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?
April 3rd, 2006 at 10:06 pm
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.
May 9th, 2006 at 7:08 am
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)
May 16th, 2006 at 2:07 pm
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