Mapping matrix to its adjugate is continuous

Question

I need help in proving that the function $f(A) =\operatorname{adj} (A)$ is continuous.

I'm not sure at all how to work with continuity and matrices.. Thanks.

Answer 1

Let $A_n$ be a sequence such that $A_k\to A$ then there exists $m$ such that $n\ge m\implies $$||A_k-A||<\epsilon $ forall $\epsilon>0$

Now $||A||=\sum _{i,j=1}^n|a_{ij}|^2$

Hence $A_k\to A\implies ||A_k-A||=\sum_{i,j=1}^n|a_{i,j}^{(k)}-a|^2<\epsilon$

Hence $\text{Adjoint} (A_n)\to \text{Adjoint} A$

Answer 2

Let $\mathrm F : \mathbb R^{n \times n} \to \mathbb R^{n \times n}$ be defined by

$$\mathrm F (\mathrm X) = \mbox{adj} (\mathrm X)$$

Note that each entry of matrix $\mathrm F (\mathrm X)$ is a cofactor of $\mathrm X$, i.e., $\pm 1$ multiplied by the determinant of a $(n-1) \times (n-1)$ submatrix of $\mathrm X$. Thus, each entry of matrix $\mathrm F (\mathrm X)$ is a polynomial function of the entries of $\mathrm X$ and, therefore, $\mathrm F$ is a continuous function.