# Linear Transformation with kernel equal to image

How can we construct a linear transformation $T : \mathbb{R}^2 \rightarrow \mathbb{R}^2$ such that $Ker(T) = Im(T)$. How about the same for a linear transformation $S: \mathbb{R}^3 \rightarrow \mathbb{R}^3$

Amakelov's answer pretty much answers everything, but let's determine all transformations $$T:\mathbb{R}^2\rightarrow \mathbb{R}^2$$ such that $$\ker(T)=\text{im}(T)$$.

Suppose you have such a transformation $$T$$. By the rank-nullity theorem you have that $$\dim(\ker(T))=1=\dim(\text{im}(T))$$. Take a basis $$\left\{v\right\}$$ of $$\text{im}(T)$$. Then $$v\in \text{im}(T)=\ker(T)$$. Hence $$T(v)=0$$. Moreover, since $$v\in \text{im}(T)$$, there exists a $$w\in \mathbb{R}^2$$ such that $$T(w)=v$$.

You can show that $$\beta=\left\{v,w\right\}$$ forms a basis of $$\mathbb{R}^2$$ (why?). Hence with respect to this basis we have that the matrix of the linear map $$T$$ is given by $$m(T)_{\beta,\beta}=\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}.$$

Conversely, any transformation defined by first fixing a basis and then using the above matrix yields such a linear transformation, henced we determined all of them.

There is a shorter method in this case. From the rank-nullity theorem we see that $$T^2=0$$. Hence $$0$$ is the only eigenvalue of $$T$$ but the matrix is not diagonalizable as $$\dim(\ker(T))\neq 2$$. Therefore, the Jordan canonical form of $$T$$ is given by the nilpotent matrix $$\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}.$$

In general, the rank-nullity theorem tells us that the dimensions of the kernel and image sum to the dimension of the domain of a linear transformation. In particular, there's no linear transformation $\mathbb{R}^3\to\mathbb{R}^3$ which has the same dimensions of the image and kernel, because $3$ is odd; and more particularly this means the second part of your question is impossible.

For $\mathbb{R}^2\to\mathbb{R}^2$, we can consider the following linear map: $(x,y)\mapsto (y,0)$. Then the image is equal to the kernel!