Is it possible to determine rank given these conditions?

Question

I stumbled upon this question in an online forum.

Let $A$ be a $3000\times 3000$ matrix with real entries such that $A^{2016}=I$.
If $\mathrm{rank}(I-A^{1008})=1000$, then $\mathrm{rank}(I+A^{1008})$ is
a) $504$
b) $2000$
c) $1992$
d) $2016$

My approach: Since $A^{2016}=I$, we can write $PQ=0$, where
$P=I-A^{1008}$ and $Q=I+A^{1008}$.
Now, applying Sylvester's rank inequality, we get:
$\mathrm{rank}(P)+\mathrm{rank}(Q)\le \mathrm{rank}(PQ)+3000\implies 1000 + \mathrm{rank}(Q) \le 3000\implies \mathrm{rank}(Q)\le 2000$

I have got an upper-bound for $\mathrm{rank}(Q)$, but I don't know how to determine its exact value.
Is it even possible to determine $\mathrm{rank}(Q)$, given this information ?
Thanks in advance.

Answer 1

We denote $k=1008$. You can observe that

$(I-A^k)(I+A^k)=I-A^{2k}=I-I=0$

$(I+A^k)(I-A^k)=I-A^{2k}=I-I=0$

Thus

$\mathrm{range}(I+A^k)\subseteq \ker(I-A^k)$

Let $v\in \ker(I-A^k)$. Then $A^kv=v$.

Then

$(I+A^k)(\frac{v}{2})=\frac{v}{2}+\frac{v}{2}=v$

This means that

$\mathrm{range}(I+A^k)=\ker(I-A^k)$

So

$\dim(\mathrm{range}(I+A^k))= 3000-\dim(\mathrm{range}(I-A^k))=2000$.