A $2\times 2$ linear matrix transformation is conformal if and only if $c=-b ,d=a$ and $a,b \neq 0$

Question

Let $T:\mathbb{R}^{2} \to \mathbb{R}^{2}$ a linear transformation represented by the matrix $A=\begin{pmatrix} a & c \\b&d \end{pmatrix}$ . Show that $T$ is conformal if and only if $c=-b ,d=a$ and $a,b \neq 0$ i.e $A=\begin{pmatrix} a & -b \\b&a \end{pmatrix}$ .

My ideas to prove this are:

(1) If $T$ is non injective then $dim(N(T)) \geq 1$ but by the dimension theorem as $dim(V)=2$ then $dim(Im(T)) =1 or 0$

$(2)$ Let $\bar{v} \in \mathbb{R}^2$ If $\beta= {v_{1},v_{2}}$ basis of $\mathbb{R}^2 \Rightarrow \bar{v}=a_{1}v_{1}+a_{2}v_{2}$ $T(\bar{v})=a_{1}T(v_{1})+a_{2}T(v_{2}) $

Can anyone help me out giving a formal proof of the original equivalence statement ? Thanks

(3) And if $T$ is conformal then T is the composition of a rotation and a homothetic transformation

Answer 1

$T$ is conformal i.e there exists $h$: $,<T(x),T(y)> =h<x,y>$. Let $u,v$ be the canonical orhtonormal basis in which is written the matrix of $T$. You have $<T(u),T(v)>=h<u,v>=0 =ac+bd=0$, you also have $<T(u),T(u)>=<T(v),T(v)>=a^2+b^2=c^2+d^2 =h$.

One coefficient is not null for example $c$.

You have $a={{-bd}\over c}$ thus $a^2+b^2 = {{b^2d^2}\over {c^2}}+b^2 = b^2(d^2+c^2)/c^2=(b^2/c^2)h=h$. Thus $b^2=c^2$. We deduce $b=c$, or $b=-c$.

If $b=c$, $ac+bd=ac+cd=c(a+d)=0$, $a=-d$. If $b=-c, a=d$.

Answer 2

If $A=\left(\begin{array}{cc} a & b \\ c & d \end{array} \right)$ is a homothetic transformation then for $X=(x,y)\in\mathbb{R}^2$ it exist $\alpha \in \mathbb{R}^*$ such that $AX=\alpha X$, it mean that : $$ AX=\left(\begin{array}{cc} a & b \\ c & d \end{array} \right)\left(\begin{array}{c} x\\ y \end{array} \right)=\left(\begin{array}{cc} ax+ by \\ cx+dy \end{array} \right)=\alpha X=\left(\begin{array}{cc} \alpha x \\ \alpha y \end{array} \right) $$ so $b=c=0$ and $a=d=\alpha$. $$ A=\left(\begin{array}{cc} \alpha &0\\ 0 & \alpha \end{array} \right) \qquad \textrm{With } \alpha\in\mathbb{R}^* $$

and If now $B=\left(\begin{array}{cc} a & b \\ c & d \end{array} \right)$ is a rotation transformation then $B^* B =I_2$ and $det(B)=1$ , it mean that : $$ I_2= \left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)=B^*B =\left(\begin{array}{cc} a & c \\ b & d \end{array} \right)\left(\begin{array}{cc} a & b \\ c & d \end{array} \right)=\left(\begin{array}{cc} |a|^2+|c|^2 & a b+ c d \\ ab+cd & |b|^2+|d|^2 \end{array} \right) $$ so $a=\cos(t)$ and $c=\sin(t)$ and $b=\cos(t')$ and $d=\sin(t')$. So : $$ I_2=\left(\begin{array}{cc} 1 & 0 \\ 0 & 1 \end{array} \right)=\left(\begin{array}{cc} 1 & \cos(t-t') \\ \cos(t-t') & 1 \end{array} \right) $$ So $t'\equiv t+\frac{\pi}{2}[2\pi]$ it mean that $$ \left\{ \begin{array}{l} a=\cos(t)\\ b=-\sin(t)\\ c=\sin(t)\\ d=\cos(t)\\ \end{array} \right. $$ So $$ B=\left(\begin{array}{cc} a & b\\ -b & a \end{array} \right) \qquad \textrm{With } a^2+b^2=1 $$ So for your solution is a product of these matrices.