$A,B,A_1,B_1$ are $n\times n$ matrices, $A\sim A_1, B\sim B_1$, $n$ is odd, $AB=0$, show that at least one of $A+A_1,B+B_1$ is singular.

Question

Question :

$A , B , A_1 , B_1$ are $n \times n$ matrices , $A \sim A_1 , B \sim B_1$ , $n$ is odd , $AB = 0$
Show that at least one of $A+A_1 , B+B_1$ is singular.

My attempt :
Similarity indicates that there exist non-singular matrices $P , Q$ such that $AP = PA_1 , BP = PB_1$.
To show the result , consider $P(A+A_1) = PA+AP $ , $Q(B+B_1) = QB+BQ$ , and $(PA+AP)(QB+BQ) = PAQB+QPQB+APBQ$.
We find the determinant of this above matrix has no reason to be zero !

What's the idea? How the assumption "$n$ is odd" helps?

Answer 1

A counterexample when $n$ is even: $$ A=\pmatrix{0&1\\ 0&0},\quad A_1=\pmatrix{0&0\\ 1&0},\quad B=\pmatrix{1&0\\ 0&0},\quad B_1=\pmatrix{0&0\\ 0&1}. $$ Now suppose $n$ is odd. Since $AB=0$, we have $\operatorname{range}(B)\subseteq\ker(A)$. Therefore $\operatorname{rank}(B)\le n-\operatorname{rank}(A)$. Since $n$ is odd, at least one of $A$ and $B$ has rank smaller than $n/2$. In case $\operatorname{rank}(A)<n/2$, we obtain $\operatorname{rank}(A+A_1)\le\operatorname{rank}(A)+\operatorname{rank}(A_1)<n$ and hence $A+A_1$ is singular. The other case is similar.