$rkA＋rkB＝n$, and A and B are diagonalizable

Question

Let $A$ and $B$ be diagonalizable $n$-dimensional square matrices. Suppose each of $A$ and $B$ has no eigenvalues other than $0, 1$.Show that such $A$ and $B$ do not exist.

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Any help would be appreciated, thank you.

P.S. Sorry, I missed important condition at first. I assume $A＋B＝E$.

Answer 1

Your question's wording is confusing, but it you really meant what is written, then the claim is false:

$$A=\begin{pmatrix}1&0\\0&0\end{pmatrix}\;,\;\;B=\begin{pmatrix}0&0\\0&1\end{pmatrix}$$

are both $\;2\times2\;$ diagonal (and thus trivially diagonalizable) matrices with only $\;0,1\;$ eigenvalues, and also

$\;A+B=E\;$ ...

Answer 2

The wording of your question is slightly unclear. When you say

And suppose both A and B has eigen value which is not 0 or 1.

I can understand this in two ways. The less likely meaning would be that none of the eigenvalues of $A$ and $B$ is equal to either $0$ and $1$.

In this case, the proof of your clame would be trivial, since if $0$ is not an eigenvalue of $A$ and $B$, then they are both nonsingular, so their ranks are $n$, and the sum of their ranks is $2n\neq n$.

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But you probably mean that at least one of the eigenvalues of $A$ and $B$ is not equal to either $0$ or $1$. In that case, simply take $$A=B=\begin{bmatrix}0& 0\\ 0 &2\end{bmatrix}$$

and you can see that:

1. Clearly, $A$ and $B$ are diagonalizable.
2. $\mathrm{rank}(A)+\mathrm{rank}(B)=1 + 1 =2 = n$
3. Both $A$ and $B$ have an eigenvalue that is not $0$ or $1$ (because they both have an eigenvalue of $2$).