Is the special linear group of order $n$ bounded in the $\mathbb{R}^{n \times n}$ metric?
Consider the set $M (n, \mathbb{R})$ equipped with the metric defined as follows:
For a matrix $A$, consider $A^T A$. Then, the distance from $0$, the zero matrix, will be given by the Euclidean distance of $A^T A$ from $0$ for $\mathbb{R}^{n^2}$. Investigate if the following sets are bounded.
(a) The set of all orthogonal matrices. This set is often denoted by $O (n, \mathbb{R})$
(b) The set $SL (n, \mathbb{R})$ of all matrices with determinant $1$. This is often termed as special linear group order $n$.
I think problem a is quite trivial since $A^{T}A=I$ for all orthogonal matrices so they're all bounded. For the special linear group, $A^{T}$ also has determinant $1$ so $A^{T}A$ is a symmetric matric with determinant $1$. However, I don't remember any theorem in linear algebra that gives us information about the entries of a matrix from the determinant, so we can have some entries arbitrarily large and we could still have a small determinant, so I think the special linear group is not bounded.
Best Answer
Yes, $SL(n,\Bbb R)$ is unbounded (if $n>1$). For instance, if $m\in\Bbb N$, then$$A(m)=\begin{bmatrix}m&0&0&\ldots&0&0\\0&\frac1m&0&\ldots&0&0\\0&0&1&\ldots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\ldots&1&0\\0&0&0&\ldots&0&1\end{bmatrix}\in SL(n,\Bbb R).$$However$$A(m)^TA(m)=A(m)^2=\begin{bmatrix}m^2&0&0&\ldots&0&0\\0&\frac1{m^2}&0&\ldots&0&0\\0&0&1&\ldots&0&0\\\vdots&\vdots&\vdots&\ddots&\vdots&\vdots\\0&0&0&\ldots&1&0\\0&0&0&\ldots&0&1\end{bmatrix}.$$