-1

The subset of all orthogonal matrices is compact .

Proof is given here

My confusion: we define $f : GLn(\mathbb{R}) \to GLn(\mathbb{R})$ by $f(A) =AA^T$

Here why $f(A)$ is continious?

My thinking :Here $f(A)$ is not a polynomial, so $f(A)$ will not continuous

If we take $A= (a_{ij}$) then $$AA^T=I=\begin{pmatrix} a_{11}^2 +...+a_{1n}^2 & 0 & .... & 0 \\ 0 & a_{21}^2 +...+a_{2n}^2 & .... & 0 \\ ... & .... & ... & 0 \\ 0 & 0 & ... & a_{n1}^2 +...+a_{nn}^2\end{pmatrix}$$

$\implies \sum_{k=1}^{n} a_{ik}^2=1$ for all $i=1,2,.....$

But in the proof it is written that $f(A)$ is continuous

I'm not getting that why $f(A)$ is continuous?

jasmine
  • 14,457

1 Answers1

2

On the contrary: each entry of $f(A)$ is a polynomial in the entries of $A$, and so it is continuous.

Robert Israel
  • 448,999
  • Here $a_{11}^2 +...+a_{1n}^2$ is not a polynomial ? im not getting that how each entry of $f(A)$ is a polynomial in the entries of $A$? Can you elaborate more – jasmine Apr 30 '21 at 19:11
  • 1
    $a_{11}^2 + \ldots + a_{1n}^2$ is a polynomial in the variables $a_{11}, \ldots, a_{1n}$. That is, it is an expression built from those variables and constants by addition, multiplication and non-negative integer powers. – Robert Israel Apr 30 '21 at 21:53