I found a thread with the same question. The answer there:
"Controllability: there does not exist a left-eigenvector for $A$ that is orthogonal for $B$.
For your problem, for any $x$ such that $xA=λx, \ xB≠0$.
Since the eigenvectors of A are also the eigenvecors of $A+αI$, i.e., $x(A+Iα)=x(λ+α)$, you have $xB≠0$."
I don't understand what is the connection here. Why is it obvious that the eigenvecors of $A$ are also the eigenvecors of $A+αI$? Even if that is true, why does it mean that $x(A+Iα)=x(λ+α)$?
I have read that replacing $A$ with $(A+αI $) can ruin stabilizability. But why can it ruin that and not controllability and observability?
Using the PBH test, it is easy to see that the controllability of the pair $(A,B)$ is equivalent to the controllability of the pair $(A+\alpha I,B)$. This is because $A$ and $A+\alpha I$ share the same eigenvectors. Indeed, if $P$ is an invertible matrix such that $J=PAP^{-1}$ where $J$ is the Jordan form of $A$, we have that
$$P(A+\alpha I)P^{-1}=J+\alpha I,$$
so the eigenvalues of $A+\alpha I$ are those of $A$ shifted by $\alpha$.
Controllability is that all modes are controllable and this will remain the same if we just shift the modes while keeping their eigenvectors.
Stabilizability requires that all the unstable modes be controllable. So, it allows for stable modes to be non-controllable. Now, the issue is that we can shift a stable uncontrollable mode to instability, and since this mode is not controllable, then the shifted system will be not be stabilizable.