Necessary and sufficient conditions for rank-$1$ update of positive definite matrix to be positive definite Given $A \succ 0$, what are necessary and sufficient conditions on $\lambda \in \mathbb{R}$ for $A + \lambda u u^\top$ to be positive definite?
What happens if we relax $A \succeq 0$ and require $A + \lambda u u^\top \succeq 0$?
 A: According to the Sherman-Morrison formula, $\det(A+\lambda uu^T)>0$  iff $1+\lambda u^TA^{-1}u>0$. Then the required condition is 
$\lambda >\dfrac{-1}{u^TA^{-1}u}$.
A: When $A\succ0$, $A+\lambda uu^T=A^{1/2}(I+\lambda A^{-1/2}uu^TA^{-1/2})A^{1/2}$ is positive definite if and only if all eigenvalues of $I+\lambda A^{-1/2}uu^TA^{-1/2}$ are positive, i.e. if and only if
$$
1+\lambda u^TA^{-1}u>0.\tag{1}
$$
If $A$ is positive semidefinite, there are three possibilities:


*

*$\lambda\ge0$. Then $A+\lambda uu^T$ is positive semidefinite.

*$v=(I-AA^+)u=0$, i.e. $u$ lies inside the column space of $A$. Then $(1)$ applies when $A$ and $u$ are replaced by their restriction or orthogonal projection on the column space of $A$. Hence $(1)$ should be modified to $1+\lambda u^TA^+u\ge0$.

*$\lambda<0$ and $v=(I-AA^+)u\ne0$. Then $A+\lambda uu^T$ is not positive semidefinite. In particular, we have $v^T(A+\lambda uu^T)v=\lambda\|v\|^4<0$.


In other words, when $A\succeq0$, $A+\lambda uu^T$ is positive semidefinite if and only if (a) $\lambda\ge0$ or (b) $(I-AA^+)u=0$ and $1+\lambda u^TA^+u\ge0$.
