1
$\begingroup$

Let $A$ be a $10×10$ matrix with complex entries so that all its eigenvalues are non negative real numbers, and at least one of them is positive. Then how to prove the following statement is always false?

There exists a matrix $B$ such that $AB-BA=A$.

*My work:*What I know is that $AB-BA\not=I$ which implies $A\not=I$.

$\endgroup$
  • 1
    $\begingroup$ What do you know about the trace of $AB-BA$? $\endgroup$ – Gerry Myerson Nov 14 '15 at 8:26
  • 1
    $\begingroup$ tr(AB-BA)=0 . ya I got it Gerry Myerson..thanks a lot. $\endgroup$ – Mathlover Nov 15 '15 at 2:32
2
$\begingroup$

Hint:

Use the facts

  1. $tr(A+B)=tr(A)+tr(B)$
  2. $tr(cA)=c.tr(A)$ for all scalars $c$
  3. $tr(AB)=tr(BA)$

to find $tr(A)$ and then use the fact

  1. Since $A$ is a complex $10\times10$ matrix, if $\lambda_i,1\le i\le10$ are the eigenvalues of $A$, then $tr(A)=\sum\limits_{i=1}^{10}\lambda_i$

and the fact given in question

  1. "all its eigenvalues are non negative real numbers, and at least one of them is positive".

...and you are done.

$\endgroup$
  • 1
    $\begingroup$ oh!!.Thanks a lot Jesse P Francis. $\endgroup$ – Mathlover Nov 15 '15 at 2:29

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.