Let $M$ be a real $n\times n$ symmetric positive definite matrix and let $M=QR$ be the QR decomposition of $M$ ($Q$ is orthogonal and $R$ is upper triangular).
Is there a connection between the smallest eigenvalue of $M$ and the smallest eigenvalue of $R$?
In other words, are there constants $c,C$ (that might depend on $\lVert{M}\rVert$ and $n$) such that
$$c\sigma_n=c\lVert{M^{-1}}\rVert^{-1}_2\le \min_{1\le i\le n} |R_{i,i}| \le C\lVert{M^{-1}}\rVert^{-1}_2=C\sigma_n$$
Where $\sigma_n$ the smallest singular value/eigenvalue of $M$.
What I have figured out so far: $c=1$ works. If $\sigma_n=0$, then equality is true for any $C$.
I will consider the case where $M$ and $R$ are invertible since you have taken care of the other case.
Note that since $M^TM = R^TR$, the singular values of $R$ are the singular values/eigenvalues of $M$.
With that, we can see that no $C$ will be large enough. For instance, consider the matrix $$ R = \pmatrix{1&t\\ 0&1}. $$ We find that the smallest singular of $R$ is given by $$ \sigma_2^2 = \frac 12 (t^2 + 2 - t\sqrt{t^2 + 4}), $$ so that $\lim_{t \to \infty} \sigma_2(R(t)) = 0$. However, the smallest eigenvalue of $R$ (for any $t$) will be $1$.