How prove this rank identity $r(A)=r(B)$ let $A_{n\to n},B_{n\to n}$  matrix,and such
$$A^2=20142014A,B^2=20142014B,$$
and $20142014I-A-B$ is invertible,
show that
$$\rm{rank{A}}=\rm{rank{B}}$$
we know 
$$A^2-20142014A=0$$ then Characteristic equation is $$\lambda^2-20142014\lambda =0\Longrightarrow \lambda=0,20142014$$
I know following  this maybe is not true.so How prove solve it
 A: If $20142014I - A - B = M$ then $AM = 20142014A - A^2 - AB = -AB$ and $MB = 20142014B - AB - B^2 = -AB$.
So, $\text{rank}(AM) = \text{rank}(-AB) = \text{rank}(MB)$. Since $M$ is invertible, $\text{rank}(A) = \text{rank}(B)$.
A: You should start reformulating this in a way to avoid all those digits shouting at you, so that your brain can focus on the actual problem. So put $c=20142014$, then it is given that $A^2=cA$, $B^2=cB$, and $c$ is not an eigenvalue of $A+B$.
Since there is not much relation between $A$ and $B$ specified (notably they need not commute), I don't see much one can do with eigenvalues. So just try multiplying by that matrix that is guaranteed to be invertible (so the multiplication will preserve the rank)
$$
  A(A+B-cI)=A^2+AB-cA=AB.
$$
Wow, that was simple. Then do the same thing with $B$, but to get a matching result exchange sides
$$
  (A+B-cI)B=AB+B^2-cB=AB.
$$
So in fact much more is true than that $A,B$ have the same rank, they are actually similar matrices, conjugate through the invertible matrix $M=A+B-cI$, that is $A=MBM^{-1}$.
Actually this is a bit disappointing. There would seem to be a bit of structure in the problem (like $A,B$ are diagonalisable with eigenvalues $0,c$ only, since they are annihilated by $X(X-c)$ which is split with simple roots), but in the end it just comes down to some simple formal multiplications. At least we can say that the "diagonalisable with eigenvalues $0,c$ only statement" shows that equal rank automatically implies similarity, which is somewhat of a consolation.

Added: on second thought there is a more structural way to see this, maybe the intended solution. Both $A,B$ are diagonalisable (since $c\neq0$ ) with eigenvalues contained in $\{0,c\}$. Let $r,s$ be the ranks of $A,B$, respectively; we need to show $r=s$. Assume for a contradiction $r\neq s$; by symmetry we may assume $r<s$. The eigenspace $E_0$ of $A$ for $\lambda=0$ has dimension$~n-r$, while the eigenspace $E'_c$ of$~B$ for $\lambda=c$ has dimension$~s$. By our assumption $(n-r)+s>n$ so
$$
  \dim(E_0\cap E'_c)=\dim(E_0)+\dim(E'_c)-\dim(E_0+E'_c)\geq(n-r)+s-n>0,
$$
in other words those eigenspaces have too large dimensions to intersect trivially. But any nonzero vector in the intersection is eigenvector of$~A+B$ for $\lambda=0+c=c$, while the hypothesis was that $c$ was not an eigenvalue of$~A+B$, which gives the contradiction we were after.
