I try to get some order into those concepts: given a matrix $A$ we can find the condition by dividing the largest by the smallest singular value:


If the singular values decay sufficiently fast, we call the A ill-conditioned.

The retrieval of the pseudo-inverse $A^\dagger$ brings the singular values into play:

$$A^\dagger y = \sum_{n\in \mathbb{N}} \sigma_n^{-1} \langle y,u_n \rangle v_n$$

If there exists an $n \in N$ such that $1/\sigma_n$ is very large, then little perturbations in the subspace of $u_n$ produces a large error in the subspace of $v_n$. But at this point we cannot say that $\kappa(A)$ is large. Because with all $\sigma_n$ of roughly the same magnitude, we can end up with $\kappa(A)$ close to $1$. Yet, we are still under strong influence of little perturbations. My conclusion would be that a large $\kappa(A)$ suffices ill-posedness, but it is no necessity.


1 Answer 1


If $\kappa (A)$ is small (close to 1), then $A$ is well-conditioned, and vice versa. This means a small (in the relative sense) perturbation in $y$ leads to a small relative change in $A^\dagger y$. In your case, if $1/\sigma_n$ is very large, then $A^\dagger y$ itself would be very large. So a small change in $y$ will lead to only a small relative change in the result.


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