Matrices and Continuity arguments
Why is it that given a matrix 
, there is always a number 
such that 
is invertible when 
? This doesn't even seem true to me, but I think that's the gist of something I saw a while back in a book-- it was stated without proof.
In general, what is a continuity argument w.r.t. matrix theory?
Possibly relevant posts:
exists if 
continuous and 
B.V. (3/17/2005)- Probability of invertibility of matrices (5/16/2006)
- The implicit and inverse function theorems (9/12/2006)
I can’t see your latex formula.
I’m not sure if it’s the same thing but I used to/still have problems when working with Lie gps / Matrix gps, where the authors just wave their hands on matrix multiplication is continuous.
Comment by tpc — 4/3/2006 @ 10:06 pm
a simple argument for invertible matrices: the determinant is a polynomial in nxn variables, continuous. For any A such that det(A) \= 0, we can perturb slightly the coefficients of A and the continuity argument gives that the perturbed matrix is inversible. Let us note that if A is not inversible, the epsilon could be though as the variable of the characteristic polynomial, so det( A+eps I ) cannot be zero except in a finite and discrete set of values of eps.
(More formal: det:\R^{n\times n} \to \R is continuous, and the invertibles matrices are the set det^{-1}(\R\setminus 0), the pre-imagen of an open set, which is open)
Comment by JuanPablo — 5/9/2006 @ 7:08 am
[…] I got the last question, where you consider all of : from an earlier problem I posted about, it is clear that the set of noninvertible matrices has empty interior, so measure zero; therefore a random matrix is invertible with probability one. […]
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