invertible matrices are dense in set of nxn matrices

If $A$ is any $n\times n$ matrix with real or complex entries, Then there are invertible matrices arbitrarily close to $A$ , under any norm for the $n\times n$ matrices.

This is easily proven as follows. Take any invertible matrix $B$ (e.g. $B=I$ ), and consider the function (for $t\in\mathbb{R}$ or $\mathbb{C}$ )

$p(t)=\det\bigl{(}(1-t)A+tB\bigr{)}\,.$

Clearly, $p$ is a polynomial function. It is not identically zero, for $p(1)=\det B\neq 0$ . But a non-zero polynomial has only finitely many zeroes, So given any single point $t_{0}$ , if $t$ is close enough but unequal to $t_{0}$ , $p(t)$ must be non-zero. In particular, applying this for $t_{0}=0$ , we see that the matrix $(1-t)A+tB$ is invertible for small $t\neq 0$ . And the distance of this matrix from $A$ is $\lvert t\rvert\,\lVert B-A\rVert$ , which becomes small as $t$ gets small.

Title	invertible matrices are dense in set of nxn matrices
Canonical name	InvertibleMatricesAreDenseInSetOfNxnMatrices
Date of creation	2013-03-22 15:38:51
Last modified on	2013-03-22 15:38:51
Owner	stevecheng (10074)
Last modified by	stevecheng (10074)
Numerical id	5
Author	stevecheng (10074)
Entry type	Theorem
Classification	msc 15A09