Find the rank of given Linear Transformation

Question

Let $T : R^2 \to R^2$ be a Linear Transformation such that $T(x,y) = 0$ when $ x^2+y^2 = 1$. Find Rank of T

I am really confused on this problem because The Null space of $T$ is given by :

$N(T) = (x,y:x^2 + y^2 = 1)$ and clearly this does Not form a vector space as $(0,0) \notin N(T)$ .So I don't think such a transformation can exist, and so rank(T) Cannot be computed.

any Ideas on how to solve this ?

Answer 1

$T(x,y)=(0,0)$ for every $(x,y)$ since $(0,1)$ and $(1,0)$ are elements of the unit circle so the rank is zero.

Answer 2

The problem is stated in a confusing matter. The only sensible interpretation I see is the following: $\{(x,y)|x^2+y^2=1\}\subset N(T)$, i.e. if $x^2+y^2=1$, then $T(x,y)=0$, not if and only if.

Then as you noted $N(T)$ is a subspace of $\mathbb R^2$ containing the unit circle, so $N(T)=\mathbb R^2$ and rank of $T$: $r(T)=0$

Answer 3

If $(x,y)\neq(0,0)$, then\begin{align}T(x,y)&=T\left(\sqrt{x^2+y^2}\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right)\right)\\&=\sqrt{x^2+y^2}T\left(\frac x{\sqrt{x^2+y^2}},\frac y{\sqrt{x^2+y^2}}\right)\\&=(0,0).\end{align}So, $T$ is the null map.

Answer 4

Let $\|(x,y)\| = \sqrt{x^2+y^2}$ be the euclidean norm on $\mathbb{R}^2$. Then for every $(x,y) \ne (0,0)$ in $\mathbb{R}^2$ we have

$$T(x,y) = \|(x,y)\|T\left(\frac1{\|(x,y)\|}(x,y)\right) = 0$$ since $\left\|\frac1{\|(x,y)\|}(x,y)\right\| = 1$. Therefore $T = 0$.