about linear transformation T with dim imT =1 I have the following question: 
Let $N \in \mathbb{N}$ with $N \geq 1.$
Consider $T: C^N \rightarrow C^N $  a linear transformation with $dim \ im \ T =1$. Here $im \ T$ denotes the image of $T$.  Then 
i) $T^2 = a T$ for some $a \in C$
ii) Define $S = T + Id$ where $Id$ denotes the identity matrix of order N. For wich values of $a$ the linear transformation $S$ is invertible? Find the inverse for such values of $a.$
My attempt:
i) Since $dim \ im \ T =1$ the by the rank $\&$ nullity theorem we have $dim \ ker \ T =N-1$ . Consider a basis $\{e_1,...,e_N\}$ with $T(e_i)=0$ for $i=1,...,N-1.$
From this it is possible to verify that the matrix of $T$ in the mentioned basis is
$$ [T]= \left[
\begin{array}{cccc}
0 & 0 & \cdots & 0 & a_{11}\\
0 & 0 & \cdots & 0 &a_{21}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 0 & a_{n1}\\
\end{array}
\right] $$
Therefore  $T^2 = a_{nn} .T$
ii) Using the matrix of $[T + Id]$ we have that $S$ is invertible if and only if $a_{nn} \neq -1$. But I don't how to find the inverse. Someone could help me?
Thanks in advance
 A: I think the condition for invertibility is $a \ne -1$.
Hint: Try looking for an inverse of the form $cT+\text{Id}$ for an appropriate $c \in C$ and be sure to use the result of part i).

 $$(T + \text{Id})(cT + \text{Id}) = cT^2 + (1+c)T + \text{Id} = (1+(1+a)c)T + \text{Id},$$ so choosing $c = -\frac{1}{1+a}$ works.


This is essentially a special case of the Sherman-Morrison formula, which in turn is a special case of the Woodbury matrix identity. Specifically, the matrix of your $T$ is rank $1$, and can be written as the outer product $uv^\top$ for some vectors $u$ and $v$. (You have already chosen a particular $u$ and $v$.) You then want to invert $[T + \text{Id}] = uv^\top + I$.
A: $S$ is invertible iff  $-1$ is not an eigen value of $T$. Using $T^{2}=aT$ you can easily see that this is true iff $a \neq -1$. If $a \neq -1$ we can compute $S^{-1}$ formally as $S^{-1} =I-T+T^{2}-T^{3}+...=I-\frac T {1+a}$. You can now verify that this is indeed the inverse.
A: A basis-dependent argument (generally considered less desirable than basis-free ones although they are more concrete) follows what you were noticing before:
For $\{e_i\}$, the basis you single out in your solution, notice that for $1\le i\le n-1$
$$ S(e_i)=T(e_i)+e_i= 1e_i$$
and
$$ S(e_n)=T(e_n)+e_n=(1+a_{nn})e_n+\sum_{k=1}^{n-1} a_{kn}e_k$$
so in this basis, $S$ is represented as the matrix
$$[S]=\begin{pmatrix} 1 & 0 & \cdots & a_{1n}\\
0 & 1 & \cdots & a_{2n}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & 1+a_{nn}\end{pmatrix}$$
Let $S'$ be the inverse of $S$. Then notice that if $S(e_i)=v_i$, then $S'(v_i)=e_i$. Using this, you can compute explicitly that, for $1\le i\le n-1$, $S(e_i)=e_i=S'(e_i)$. The only thing remaining is to compute is $S'(e_n)$, but to find this, notice
$$e_n=S'S(e_n)=S'\left((1+a_{nn})e_n+\sum_{k=1}^{n-1} a_{kn}e_k\right)=(1+a_{nn})S'(e_n)+\sum_{k=1}^{n-1} a_{kn}e_k$$
by the linearity of $S'$. Doing some algebra gets us that
$$(1+a_{nn})S'(e_n)=e_n+\sum_{k=1}^{n-1} a_{kn}e_k$$
and since the right hand side can't be zero (the $e_i$ form a linearly independent set) this forces $a_{nn}\ne -1$ and so we can divide through to get
$$S(e_n)=\frac{1}{1+a_{nn}}e_n+\sum_{k=1}^{n-1}\frac{a_{kn}}{1+a_{nn}}e_k$$.
Thus 
$$[S']=\begin{pmatrix} 1 & 0 & \cdots & \frac{a_{1n}}{1+a_{nn}}\\
0 & 1 & \cdots & \frac{a_{2n}}{1+a_{nn}}\\
\vdots & \vdots & \ddots & \vdots\\
0 & 0 & \cdots & \frac{1}{1+a_{nn}}\end{pmatrix}$$
And such a matrix corresponds uniquely to the linear transformation $S'$, which we have cooked up to be the inverse of $S$ on the basis $e_i$, thus is the inverse on  all of $\mathbb C^n$.
