Singular linear transformation and its eigenvalue Suppose $T : \mathbb{C}^4 \to \mathbb{C}^4$ is a linear map and $\operatorname{ker}(T)\neq \{0\}$, $\dim \operatorname{Im}(T+I)=3$ (and some other additional information which isn't necessary). I had to find some eigenvalues, and i didn't understand the following statement in the solution:

Because $\dim\operatorname{ker}(T+I)=1$, the transformation $-(T+I)$ is singular and thus $(-1)$ is an eigenvalue.

I understand that $\dim\operatorname{ker}(T+I)=4-\dim \operatorname{Im}(T+I)=1$. I also understand that $(T+I)$ is singular (because $\dim\operatorname{ker}(T+I)>0$) and I also understand that $-(T+I)$ is singular. But how they immediately deduced that $-1$ is an eigenvalue?
My initial approach was:
Suppose $0 \neq v \in \ker (T+I)$, then $(T+I)(v)=T(v)+v=0$ which leads to $T(v)=(-1)v$ and thus $-1$ is an eigenvalue.
Is this correct? Is there a more elegant solution to this? Can we see it directly from the fact that $(-I-T)$ is singular like stated in the solution? And why they mentioned $(-I-T)$ instead of $(I+T)$?
 A: Looks good, and is a good way to approach eigenvalues/eigenvectors. If you have a linear map $T$ with eigenvalue $\lambda$, then there is an eigenvector $v$ such that
$$ Tv = \lambda v. $$
This means $Tv - \lambda v = 0$, and so $(T - \lambda I)v = 0$, and so the eigenvectors associated to $\lambda$ are the non-zero elements of $\ker(T-\lambda I)$.
In your case you have $(T + I)v = (T - (-1)I)v = 0$ if $v\in \ker(T+I)$ and so the assumption that $T+I$ has non-zero kernel is quite correctly what you want to use! (I don't know if one can get too much more elegant than a short one-line solution, and would say your solution does directly use that $\pm (T+I)$ is singular.)
I don't really see the need to look at $-(T+I)$ though, as any vector $v$ with $(T+I)v = 0$ certainly satisfies $-(T+I)v = 0$, and vice versa, so passing from $T+I$ to $-(T+I)$ doesn't actually give you anything else... Maybe the author prefers to move the $-Tv$ to the other side, already having $-v$ in place, hinting at the eigenvalue $-1$?
