0
$\begingroup$

Let $ M $ be an $ n\times n $ Hermitian matrix of rank $ k $, $k\neq n$. If $\lambda $ $\neq 0$ is an eigenvalue of $ M $ with corresponding unit column vector $ u $, with $ Mu= \lambda u$,then Which of the following are true?

  1. rank ($ M-\lambda $ $uu^* $) =$ k-1 $

  2. $ ( M- \lambda uu^* )^n = M^n- \lambda^n uu^* $

I think both are true, I considered some examples which is obviously not the right way to solve the above problem. I tried to use the fact that any Hermitian matrix is unitarily diagonalizable to show the 2nd option , but I couldn't do it.

please give me some hints . Thanks in advance.

$\endgroup$
  • 1
    $\begingroup$ \lambda not \lamda $\endgroup$ – kimchi lover Apr 28 at 15:58
  • 1
    $\begingroup$ Diagonalisation works perfectly here. At what step did you fail? $\endgroup$ – A.Γ. Apr 28 at 16:06
  • $\begingroup$ Thank you, I am trying again ,If I fail again I will edit my post . $\endgroup$ – suchanda adhikari Apr 28 at 16:09
  • 1
    $\begingroup$ Pay attention that it is a unitary diagonalisation, i.e. the columns (eigenvectors) of a transformation matrix are mutually orthogonal (including $u$, in particular). $\endgroup$ – A.Γ. Apr 28 at 16:12
  • $\begingroup$ Thank you sir, I am trying to solve it $\endgroup$ – suchanda adhikari Apr 28 at 16:20

Your Answer

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

Browse other questions tagged or ask your own question.