5
$\begingroup$

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.

$\endgroup$
5
  • 2
    $\begingroup$ Does your given proof account for multiplicity? Specifically, that $A$ and $B$ have the same eigenvectors but with differing multiplicity? $\endgroup$
    – DanielV
    Jan 22 '15 at 5:06
  • $\begingroup$ What you are working on seems very similar to mathoverflow.net/questions/65424/… $\endgroup$
    – DanielV
    Jan 22 '15 at 5:09
  • $\begingroup$ @DanielV Yes you need to account for multiplicity. You need to prove that $\lambda$ is an eigenvalue of $B$ iff $\lambda$ is an eigenvalue of $A+B$. If there is only one eigenvalue, then the eigenvalue must have multiplicity 2, in which case the determinant is equal. If there are two eigenvalues, then the determinant is equal as well. $\endgroup$ Jan 22 '15 at 5:53
  • $\begingroup$ Apologies, I missed the part where it was a 2 by 2 matrix, someone should write a book with all the on-the-fly notations that people use to write out declarations of matrices. $\endgroup$
    – DanielV
    Jan 22 '15 at 5:55
  • $\begingroup$ @Tien Kha Pham : From $B(Ax)=\lambda (Ax)$ , how do you conclude $\lambda$ is an eigenvalue of $B$ ? $Ax$ could be null-vector .. $\endgroup$
    – user228168
    Dec 23 '15 at 13:43
5
$\begingroup$

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 $A^2 - \text{tr}(A)A + \det(A)=0 \to A^2=0$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.