TITLE: Eigenvalue test faster than $O\left(n^3\right)$?
QUESTION [13 upvotes]: Given a real $n\times n$ matrix $A$, one can find the eigenvalues in $O\left(n^3\right)$ by using say, the $QR$ algorithm.
Now, what if we guess an eigenvalue $\lambda_0$, and we want to know if it's actually an eigenvalue of the matrix $A$? Intuitively, this should be significantly faster than actually finding all of the eigenvalues. We can of course check if $\lambda_0$ is actually an eigenvalue by calculating $\det(A-\lambda_0 I)$, but calculating the determinant is also $O\left(n^3\right)$.
So: 

Is there a faster than $O\left(n^3\right)$ method for testing the
"eigenvaluedness" of a specific number $\lambda_0$, without solving
for the rest of the $\lambda_i$'s? Can one prove there isn't?
Does it help if $A$ is orthogonal, symmetric or has other special
(non triangular) form?

Bounty update:
Bounty goes to whoever shows either of the following: 

A method of checking a possible eigenvalue in less than $O(n^3)$.
A proof or sufficiently convincing heuristic argument that no such method exists.

REPLY [3 votes]: It is not necessary to actually calculate the determinant, only to confirm whether it's zero.
As such, it suffices to attempt a matrix inversion for $\mathbf{A}-\lambda\mathbf{I}$... if it fails, then the determinant is zero, and therefore we have confirmed $\lambda$ to be an eigenvalue.
It can be shown that, through a divide-and-conquer approach, matrix inversion has the same complexity as matrix multiplication. And because Williams algorithm exists with $O(n^{2.3727})$, this is also an achievable complexity for confirmation of an eigenvalue (up to a certain numerical accuracy, of course).