show set of matrices A such that I-A is invertible is open If $U = \{A\in Mat(2,2) : I-A \text{ is invertible} \}$, how can I show that $U$ is open?
I know that the set, say $V$, of $n\times n$ invertible matrices is open. Can I use this fact with the linear transformation $T:U\rightarrow V$ defined by $T(A) = I-A$ to show that $U$ is open? ($U$ would be the preimage of the open set $V$ and linear maps are continuous)
 A: Note that the function $f(A) = det(I - A)$ is continuous. So $U = f^{-1}(\mathbb{R} - \{0\})$ is open as a preimage of an open set under a continous map.
A: Since the set of matrices over a finite dimensional vector space is a finite dimensional vector space, all topologies coming from norms are equivalent. Usually people use the so called operator norm. This is defined on a matrix A as the supremum over the norm of
Av where v  is a vector with length less than or equal 1.
In this case it is easier to see the continuity by using the usual topology coming from viewing the set of n by n matrices as n^2 tuples of the real numbers. Then det(I-A) is just a polynomial in n^2 variables, which is a continuous function in the product topology.
A: Here is a proof that does not use determinants. Given a matrix $A$ such that $I-A$ is invertible, we explicitly find a neighborhood of invertible matrices $(I-A)+H$ around $A$.
Choose
$$|H| < \frac{1}{|(I-A)^{-1}|},$$
where $|A|$ denotes the Frobenius norm of $A$. Now, notice that since $|-H(I-A)^{-1}| < 1$, we have that $I+H(I-A)$ is invertible. Therefore,
$$(I-A)^{-1} (I+H(I-A)^{-1})^{-1}
= ((I+H(I-A)^{-1})(I-A))^{-1}
= (I-A+H)^{-1}$$
is invertible as well. Hence, $U$ is open.
