TITLE: Actual computational complexity of solving a linear system accounting for numerical accuracy (digit)
QUESTION [6 upvotes]: Solving a system of linear equations is solving for $n \times 1$ vector $x$ out of $Ax = b$, where $A$ is $n \times n$ matrix. Suppose that $A$'s entries have $k$ digits at maximum, in binary or decimal. Similarly, entries of $b$ also has $k$ digits at maximum. 
It is said that computational complexity of solving lienar equation is $O(n^3)$, but this does not account for numerical issues - such as actually turning out $x$ to be accurate up to $y$ digits. 
What would be actual computational complexity accounting for $k$ and $y$?

REPLY [5 votes]: I assume that you want to solve only one system $Ax=b$. We don't consider iterative methods but only the standard method of using the LU decomposition of $A$.
The complexity of calculating $x$ is essentially the number of used couples of operations $(+,\times)$ to obtain $L,U$, that is, $\sim n^3/3$.
How to choose $k$ ? If the condition number of $A$ is $10^r$, then choose $k=y+r$ and you will get $x$ with $y$ significant digits.
The complexity of a multiplication with $k$ digits is $\approx 2\log_2(10)k^2$ (a computer works in binary) elementary operations.
Finally the complexity is $\approx (2.2)n^3k^2=(2.2)n^3(y+r)^2$ elementary operations.