Linear transformations satisfying $T^3=T$ and $T^2\neq I$ 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.
 A: 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 $b=0$.
A: How about $T=\begin{pmatrix}&1&\\1&&\\&&&\end{pmatrix}$?
A: Another solution :
$$T_0=compan(\underbrace{[1,\color{red}{0,-1,0}]}_{\text{coeff. of } P})=\begin{pmatrix}0&0&\color{red}{0}\\\color{blue}{1}&0&\color{red}{1}\\0&\color{blue}{1}&\color{red}{0}\end{pmatrix}$$
obtained by taking the so-called companion matrix to the polynomial equation:
$$P(x)=x^3+0x^2-x+0=0$$
verified by matrix $T_0$.
The companion matrix is obtained by replacing into the zero matrix

*

*the first subdiagonal by $\color{blue}{1}$s.


*the last column by the (non dominant) opposites $\color{red}{0,1,0}$ of coefficients of polynomial $P$.
Remark: the transpose $T_0^T$ and the opposite $-T_0$ of $T_0$ are other solutions.

Yet other solutions:
$$T_1=\begin{pmatrix}0&1&0\\1&0&1\\0&0&0\end{pmatrix}$$
and its transpose... and its opposite.

By the theory of the companion matrix (recalled in the answer by @Q. Zhang below), one can say that the general solution to your issue has the form:
$$T=PT_0P^{-1}$$
where $P$ is any invertible matrix.
A: The condition says that the minimal polynomial of $T$ is exactly $x^3-x$. Since $dim(R^3)=3$, the characteristic polynomial is also $x^3-x$. The cyclic decomposition theorem says that such $T$ is unique and $T$ has a cyclic vector $\alpha$. If we take $\mathcal{B}=\{\alpha, T\alpha, T^2\alpha\}$ as an ordered basis, then $[T]_{\mathcal{B}}$ is just the companion matrix, namely,
$$\begin{bmatrix}0&0&0\\ 1&0&1\\ 0&1&0 \end{bmatrix}.$$
