Continuity of a matrix valued function

Question

Let $f:\mathbb{R}^{(n,n)}\to\mathbb{R}^{(n,n)}$ be a function such that $A\mapsto CAC^{-1}$ for some regular $C\in\mathbb{R}^{(n,n)}$. How can I show that $f$ is continuous? How exactly is continuity of a function from matrix spaces to matrix spaces defined?

Answer 1

Let $X=\mathbb{R}^{(n,n)}$. The standard metric of $X$ is obtained by identifying $\mathbb{R}^{n^2}$, for example, let $n=2$ and $$ A=\begin{pmatrix} a & b \\ c & d \end{pmatrix} $$, then metric of $A$ is $\sqrt{a^2+b^2+c^2+d^2}$. Hence the continuous fuctions between $X$ is just a continuous function of this induced topology.

Now let $f=(f_1,\cdots,f_{n^2})$ (Identifying $X$ and $\mathbb{R}^{n^2}$.) You can see that $f_i : \mathbb{R}^{n^2} \rightarrow \mathbb R$ is just a polynomial of several variable, so is continuous. Therefore $f$ is also continous.

Answer 2

Continuity of a function depends on the metrics of your target space and your domain space. Usually, the norm of a matrix $A \in \mathbb{R}^{n \times n}$ is given by

$$\Vert A \Vert = \sup\{ \Vert Au \Vert : u \in \mathbb{R}^n \text{ and } \Vert u \Vert = 1\}.$$

This gives us the metric $d(A, B) = \Vert A - B \Vert$ on a space of matrices.

A useful property for this problem is that $\Vert AB \Vert \leq \Vert A \Vert \Vert B \Vert$ for square matrices. This is not too hard to show:

Let $u \in \mathbb{R}^n$ with norm $1$. Then

$$\Vert ABu \Vert = \Vert A(Bu) \Vert \leq \Vert A \Vert\Vert Bu \Vert \leq \Vert A \Vert \Vert B \Vert \Vert u \Vert = \Vert A \Vert \Vert B \Vert \implies \Vert AB \Vert \leq \Vert A \Vert \Vert B \Vert.$$

It follows that if $C$ is invertible, then $\Vert CC^{-1} \Vert = \Vert I_n \Vert = 1.$

Now we're ready to prove continuity.

Let $\epsilon > 0$. If we choose $\delta = \epsilon$, then we that whenever $\Vert A - B \Vert < \delta$,

$$\Vert f(A) - f(B) \Vert = \Vert C(A - B)C^{-1} \Vert \leq \Vert C \Vert \Vert A - B \Vert \Vert C^{-1}\Vert< \delta = \epsilon$$

so $f$ is continuous.