# 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?

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

$$S$$ 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 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 $$\mathbb C^n$$.