can be transformed to unit length vectors by a single affine transformation.

This is just a bonus question that the teacher gave us but i dont know even know whats it talking about or how to go about it

-
When you say affine do you mean linear or do you mean affine as in linear plus a shift? – Jim Feb 4 '13 at 16:56

(I assume you mean an invertible affine transformation, otherwise it is trivial. The affine mapping $\phi(x) = (1,0)^T$ maps any three vectors into unit length.)

$a_i$, $i \in \{1,2,3\}$ can be transformed by an invertible affine transformation into unit vectors iff either (i) $a_i$ are affinely independent (ie, not collinear) or (ii) at least two of the vectors are the same.

($\Leftarrow$): If $a_i$ are linearly independent, then $\binom{1}{a_i}$ are linearly independent, and so the matrix $\overline{A} = \begin{bmatrix} 1 & 1 & 1 \\ a_1 & a_2 & a_3 \end{bmatrix}$ is invertible. Now choose any three $x_i \in \mathbb{R}^2$, and similarly form $\overline{X} = \begin{bmatrix} 1 & 1 & 1 \\ x_1 & x_2 & x_3 \end{bmatrix}$. Then consider the affine transformation $\phi(x) = [\overline{X}\,\overline{A}^{-1} \binom{1}{x}]_{2,...,n+1}$, where $[\cdot]_{2,...,n+1}$ means all components except the first. By construction $\phi(a_i) = x_i$, hence we can choose the $x_i$ to lie on the unit circle.

If all the vectors are the same, the invertible mapping $\phi(x) = x-a+(1,0)^T$ will suffice, if $a=b$ and $a\neq c$, then the invertible mapping $\phi(x) = \frac{2}{\|a-c\|}(x-\frac{1}{2}(a+c))$ will do.

($\Rightarrow$): Suppose the vectors are all distinct and affinely dependent, and let $\phi$ be an invertible affine transformation that maps the $a_i$ to unit vectors. Since the vectors are distinct and affinely dependent, they line on a line. Since $\phi$ is affine, the points $\phi(a_i)$ also lie on a line. Since a line intersects the unit circle in at most two places, and $\phi(a_i)$ all have unit norm, then at least two vectors map to the same point, which contradicts invertibility of $\phi$.

-

Assume that $a,b,c$ are different. (If not, they will lie on a circle and that circle could be affine transformed to the unit circle around the origo.)

Prop.: $a,b,c\in\Bbb R^2$ are not collinear iff there is an affine transformation $\phi(x)=Mx+v$ such that $\phi(a),\phi(b),\phi(c)$ are unit vectors.

$\Rightarrow$: If $a,b,c$ points on plane are not collinear, then there is a circle that contains them, let's call its center $q$, and its radius $\rho$. Then the affine transformation $\ x\mapsto x-q\$ followed by $\ x\mapsto x/\rho$ will map each $a,b,c$ to the unit circle.

$\Leftarrow$: If $a,b,c$ are collinear, then they will stay collinear under any affine transformation, hence they are not all going to fit the unit circle for sure, as a line can meet a circle at most in $2$ points.

-
So $q$ is called the circumcenter, right? You might want to consider cases where $a=b=c$ or $a=b$, etc... We can still find an affine transformation that does the job then. – 1015 Feb 4 '13 at 17:16
Yes, we only need a circle that contains all $a,b,c$. And the only case there not exist such is when they are $3$ different points on a line. – Berci Feb 5 '13 at 10:27