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 justremovingrows(i) and colums(i) from A a valid option??

Sorry if it sounds "#&$%#*&~!". It's *decades* I'm not touching this.

And... Thanks!

P.S. The question could be restated as: does the eigenvector i correspond to the cartesian vector base i?