Conjugate gradient convergence

Question

Suppose a symmetric positive definite matrix $A$ has one thousand eigenvalues uniformly distributed between $1$ and $10$, and one eigenvalue of $10^4$. Suppose another symmetric positive definite matrix $B$ has an eigenvalue of $1$ and has one thousand eigenvalues uniformly distributed between $10^3$ and $10^4$.

I need to make an educated guess for which matrix the conjugate gradient algorithm converge more rapidly.

Since the ratio between the largest eigenvalue and the smallest eigenvalue is the same for $A$ and $B$ I could not figure out how to approach this question.

Answer 1

Straight from Wikipedia: The most widely known and taught convergence result is formulated in terms of the condition number $$ \big \Vert\boldsymbol e^k \big\Vert_A \leq f\big(\kappa(A)\big) \big\Vert\boldsymbol e^0 \big\Vert_A$$where the condition number $\kappa(A)$ is for the considered symmetric (pos. def.) and most likely real matrices => normal matrices $A, B$ given by $$\kappa(A) = \frac{\vert \lambda_\max(A) \vert}{\vert \lambda_\min(A) \vert} \leq \frac{10^4}{1} = 10^4.$$ Also, $$\kappa(B) \leq \frac{10^4}{1} = 10^4.$$

So unless you can refine the bounds, there is without taking a look at more sophisticated theories (if they exist) not much more you can say.