Why is it that $\mathbb{C}[\{m_{ij}\}]^{G} \subseteq \mathbb{C}[\textrm{tr}(m),\ldots, \det(m)]$?

Question

Let $G= GL(n,\mathbb{C})$ act on the set of all matrices $M_n$ by conjugation, i.e., for $g\in G$ and $m \in M_n$, $g\circ m = gmg^{-1}$.

Then if $m=(m_{ij})$, then the $G$-invariant ring $\mathbb{C}[\{m_{ij}\}]^{G}$ contains $\mathbb{C}[\textrm{tr}(m),\ldots, \det(m)]$ since the coefficients of the characteristic polynomial for $m$ are invariant under the action of $G$.

How does one prove the other containment? I saw an argument in some book awhile back but I don't remember where I saw it.

Thank you.

$\textit{Update}$: Rather than answering the question for any $n$,

how do you know that the only invariants for $GL(2,\mathbb{C})$ acting on $M_2(\mathbb{C})$ by conjugation are
$\mathbb{C}[\textrm{tr}(m),\det(m)]$? Does one use some sort of a dimension or stabilizer argument?

Answer 1

T. Springer, Invariant Theory, Theorem 1.5.7.