Please could someone check my work on this exercise (from book I am reading). Thanks!
Exercise:
Prove that $GL_n (\mathbb K)$ is non-compact when $n \ge 1$. Prove that $SL_n (\mathbb K)$ is non-compact when $n \ge 2$. What about $SL_1 (\mathbb K)$?
My solution:
Note that $nI \in GL_n$ and $\|nI\| = n \cdot \sqrt{n}$ where $\|\cdot\|$ denotes the Euclidean metric hence $GL_n$ is unbounded.
Let $D_n$ be the diagonal matrix with entries $d_{11} = n, d_{22}={1\over n}$ and $d_{ii}=1$ otherwise. Then $\det D_n = 1$ and $\|D_n\| \ge n$ hence $SL_n$ is unbounded.
$SL_1 (\mathbb R) = \{-1,1\}$ hence is bounded. $SL_1 (\mathbb C) = S^1$ hence is bounded. For the quaternions $\mathbb H$, $SL_1 (\mathbb H) = S^3$ hence is bounded.