$\psi:A\rightarrow A^{-1}$ is continuous Define a map that takes a matrix to it's inverse. Give $A\in M(n\times n)$ over reals field, define:
$$\psi:A\rightarrow A^{-1}$$
Is it always continuous, where defined? How do I prove this?
Thanks.
 A: You could argue that, according to Cramer's rule, the matrix $A^{-1}$ is given by $$\frac{1}{\det A} Adj(A)$$ where Adj(A) is the adjugate matrix of A. Now, every element of Adj(A) is just a polynomial of the coefficients of the matrix A, and therefore continuous. Similarly, $\det(A)$ is also a polynomial in the coefficients of the matrix A. This should be enough to conclude what we need to.
A: Here is another approach that does not rely on Cramer's rule or properties of the determinant.
If $\|\cdot\|$ satisfies $\|AB\| \le \|A \| \|B\|$, then if $\|X\| < 1$, we have $(I+X)^{-1} = \sum_{k=0}^\infty (-1)^kX^k$. 
So we have $(A+H)^{-1} = (A(I+A^{-1}H))^{-1} = (I+A^{-1}H)^{-1} A^{-1}$. Using the above identity, and assuming that $\|H\| < \frac{1}{\|A^{-1} \|}$ (which imples $\|A^{-1}H\| < 1$), we have
$(A+H)^{-1} = (\sum_{k=0}^\infty (-1)^k (A^{-1}H)^k) A^{-1}$.
Hence we have 
\begin{eqnarray}
\|A^{-1} - (A+H)^{-1} \| &\le& \sum_{k=1}^\infty \|H\|^k \|A^{-1}\|^{k+1} \\
&=& \|H\| \|A^{-1}\|^{2} \sum_{k=1}^\infty (\|H\| \|A^{-1}\|)^{k-1} \\
&\le& \|H\| \frac{\|A^{-1}\|^{2}}{1-\|H\| \|A^{-1}\| }
\end{eqnarray}
If we assume $\|H\| < \frac{1}{2\|A^{-1} \|}$, the above estimate simplifies to 
$\|A^{-1} - (A+H)^{-1} \| \le 2 \|A^{-1}\|^{2} \|H\|$, from which it follows that the operator $A \mapsto A^{-1}$ is continuous.
The above shows that, in fact, the operator $A \mapsto A^{-1}$ is smooth.
