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 $S: \mathbb{R}^3 \rightarrow \mathbb{R}^3$


2 Answers 2


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!

  • $\begingroup$ I am trying to explain why the image is the kernel. Is this a valid explanation. The idea of the kernel is still confusing to me. Our image is simply the x-axis and then since ker$(T):=\{X\in V:T(X)=0\}$ we have that ker$(T)$ is when $y$ is zero, exactly our image? $\endgroup$ Commented Oct 14, 2020 at 18:59
  • 1
    $\begingroup$ yes, this is valid! $\endgroup$
    – amakelov
    Commented Oct 15, 2020 at 10:28

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 $$\begin{pmatrix} 0&1\\ 0&0 \end{pmatrix}.$$


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