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

  • $\begingroup$ 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

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 $f(x,y)=xf(1,0)+yf(0,1)$.

  • $\begingroup$ 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

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