I have been asked to show that controllability and observability are not affected by replacing $A$ with $(A+αI)$. And also to show that this is not necessarily true for stabilizability.
I have thought about different approaches:
Popov-Belevitch-Hautus test
If the system is controllable must be true that $$Rank [(A - \lambda I) B] = n$$
If we replace $A$ with $(A+αI)$ then $$Rank [((A +αI) - \lambda I) B] $$ $$Rank [(A - (\lambda -α)I) B] $$ $$Rank [(A - gI) B] $$ where $g = (\lambda -α) \in C$
$(A - gI)$ has already $rank = n$ for all $g \in C$ except for those which are eigenvalues for $A$.
For those values, we need to prove that the concatenation of $B$ will guarantee the rank to be n.
The issue is that since B is the same we can't guarantee it.
Controllability Matrix
Given an LTI if and only if the control matrix $C$ has full column rank, then the system is controllable. $$ C = [B ,AB, A²B ... A^{n-1} B]$$ but then altering the diagonal of $A$ can affect the rank of C, or if not, how come?
Controllability Gramian
I tried to plug $(A+αI)$ in the integral, but doing that then I don't know how to prove that the matrix $W$ is still nonsingular for any t > 0.
Controllability: there does not exist a left-eigenvector for $A$ that is orthogonal for $B$.
For your problem, for any $x$ such that $xA = \lambda x$, $xB\neq0$.
Since the eigenvectors of $A$ are also the eigenvecors of $A+\alpha I$, i.e., $x(A+I\alpha) = x(\lambda+\alpha)$, you have $xB\neq0$.