# How to prove the following problem on Hermitian matrix.

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.

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