# 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 $\u2102$)

$$p(t)=det\left((1-t)A+tB\right).$$ |

Clearly, $p$ is a polynomial function. It is not identically zero, for $p(1)=detB\ne 0$.
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 $t\ne 0$.
And the distance of this matrix from $A$ is $|t|\parallel B-A\parallel $,
which becomes small as $t$ gets small.

Title | invertible matrices are dense in set of nxn matrices |
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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 |