Given a singular matrix, I am tring to find an invertible matrix... (Finite Dimensional Space)

Question

In coordinates and in a finite-dimensional space, how would I prove that given any singular $n$x$n$ matrix $A$, any $\epsilon\gt0$ and any matrix norm $||.||$, there is an invertible $n$x$n$ matrix $B$ such that $||A-B|| \lt \epsilon$ ?

Is it worthwhile to consider

$spectrum(A)=\{\lambda | A-\lambda I$ is singular$\}$ ?

@StevenTaschuk: Not a duplicate. This question asks for proof that there's always an invertible operator in $\mathbb R^n$; the other for an example where there isn't. — hmakholm left over Monica, May 02 '15 at 06:34
I would like to hear, from those who voted to close, in what sense this question is a duplicate of the one mentioned above. — Martin Argerami, May 03 '15 at 02:58

score 3 · Answer 1 · answered May 02 '15 at 06:36

3

Hint: The determinant is a polynomial in the matrix entries. If it was identically zero in any open subset of $\mathbb R^{n\times n}$, it would need to be the zero polynomial, but it isn't.

answered May 02 '15 at 06:36

hmakholm left over Monica

291,611

Given a singular matrix, I am tring to find an invertible matrix... (Finite Dimensional Space)

1 Answers1