$(1)$ Let's suppose $A,B$ are $n-$order matrix, satisfying $AB=BA=O$, $r(A)=r(A^2)$, prove that$$r(A+B)=r(A)+r(B)$$
To prove this, I construct the following block matrix $$ \begin{pmatrix} A & O \\ O & B \\ \end{pmatrix} $$
Then I want to use the fact that "addition transformation does not change the rank of the matrix", I hope after this type transformation, the block matrix look like $$ \begin{pmatrix} A+B & O \\ O & O \\ \end{pmatrix} $$
The difficulty is that I can't use $r(A)=r(A^2)$ properly. So can someone give me some hints ? Thanks a lot.
Hints.
Let $f$ be a linear operator of a linear space $L$ with matrix $A$ in some basis of $L$.
If $M=f(L)$, then $f(M)=M$. (Use here the equality $r(A)=r(A^2)$.)
It follows that the matrix $A$ is conjugate to a block matrix of the form
$$ \begin{pmatrix} A_{11} & A_{12} \\ 0 & 0 \\ \end{pmatrix}, $$ where $|A_{11}|\neq0$. (What are the dimensions of the blocks?)
Let $$ B= \begin{pmatrix} B_{11} & B_{12} \\ B_{21} & B_{22} \\ \end{pmatrix}. $$ Since $AB=BA=0$, then $B_{11}=0$, $B_{12}=0$, $B_{21}=0$.
Now you can easily prove your equality.