cube root of positive definite matrix Suppose that $A$ is a real symmetric positive definite $20\times 20$ matrix with condition number $\kappa\le 1000$. I want to solve the system of linear equations $$A^{1/3}x=b$$
with $10$-digit precision.
Since $A$ is symmetric, I can diagonalize it such that $A=PDP^{-1}$ where $A$ and $D$ have the same condition number. Then $A^{1/3} = PD^{1/3}P^{-1}$ where $D^{1/3}$ has condition number $\kappa\le 10$ (not sure for this). Thus alternatively, we are trying to solve
$$PD^{1/3}P^{-1}x=b\implies D^{1/3}P^{-1}x=P^{-1}b$$
I think the matrix is not so big ($n=20$), so we can use direct method like Gaussian elimination with partial pivoting. Is there anything that I need to be aware for the $10$-digit precision?
 A: Here are some experiments using Maple and matrices $A$ with condition number $<10^3$. We calculate $P,D,D^{1/3}$ and $X=PD^{1/3}P^{-1}$. The test is $\dfrac{||X^3-A||_2}{||A||_2}<10^{-10}$, that is the relative error on $X^3$ is $<10^{-10}$, that is $X^3$ has $10$-digits precision. Let $\delta$-digits denotes the accuracy of intermediate computations and $n$ denotes the dimension of the matrix $A$. The correspondance between $n$ and $\delta$ is as follows:
$n=4\rightarrow \delta=15,n=10\rightarrow \delta=68,n=15\rightarrow \delta=170,n=20\rightarrow \delta=??$.
EDIT 1: ah yes, I am stupid. I change my second $D$ with $\delta$. $\delta$ denotes the value of the option "Digits" in Maple, that is, Maple must work with $\delta$ digits so that $X^3$ has $10$ significant digits.
For solving $A^{1/3}x=b$, you can directly calculate $A^{-1/3}$ ( with $D^{-1/3}$).
EDIT 2: There is a much better method. We choose $\delta=35$ and $n=20$. The total time of the calculation of $A^{-1/3}$ is $5"$.
STEP 1: Calculate, with Maple, $spectrum(A)=(\lambda_i)_i$.
STEP 2: Use the interpolation Lagrange polynomial method that associates $\lambda_i$ with ${\lambda_i}^{-1/3}$ and deduce an approximation $R$ of $A^{-1/3}$. Then the test $||A.R^3-I||_2<10^{-10}$ is satisfied.
