Matrices: Rank Nullity theorem Let $A$ and $B$ be $n\times n$  real matrices such that $AB=BA=$ $The$ $zero$ $matrix$ and $A+B$ is invertible. Which of the following are always true?
(a) Rank($A$) = Rank($B$)
(b) Rank($A$) + Rank($B$) = $n$
(c) Nullity($A$) + Nullity($B$) = $n$
(d) $A-B$ is invertible
I have no  idea other than $A, B$ are simultaneously diagonalizable and $0$ is an eigen value of either $A$ or $B$
 A: (a) is not generally true.
Since $AB=0=BA$, you know that $C(B)\subseteq N(A)$ (column space and null space) and that $C(A)\subseteq N(B)$. Therefore $\dim C(B)\le\dim N(A)$ and $\dim C(A)\le\dim N(B)$.
By the rank nullity theorem,
$$
n=\dim C(B)+\dim N(B)\le \dim N(A)+\dim N(B)
$$
On the other hand, $N(A)\cap N(B)=\{0\}$, because $A+B$ is invertible, so $\dim N(A)+\dim N(B)=\dim(N(A)+N(B))\le n$.
Therefore $\dim N(A)+\dim N(B)=n$. Again by the rank nullity theorem,
$$
n=\dim N(A)+\dim N(B)=\dim C(A)+\dim N(A)
$$
so $\dim N(B)=\dim C(A)$ and similarly $\dim N(A)=\dim C(B)$. Hence $\dim C(A)+\dim C(B)=n$.
Thus (b) and (c) are true.
For (d), consider $(A-B)^2=A^2-AB-BA+B^2=A^2+B^2$; similarly, $(A+B)^2=A^2+B^2$.
A: a) is right out.
A could be full rank, and B be the zero matrix.
b) is true if c) is true by the rank-nullity theorem.
c) is true.  $AB = 0$ implies that Nullity $A$ + Nullity $B \ge n$
and if Nullity $A$ + Nullity $B > n$, there would be a non-zero intersection of the two null spaces.  Let $v$ be a vector in this intersection.  Then $(A+B)v = 0$ and $(A+B)$ is singular.
d) seems plausible. 
We haven't yet touched on the commutativity of $AB.$
$A = PD_1P^{-1}, B = PD_2P^{-1}$
Along the main diagonals $D_1$ is $0$ where $D_2$ is not-zero and vice versa.
$A - B$ is invertable.
