Given a matrix $A$, what properties must $A$ have to satisfy a bound like
$$ y^T A y \geq \mu \| y \|_2^2 $$
For some positive constant $\mu$?
My instinct is that $\mu$ must be smaller than the smallest eigenvalue of $A$, but I am not sure if this is correct or if my intuition is wrong.
Furthermore, how could I find $\mu$ given a specific $A$? (Assuming $A$ can satisfy the bound).
Edit:
For the specific problem I am interested in, we can assume $A$ is symmetric and positive definite.
Equivalently, you want to find $\mu$ where $$ \mu := \min_{y\neq 0 } \frac{y^TAy}{y^Ty}. $$
Now take the eigenvalue decomposition of $A = U\Lambda U^T$, where $U^TU = UU^T = I$. Then we can do a change of variables $v = U^Ty \iff y = Uv$ and equivalently arrive at $$ \mu := \min_{v\neq 0 } \frac{v^T\Lambda v}{v^Tv}. $$ where $v^T\Lambda v = \sum_{i=1}^n \Lambda_{ii}v_i^2$. Now this becomes kind of a "weight placement" problem: how do we design $v$ so that it puts the most weight on the smallest value $\Lambda_{ii}$? To make this even more obvious, I'm going to introduce another change of variables: $z_i = (v_i^2 / \|v\|_2^2)$. Now the problem becomes $$ \mu := \min_{z > 0, \sum_i z_i = 1 } \sum_{i}\Lambda_{ii} z_i. $$ The solution to this "weight placement" problem is to put all the "mass" of $z_i$ on the smallest value of $\Lambda_{ii}$. That is, if all the values of $\Lambda_{ii}$ are distinct, then $$ z_i = \begin{cases} 1, & \Lambda_{ii} = \min_j\Lambda_{jj}\\ 0, & \text{else.} \end{cases} $$ (If the values are not distinct then you can just pick one such minimal $\Lambda_{ii}$ value and get the same result.)
Thus, $\mu = \min_i\Lambda_{ii}$.
Note that as pointed out, $\mu \geq 0$ only if $A$ is PSD, e.g. $\min_i\Lambda_{ii} \geq 0$.