The (n x n) singular matrix A has all real eigenvalues, say L1, L2, ..., Ln.
Being A singular, Li = 0 for some i (assume that all Li<>0 are distinct).
I want to make A non singular, fully ranked.
Is just removing rows(i) and colums(i) from A a valid option??
Sorry if it sounds "#&$%#*&~!". It's *decades* I'm not touching this.
P.S. The question could be restated as: does the eigenvector i correspond to the cartesian vector base i?