Consider an $n\times n$ matrix $A$ that is positive semidefinite and has rank $n-1$, so there exists exactly one eigenvector $v$ such that $Av=0$. Let now $B$ be a symmetric matrix such that $v^TBv=0$. I'd like to argue that, if the norm of $B$ is small enough, $A+B$ is always positive semidefinite.
1 Answer
$\begingroup$
$\endgroup$
Converted from (now-deleted) a comment by Christian Remling:
This is false: $$ A=\begin{pmatrix} 0 & 0\\ 0& 1\end{pmatrix},\quad B=\begin{pmatrix} 0 & 1\\ 1 & 0\end{pmatrix}$$ Then $A+\epsilon B$ is not positive definite for any $\epsilon\not= 0$.