Suppose there is a Positive definite matrix A $\in \mathbb{R^{d\times d}}$, and a non-zero row vector $\phi \in \mathbb{R^{1\times d}}$. Obviously, the matrix $\phi^T\phi$ is semi-positive definite. Denote $n$ as a parameter. My question is, how should we determine the value of $n$, so that $$ \lambda_{\min}(A - n\phi^T\phi) = 0 $$ where $\lambda_{\min}(X)$ means the lowest eigenvalue of matrix X.
I guess there should be a solution to this problem, for example, $n$ should be a function of A (or A's norm), and $\phi$ (or $\phi$'s norm).
It gives me a feeling about solving a optimize problem with some constraints (but I'm not sure it's the correct direction):
Maximize $n$, so that $$ \inf_{\psi:\|\psi\|=1} \frac{\psi(A-n\phi ^T\phi)\psi^T}{\|\psi\|^2} \geq 0 $$
Moreover, is it possible to find a solution for other values besides $0$? For example,
Maximize $n$, so that $$ \inf_{\psi:\|\psi\|=1} \frac{\psi(A-n\phi ^T\phi)\psi^T}{\|\psi\|^2} \geq \lambda $$ where $\lambda>0$
A different perspective is that we're looking for the lowest $n > 0$ for which $\det(A - n \phi^T\phi) = 0$. The matrix determinant lemma tells us that $$ \det(A - n\phi^T\phi) = (1 - n\phi A^{-1}\phi^T)\det(A). $$ So, we're looking for the lowest value of $n$ such that we will have $1 - n\phi A^{-1}\phi^T = 0$. Solving this for $n$ yields $$ n = \frac{1}{\phi A^{-1}\phi^T}. $$
Edit: Here's a way to generalize the result for arbitrary $\lambda$. Note that $$ \inf_{\psi:\|\psi\|=1} \frac{\psi(A-n\phi ^T\phi)\psi^T}{\|\psi\|^2} \geq \lambda = \frac{\psi(\lambda I)\psi^T}{\|\psi\|^2} \iff \\ \inf_{\psi:\|\psi\|=1} \frac{\psi((A - \lambda I)-n\phi ^T\phi)\psi^T}{\|\psi\|^2} \geq 0. $$ So, it suffices to repeat the above analysis, replacing $A$ with $A - \lambda I$.