I am looking for linear transformations $T:R^3\rightarrow R^3$, such that $T^3=T$ but $T^2\neq T$ and $T^2\neq I$.
I've been playing around with this for a while, but can't see how any such transformation is possible. For example:
Take $u\in \text{Im }T$, so $u=Tv$ for some $v$.
Then $T^2u=T^3v=Tv=u$. Wouldn't this imply that we must have $T^2=I$?
I have asked my professor and been told that there are indeed transformations satisfying these conditions, and that I should just think harder about it. Can anyone help me out? I've also tried to think geometrically about what transformations could do this but am not getting anywhere with that either.
Best Answer
To your first question: You have shown $T^2u=u$ for all $u\in \operatorname{Im}(T)$. This does not imply $T^2=I$ because for this you need $T^2u=u$ for all $u\in \mathbb{R}^3$.
From $T(T^2-I)=0$ and $T^2-I \neq 0$ you get that $0$ is an Eigenvalue of $T$. Furthermore if $a$ is an Eigenvalue of $T$, we get $a^3=a$ out of $T^3=T$. Hence $T$ is similar to a matrix of the form $$ \begin{pmatrix} a&0&0\\ 0&b&0\\ 0&0&0\\ \end{pmatrix}, \begin{pmatrix} a&1&0\\ 0&a&0\\ 0&0&0\\ \end{pmatrix}, \begin{pmatrix} a&0&0\\ 0&0&1\\ 0&0&0\\ \end{pmatrix} \text{ or } \begin{pmatrix} 0&1&0\\ 0&0&1\\ 0&0&0\\ \end{pmatrix} $$ for $a,b\in \{-1,0,1\}$. The last three cases are not possible due to $T^3=T$. For the first case, one just has to check the possible 9 cases for $a,b\in \{-1,0,1\}$. An get that $T$ fulfils the properties from above, if and only if it there is a basis $\mathcal{B}$ such that the transformation matrix of $T$ in $\mathcal{B}$ is $$\begin{pmatrix} a&0&0\\ 0&b&0\\ 0&0&0\\ \end{pmatrix} $$ where $a,b=-1$, $a=1$ and $b=-1$ or $a=-1$ and $b=0$.