All finite-dimensional vector spaces over R are complete.

I think the proof for this needs to come in two stages. (i) show that ; (ii) show that if then is nonsingular.

I don't like that asterisk notation for the operator norm, so I'll write instead of ; and I'll use for the sum of the absolute values of all the entries of A.

To show (i), use the fact that ( is the transpose of A), and check that . Then is a positive matrix, so its operator norm is equal to its largest eigenvalue. This is less than (or equal to) the sum of the eigenvalues, which in turn is the sum of its diagonal elements and therefore less than the sum of the absolute values of all its elements.

Putting all that together, you see that .

For (ii), use the Neumann series , which converges if .

Suppose that A is invertible and that . If then . This is the product of two invertible matrices (because ) and is therefore invertible. So any matrix sufficiently close to A is invertible, and that shows that the set of invertible matrices is open.

is the larger of the two eigenvalues of , which you can easily calculate. I get .