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A matrix cannot be expressed as a sum $B+C$ with $B^2=B$ and $C^3=0$

Mathematics Asked by Mashed Potato on February 21, 2021

Let $$A=begin{bmatrix} 0 & 0 & 0 & 1 \
1 & 0 & 0 & 0 \
0 & 1& 0 & 0 \
0 & 0 & 1 & 1
end{bmatrix} in M_4(mathbb{Z}_2).$$

Prove that the direct sum $A oplus A$ cannot be written as a sum $B+C$ such that $B^2=B$ and $C^3=0$ for some $B,C in M_8(mathbb{Z}_2)$.

I want to prove this by contradiction, so I assume that there exist matrices $B$ and $C$, but I don’t know how to move forward after that.

I’m not looking for an exact answer as I want to prove this on my own, but since I’m stuck on this problem for days I really need help. I just want to ask for ideas about properties, operations, or theorems of matrices that can possibly lead to a contradiction.

Any help would be appreciated.

One Answer

Not sure if this is useful or not, but $operatorname{rank}(B)$ must be equal to $4$.

Let $K=Aoplus A$. Since $K$ has zero trace, $B$ has an even rank. Clearly $Bne0$ or $I_8$. Therefore the rank of $B$ can only be $2,4$ or $6$.

Suppose $operatorname{rank}(B)=2$. Since $C^3=0$, the largest Jordan block in the Jordan form of $C$ is at most $3times3$. As $C$ is $8times8$, its Jordan form has at least three Jordan blocks. Therefore $dimker(C)ge3$, i.e. $operatorname{rank}(C)le5$. But then $operatorname{rank}(K)=operatorname{rank}(B+C)leoperatorname{rank}(B)+operatorname{rank}(C)le7$, which is a contradiction to the assumption that $K$ is nonsingular.

Therefore $operatorname{rank}(B)ne2$. Since the characteristic polynomial of $A$ is $x^4+x^3+1$, $1$ is not an eigenvalue of $A$. Hence $K+I$ is a traceless nonsingular matrix, $B+I$ is idempotent and $K+I=(B+I)+C$. So, by applying our previous argument to $(K+I,B+I,C)$ instead of $(K,B,C)$, we also have $operatorname{rank}(B+I)ne2$, i.e. $operatorname{rank}(B)ne6$. Therefore the rank of $B$ must be $4$.

Answered by user1551 on February 21, 2021

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