matrices rank inequality

Question

Let $A \in \mathscr{M}_n(\mathbb{C})$, $n \ge 3$, be a matrix, such that $rank(2I_n+3A)\ge rank(4I_n+5A)$. Prove that $rank(2I_n+3A) \ge 1 + \lfloor{\frac{n}{2}}\rfloor$.

Answer 1

We have two possible eigenvalues of $A$, namely $s = -2/3$ and $t = -4/5$. Now, the inequality says that the dimension of the eigenspace w.r.t. $s$ is not larger than that of the eigenspace w.r.t. $t$, i.e., $n_s = \dim eig(s)\le\dim eig(t) = n_t$. Now, $eig(s)\cap eig(t) = \{0\}$, which implies $n_s\le n-n_t$. Hence, $n_s\le\min\{n_t,n-n_t\}$. If $n$ is even, this minimum is not larger than $n/2$. If $n$ is odd, it is not larger than $\lfloor n/2\rfloor$ (the largest integer below or equal to $n/2$). In both cases, $n_s\le\lfloor n/2\rfloor$. This implies $rank(2I + 3A) = rank(A-sI) = n - n_s\ge n - [n/2] = \lceil n/2\rceil$ (the smallest integer above or equal to $n/2$). In the case where $n$ is odd this equals $1 + \lfloor n/2\rfloor$, but not in the case, where $n$ is even. And in this case, the claim is actually false. Just consider $$ A = \operatorname{diag}\{s,s,t,t\}. $$ Here, $n=4$ and $rank(A-sI) = rank(A-tI) = 2$, which is not equal to $1+\lfloor 4/2\rfloor = 3$.