# 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 InvertibleMatricesAreDenseInSetOfNxnMatrices 2013-03-22 15:38:51 2013-03-22 15:38:51 stevecheng (10074) stevecheng (10074) 5 stevecheng (10074) Theorem msc 15A09