1
$\begingroup$

We are considering a transformation $f: E^2 \rightarrow E^2$. I need to determine if f is linear and if so, I need to describe its null space and range, and compute its nullity and rank.

$f$ is defined such that it maps each point $(x,y)$ onto its reflection with respect to a fixed line through the origin.

I think I know how to show that is a linear transformation, however, I am having trouble expressing $f(x,y)=\space ?$

Can someone just express for me $f(x,y)$ and explain to me why it has this expression. I can continue from there.

Thank you

$\endgroup$
  • $\begingroup$ Hint: choose co-ordinates carefully. $\endgroup$ – Chris Eagle Feb 16 '13 at 18:32
  • $\begingroup$ If it is linear, you need only work out the behaviour on a basis. Try $(1,0)^T, (0,1)^T$. $\endgroup$ – copper.hat Feb 16 '13 at 18:32
  • $\begingroup$ I don't understand your notation. What is $(1,0)^T$ and $(0,1)^T$ $\endgroup$ – user43418 Feb 16 '13 at 18:37
  • $\begingroup$ @user43418, those are column vectors... $\endgroup$ – DonAntonio Feb 16 '13 at 18:43
  • $\begingroup$ Oh ok so its f(x,y)=(0,1) right ? $\endgroup$ – user43418 Feb 16 '13 at 18:43
3
$\begingroup$

Let $r$ be the line whith respect to which you're making the reflection. Simple choose a non-zero vector $v$ that spawns $r$ (for example, if $r$ is given by some equation $r=\left\{(x,y)\in E^2: ax+by=0\right\}$, with $a^2+b^2\neq 0$, then you can choose $v=(-b,a)$)

Now you should be convinced that:

(1) The points of $r$ are fixed by $f$, so $f(v)=v$.

(2) if $w$ is perpendicular to $r$, then $f(w)=-w$.

Now, let $v=(a,b)$, a non-zero vector spanning $r$, and $w=(-b,a)$, which is perpendicular to $v$, and hence is perpendicular to $r$. From (1) and (2) above, and by linearity of $f$, you have the system of equations: \begin{equation} \begin{cases} af(1,0)+bf(0,1)=(a,b)\\ -bf(1,0)+af(0,1)=(b,-a) \end{cases} \end{equation}

It's easy to solve that system for $f(1,0)$ and $f(0,1)$, giving you the solutions $f(1,0)=\dfrac{(a-b,b+a)}{a^2+b^2} $and $f(0,1)=\dfrac{(a+b,b-a)}{a^2+b^2}$.

Now to have a formula for $f$, you can use linearity again and expand $f(x,y)=xf(1,0)+yf(0,1)$.

$\endgroup$
  • $\begingroup$ It should be f(x,y)=xf(1,0)+yf(0,1) no? $\endgroup$ – user43418 Feb 16 '13 at 20:08
  • $\begingroup$ So we get: $f(x,y)=\frac{((a-b)x+(a+b)y,(a+b)x+(b-a)y)}{a^2+b^2}$ $\endgroup$ – user43418 Feb 16 '13 at 20:17
  • $\begingroup$ the expression is correct right ? $\endgroup$ – user43418 Feb 16 '13 at 20:35
  • $\begingroup$ In addition to that, I just wanted to check the answers for the final question. I foud Ker(f)={(0,0)}, the range is $\mathbb{R}^2$, the nullity and rank of T are both equal to 2 $\endgroup$ – user43418 Feb 16 '13 at 20:41
  • $\begingroup$ The nullity is $dim(Ker(f))=0$ and the rank is $dim(f(\mathbb{R}^2))=dim\mathbb{R}^2=2$... $\endgroup$ – Luiz Cordeiro Feb 16 '13 at 20:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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