# Thread: Eigenvalues & Singularity

1. ## Eigenvalues & Singularity

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 "#&$%#*&~!". It's *decades* I'm not touching this. And... Thanks! P.S. The question could be restated as: does the eigenvector i correspond to the cartesian vector base i? 2. Originally Posted by paolopiace The (n x n) singular matrix A has all real eigenvalues, say L1, L2, ..., Ln. Being A singular, Li = 0 for some i (assume that all Li<>0 are distinct). I want to make A non singular, fully ranked. Is just removing rows(i) and colums(i) from A a valid option?? Sorry if it sounds "#&$%#*&~!". It's *decades* I'm not touching this.

And... Thanks!

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

No. Rather you'd better bring your matrix to echelon form (Gauss or Gauss-Jordan method) and then check which row(s) become all zeroes while doing this: these are the rows you'll have to change for ones that won't vanish.
Now, during the very process of echeloning your matrix you can usually see what must be the conditions the vanishing rows must fulfill so that they won't vanish.

Tonio

3. The real question is, what do you mean by "make A non singular, fully ranked." Since A is singular, you would have to change A to some other matrix. What restrictions are you placing on the changes? If none, the replacing A by the identity matrix would work!