Lets say I've given a sum of matrices $$ \sum_{i\in[n]}A_i = A$$ where I know that all the $A_i$ are positive semidefinite.
Now given a single vector $v$, I'd like to know if I can make any statement in terms of the individual $A_i$ about the possible range of a scalar $t$ such that $$ t\cdot A - vv^T$$ is positive semidefinite. Ideally, this would break down to some evaluations of the $A_i$ considered as quadratic and / or linear forms.
Let $M \geq 0$ mean that $M$ is positive semidefinite. Note that $$ tA - vv^T \geq 0 \iff\\ x^T(tA - vv^T)x \geq 0 \quad \forall x \iff\\ t(x^TAx) \geq (v^Tx)^2 \quad \forall x \iff\\ \frac{(x^TAx)}{(v^Tx)^2} \geq \frac 1t \quad \forall x $$
Now, for a matrix $A_i$, we can define $$ M_i = \max_{v^Tx \neq 0} \frac{(x^TA_ix)}{(v^Tx)^2} $$ Then $t$ can only be such that $tA - vv^T \geq 0$ if $$ \frac 1t \leq \sum_{i=1}^n M_i $$ note that this condition is necessary, but not sufficient. Now, define $$ m_i = \min_{v^Tx \neq 0} \frac{(x^TA_ix)}{(v^Tx)^2} $$ Then $t$ will be such a value if $$ \frac 1t \leq \sum_{i=1}^n m_i + \max_{i} (M_i - m_i) $$ Note that this condition is sufficient, but not necessary.