About linear transformation T with dim imT =1

linear algebra

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

Best Answer

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 $[T + \text{Id}] = uv^\top + I$.

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