Let $T: V \rightarrow V$ be a linear map, where $nullity(T) = dim(V) - 1$. Prove there is a $\lambda$ such that $T^{2}(v) = \lambda T(v)$. Let $T: V \rightarrow V$ be a linear map, where $nullity(T) = dim(V) - 1$. 
Let $w$ be a vector from the image of $T$. If $T(w) \neq 0$, prove there is a non-zero number $\lambda$ such that $T^{2}(w) = \lambda T(w)$.
Prove that $T^{2}(v) = \lambda T(v)$ for all $v \in V$.
State any results you use.
Edit: Forgot to mention, $V$ is a vector space over $\mathbb{R}$.
I started this by using the rank-nullity theorem to find that $rank(T) = 1$, but I'm not quite sure where to go from there. I understand the rank of a linear map to be sort of the "degrees of freedom", so the image only having one dimension means that any vector $w$ that is a result of putting a vector $v$ into the linear map (in the image of T) can be represented by just one number, so if this vector $w$ is non-zero then applying T again will just change this one number by a factor of $\lambda$.
That's the general idea I get from this question, however I'm not sure how to eloquently and rigourously prove it.
 A: We are assuming that $V$ is a finite-dimensional vector space over a field $F$. The presence of the linear map $T$ with $\mathrm{nullity}(V)=\dim\ker T = \dim(V)-1$ testifies that $\dim V\geq 1$.
The image of $T$, $\mathrm{im}\mspace{2mu}T = T\mspace{1mu}V$, is a one-dimensional subspace $Fa$ of $V$, where $a$ is a nonzero vector.
If $v$ is any vector in $V$, then $Tv=\varphi(v)a$ for a unique $\varphi(v)\in F\,$; this defines the mapping $\varphi\colon V\to F$ which is easily seen to be a linear functional (because $T$ is a linear transformation).
Now, again for an arbitrary $v\in V$, we have
$$
T^2v = T(Tv) = T(\varphi(v)a) = \varphi(v)Ta=\varphi(v)\varphi(a)a=\varphi(a)\varphi(v)a=\varphi(a)Tv~,
$$
therefore $T^2v=\lambda\mspace{2mu} Tv$ with $\lambda=\varphi(a)$. Done.
Remark. Commutativity of $F$ is essential. The reasoning above breaks down, near to its end,
if $F$ is a non-commutative division ring.
A: $\DeclareMathOperator{im}{im}\DeclareMathOperator{rank}{rank}$
We have $\rank(T) = 1$ and $n = \dim(V)$. This means 
$$
T 
= 
A 
\left(
\begin{array}{r}
t_1 & \dots & t_n \\
0 & \dots & 0 \\
\vdots & & \vdots \\
0 & \dots & 0 \\
\end{array}
\right)
= (t_1 a_1, t_2 a_1, \dotsc, t_n a_1) \\
$$
for some vector $t \in V$, $t \ne 0$ and some invertible $n \times n$ matrix 
$A = (a_1, \dotsc, a_n)$, where the $a_i$ are the colum vectors of $A$.
The matrix $A$ is the inverse of the product of the elementary matrices representing the transformations used by the Gauss elimination to reach the row echelon form.
So for any $v \in V$ we have
$$
T v 
= (t_1 a_1, t_2 a_1, \dotsc, t_n a_1) v
= (t^\top v) \, a_1
$$
which lies on the line $\alpha a_1$, $\alpha \in \mathbb{R}$ and
where $\top$ means matrix transposition. 
Further because $T$ is linear we have
\begin{align}
T^2 v 
&= T(Tv) \\
&= T((t^\top v) a_1) \\
&= (t^\top v) \, T(a_1) \quad (*) \\
&= (t^\top v) (t^\top a_1) a_1 \\
&= (t^\top a_1) (t^\top v) a_1 \\
&= (t^\top a_1) T(v) \\
&= \lambda T(v)
\end{align}
with $\lambda = t^\top a_1$. We have $t\ne 0$ because of the rank, and $a_1 \ne 0$ because $A$ is invertible.
What is left is an argument, why $t$ and $a_1$ are not orthogonal:
Suppose $t^\top a_1 = 0$ then $T(a_1) = (t^\top a_1) a_1 = 0$. Equation $(*)$ would then imply that $T^2 = 0$. 
However the task assumes there is a $w \in \im(T)$, $w \ne 0$ with $T(w) \ne 0$. Because $w \in \im(T)$ there must be a $x \in V$ with $T(x) = w$, so $T^2(x) = T(w) \ne 0$ by the assumption, so $T^2 \ne 0$.
This means $t$ and $a_1$ are not orthogonal, and thus $\lambda \ne 0$.
