# Linear operator $T:\mathbb{C}^n\rightarrow \mathbb{C}^n$ of rank 1

I apologize if my question is a duplicate, but I looked for a similar one and could not find it. I was given the following question in a recent Linear Algebra exam:

Let $T:\mathbb{C}^n \rightarrow \mathbb{C}^n$ be a linear transformation of rank 1.

(a) Show that there exists $a\in \mathbb{C}$ such that $T^2=aT$;

(b) If $S:= T + I$,where $I$ denotes the identity operator, find all values of $a$ such that $S$ has an inverse.

My main problem is with item (a). Using the fact that $\dim \text{Im}\, T = 1$, it is easy to determine $\dim \ker T = 2n-1$. But I have not been able to come up with any kind of strategy to use this (if it helps at all).

Any hints or suggestions?

PS: There is no information about if $\mathbb{C}^n$ is a linear space over $\mathbb{C}$ or $\mathbb{R}$. I assume it is $\mathbb{R}$ because the exam was supposed to be about spaces over $\mathbb{R}$.

Suppose that $T^2=0$, $T^2=0.T$ done.
If $T^2\neq 0$, you know that $dimkerT+dimImT=n$, let $u\in kerT\cap Im(T), u\neq 0$ for every $x\in\mathbb{C}^n$, $T(x)=au$, this implies that $T^2(x)=T(au)=0$ contradiction since $T^2\neq 0$, we deduce that $kerT\cap ImT=0$ and $\mathbb{C}^n=kerT\oplus ImT$. Let $(e_1,...,e_{n-1})$ a basis of $kerT$ and $e_n$ a basis of $Im(T)$, $T(e_n)=ce_n$ implies that $T^2=cT$.
• That's true for $T^2 = 0$ but not necessarily for any general $T$ of rank 1. Commented May 25, 2018 at 0:44
First, notice that in order to prove $(a)$ it suffices to prove that there exists $a \in \mathbb{C}$ such that $T(w)=aw$ for all $w \in \operatorname{Im}(T)$.
Since the image is rank $1$, it consists of eigenvectors. This together with the description above solves your problem.