gl(n,C) as a direct sum?

Question

I've been trying to prove this:

$gl_n(\mathbb{C})=Sl_n(\mathbb{C})\oplus \mathbb{C}I_n $

where $gl_n(\mathbb{C})$ is the General Linear Lie Algebra and $Sl_n(\mathbb{C})$ is the Special Linear Lie Algebra.

Thanks!

Answer 1

It's trivial that both subspaces are ideal of $\mathfrak{gl}_n(\mathbb{C})$.In addition the direct sum is true in vector space sense.Thus we just need to show that $[\mathfrak{sl}_n(\mathbb{C}),\mathbb{C}I_n]=0$.It's obivious.

Answer 2

You can use the trace map $tr:gl_n(\mathbb{C}) \rightarrow \mathbb{C}I_n$ sending an endomorphism $x \in gl_n(\mathbb{C})$ to $tr(x)I_n$. This is linear, surjective, and the kernel will be $sl_n(\mathbb{C})$ so you can use the first isomorphism theorem to get your vector space direct sum.