1
$\begingroup$

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

$\endgroup$
3
$\begingroup$

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

$\endgroup$
7
$\begingroup$

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!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.