Give an example of a $T\in\mathcal L\left(\mathbb R^2\right)$ s. t. $Ker(T) = Im(T)$.
MY APPROACH
According to the rank-nullity theorem, $\dim Ker(T) = \operatorname{rank}(T) = 1$.
Considering that $\text{span}\{v\} = Ker(T) = Im(T)$, one has that $T(v) = 0$ and $T(w) = v$, for some $w = (c,d)\in\mathbb R^{2}\setminus\{(0,0)\}$.
Consequently, we have the following system of equations to solve
\begin{align*}
\begin{cases}
T(v) = aT(1,0) + bT(0,1) = (0,0)\\\\
T(w) = cT(1,0) + dT(0,1) = (a,b)
\end{cases}
\end{align*}
whose solution is given by
\begin{align*}
\begin{cases}
\displaystyle T(1,0) = -\left(\frac{ab}{ad-bc},\frac{b^{2}}{ad-bc}\right)\\\\
\displaystyle T(0,1) = \left(\frac{a^{2}}{ad-bc},\frac{ab}{ad-bc}\right)
\end{cases}
\end{align*}
For a particular case, it suffices to consider $v = (1,0)$ and $w = (0,1)$, from whence we get
\begin{align*}
T(x,y) = xT(1,0) + yT(0,1) = x(0,0) + y(1,0) = (y,0)
\end{align*}
It just remains to double-check the proposed answer.
In fact, it solves the problem.
This is because $Ker(T) = \text{span}\{(1,0)\}$ and $Im(T) = \text{span}(\{T(1,0),T(0,1)\}) = \text{span}\{(1,0)\}$.
Could someone verify my reasoning or provide another approach to solve it?
Best Answer
The rank-nullity theorem tells you that the sum of the dimensions of the range and nullspace (also called the kernel) must equal the dimension of the domain, i.e. 2. Thus in your example, you are looking for 1D subspaces of $\mathbb{R}$. You might as well just take that be the $x$-axis, i.e. $span (1,0)$. So we want a matrix that has range and kernel of the $x$-axis. We just define it on that basis $(1,0),(0,1)$. We want: $$ M\begin{pmatrix} 1 \\ 0 \end{pmatrix} = 0$$ so we have that the first column of $M$ must be $\begin{pmatrix} 0 \\ 0 \end{pmatrix}$. Then we want to take $(0,1)$ to $(1,0)$ so the forces the second column of $M$ to be $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$. Thus we take: $$M = \begin{pmatrix} 0 & 1\\ 0 & 0 \end{pmatrix}.$$