TITLE: $|\det (A+B)|\ge |\det B|$ for all $B$ such that $AB=BA$ iff $A^2=O$
QUESTION [5 upvotes]: Let $A\in M_2(\mathbb{C})$. $Z(A)$ is the set of all $B\in M_2(\mathbb{C})$ such that $AB=BA$. Prove that $|\det(A+B)|\ge |\det B|$ for all $B\in Z(A)$ if and only if $A^2=O$.

If $A^2=O$ and $A\neq O$, suppose $\lambda$ is an eigenvalue of $A+B$. Then $(A+B)x=\lambda x$, then $A(A+B)x=\lambda(Ax)$, or $ABx=\lambda (Ax)$, or $B(Ax)=\lambda(Ax)$. Then $\lambda$ is also an eigenvalue of $B$. Thus $\det (A+B)=\det B$.
The problem is I don't know how to deal with the converse, that is if $|\det (A+B)|\ge |\det B|$ for all $B\in Z(A)$, then $A^2=O$.
Thanks in advance.

REPLY [5 votes]: Claim: Both eigenvalues of $A$ are zero.
Proof: If $A$ has a non-zero eigenvalue $\lambda \neq 0,$ then take $B=-\lambda I$ resulting $0= |\det(A - \lambda I)| \geq |\det(-\lambda I)|=|\lambda|^2,$ thus, $\lambda =0 $ also implying $\text{tr}(A)=\det(A)=0$
Now, by Cayley-Hamilton (for $2$ by $2$ matrices), we have $A^2 - \text{tr}(A)A + \det(A)=0 \to A^2=0$