Linear Transformation $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $T^3=T$, $T^2 \neq T$, $T^2\neq I$

Question

I need to show that the linear transformation $T:\mathbb{R}^3 \rightarrow \mathbb{R}^3$, $T^3=T$, $T^2 \neq T$, $T^2\neq I$ is such that

(1) $\text{null}\,T=1$ or $2$

(2) $\text{null}\,T=2$ $\Rightarrow$ (using the Characteristic polynomial $\chi_T(x)\,)$ there is a basis for $\mathbb{R}^3$ such that $$M(T)= \begin{pmatrix} 0&0&0\\ 0&0&0\\ 0&0&-1\\ \end{pmatrix} $$

(1) Since $T^3=T$, we have $T(T^2-I)=0$, which gives $\text{Im}(T^2-I)\subseteq \text{ker}\,T $and since $T^2\neq I$, this says that the kernel of $T^2-I$ is not $\mathbb{R}^3$ and so by rank-nullity $\text{rank}(T^2-I)\geq1$, and so $\text{null}\,T\geq1$.

Also if $\text{null}\,T=3$, then $T=0 \Rightarrow T^2=T$ which is a contradicition.

So $\text{null}\,T=1$ or $2$.

(2) $\text{null}\,T=2$ means $\text{dim}\,E_0=\text{dim ker}\,T=2$.

Suppose $\lambda\neq 0$ is an eigenvalue of $T$. Then $\exists v\neq 0$ such that $T^3v=Tv \iff \lambda^2v=v \iff \lambda=1,-1$. $\color{red}{This}$ $\color{red}{is}$ $\color{red}{where}$ $\color{red}{I}$ $\color{red}{am}$ $\color{red}{stuck}$

I want to say that $\chi_T(x)=x^2,x^3, x^2(x-1),$ or $x^2(x+1)$ and I can see that if it is $x^2(x+1)$ the basis can be found easily using the fact $\mathbb{R}^3=E_0\oplus E_{-1}$. But I do not know how to rule out the other three possibilities.

Any hints would be great!

Answer 1

If the dimension of the nullspace is $2$, then the algebraic multiplicity of the eigenvalue $\lambda=0$ is either $2$ or $3$.

Let $v_1,v_2$ be a basis for the nullspace, and let $v_3$ be any vector not in the nullspace. Then $v_1,v_2,v_3$ form a basis for $\mathbb{R}^3$, and the coordinate matrix of $T$ relative to this basis will be of the form $$\left(\begin{array}{ccc} 0 & 0& a\\ 0 & 0 &b\\ 0 & 0 &c \end{array}\right)$$ for some real numbers $a$, $b$, and $c$. The characteristic polynomial of this matrix is $x^2(x-c)$; I claim that $c\neq 0$.

Indeed, if $c=0$, then $T(v_3)=av_1+bv_2$, so $T^2(v_3) = aT(v_1)+bT(v_2) = a\mathbf{0}+b\mathbf{0}=\mathbf{0}$. That means that $T^2(v_i)=\mathbf{0}$, so $T^2$ is the zero linear transformation. That means $T=T^3$ is also the zero linear transformation, which implies $T=T^2$.

So $c\neq 0$.

Tha means the algebraic multiplicity of $\lambda=0$ is two, then the characteristic polynomial must be of the form $x^2(x-c)$ for some nonzero $c$; and the algebraic and geometric multiplicities of the eigenvalue $0$ match, so the matrix is diagonalizable.

You've already determined that in this case, you must have $c=1$ or $c=-1$.

If $c=1$, then you can find a basis $v_1,v_2,v_3$ where $T(v_1)=T(v_1)=\mathbf{0}$, and $T(v_3)=v_3$. But then, how do $T$ and $T^2$ compare?

What can you conclude?