Let $A$ and $B$ be diagonalizable $n$-dimensional square matrices. Suppose both $A$ and $B$ has eigenvalues other than $0, 1$, and and $rkA＋rkB＝n$. Show that such $A$ and $B$ do not exist.
Any help would be appreciated, thank you. In my original problem, $A＋B＝E$ is also condition, but I guess this condition is not needed.
The requirement that $A + B = E$ is necessary. For instance, taking $$ A = \pmatrix{2&0\\0&0}, \quad B = \pmatrix{3&0\\0&0} $$ gives us an example where both $A$ and $B$ have an eigenvalue not equal to $0$ or $1$ ($2$ for $A$ and $3$ for $B$), but the sum of the ranks is $1 + 1 = 2$.
On the other hand, if $A$ and $B$ satisfy $A + B = E$, and $A$ is diagonalizable, then we can assume without loss of generality that $A$ is diagonal so that $$ A = \pmatrix{a_1 \\ & \ddots \\ && a_n}, \quad B = E - A = \pmatrix{1 - a_1 \\ & \ddots\\ && 1 - a_n}. $$ The rank of a diagonal matrix is the number of non-zero values on its diagonal. We see that the sum of the ranks of $A,B$ must be at least $n$, and the sum can only be exactly $n$ if we have either $a_i = 0$ or $1 - a_i = 0$ for $i = 1,\dots,n$.
However, if $A$ has an eigenvalue not equal to $0$ or $1$, then there is an $i$ for which $a_i$ is neither $0$ nor $1$, which means that it is not true that $a_i = 0$ or $1 - a_i = 0$. So, the sum of the ranks is not $n$.