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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.

Davide Giraudo
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ChikChak
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    Can you first show that the inverse operator is continuous? Have a look at -http://math.stackexchange.com/questions/810670/continuity-of-the-inverse-matrix-function – Shraddheya Shendre Nov 25 '16 at 15:21
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    An $m \times n$ matrix of real numbers is essentially the same as a vector in $\mathbb{R}^{m \times n}$. So a map from $m \times n$ matrices to $r \times s$ matrices is a map from $\mathbb{R}^{m \times n}$ to $\mathbb{R}^{r \times s}$, and it is continuous provided it is a continuous as a map from $\mathbb{R}^p$ to $\mathbb{R}^q$ where $p=m\times n$ and $q=r\times s$. – kccu Nov 25 '16 at 15:23
  • @ShraddheyaShendre Well,the initial questions was to prove that the inverse operator in sontinuous,so I used the equation of inverse matrices and adj.I know that det mapping is continuous,but I don't understand why adj mapping is continuous... – ChikChak Nov 25 '16 at 15:30
  • @WakaWaka123 - See the question in my previous comment. In that sense, your question is duplicate. You should have searched before asking. – Shraddheya Shendre Nov 25 '16 at 15:37
  • @ShraddheyaShendre And yet,They don't explain there why the mapping to adj is continuous... – ChikChak Nov 25 '16 at 15:41
  • @kccu But how can I know that the mapping from $R^p$ to $R^q$ is continuous? – ChikChak Nov 25 '16 at 15:43
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    @WakaWaka123 If $f: \mathbb{R}^p \to \mathbb{R}^q$, then $f$ is continuous provided that for all $x \in \mathbb{R}^p$ and all $\epsilon>0$, there exists $\delta>0$ such that whenever $||x-y||< \epsilon$, $||f(x)-f(y)||< \delta$. Equivalently, whenever ${x_n}{n \geq 0}$ is a sequence in $\mathbb{R}^p$ such that $\lim{n\to\infty} x_n=x$, we have $\lim_{n \to \infty} f(x_n)=f(x)$. Just the usual notion of continuity for functions between metric spaces. – kccu Nov 25 '16 at 16:19

2 Answers2

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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.

Rodrigo de Azevedo
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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 $\text{Adjoint} (A_n)\to \text{Adjoint} A$

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