Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question
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 $\phi$.

share|cite|improve this answer

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.

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.