Let us say that we know the following things about an $n\times n$ matrix $A$:
$A$ is symmetric and strictly positive definite.
For some particular vector $v$, we have $\lambda |v| \leq |Av|\leq \Lambda |v|$. (Here, $\Lambda > \lambda >0$.)
For all vectors $w$ that are perpendicular to $v$, we have $\lambda |w|\leq |Aw| \leq \Lambda |w|$.
Can we conclude that $\lambda |e| \leq |Ae| \leq \Lambda |e|$ for all $e\in \mathbb R^n$? If not, can we say $m|e| \leq |Ae| \leq M |e|$ for some other constants depending on $\lambda$ and $\Lambda$? Such as $\lambda/2 |e| \leq |Ae| \leq 2\Lambda|e|$? This upper bound is obvious, but I am not sure about the lower bound.
No such luck. Consider $$ A = \pmatrix{1&0\\0&\epsilon} $$ for (a small value of) $\epsilon > 0$. For $v = (1,1)$, we have $\frac 1{\sqrt{2}} \|v\| < \|Av\| < \|v\|$. This similarly holds for the vectors perpendicular to $v$, which are parallel to $(-1,1)$. However, $\min_{e \in \Bbb R^2} \frac{|Ae|}{|e|} = \epsilon$.
So, our actual lower bound could be arbitrarily close to $0$.