Why is conjugation by a diagonal matrix a semisimple automorphism of $\textrm{GL}_n$? Let $$s = \begin{pmatrix} \lambda_1 & & & 0\\ & \lambda_2 & \\ & & \ddots \\ 0& & & \lambda_n \end{pmatrix}$$ be a diagonal invertible matrix.  Let $G = \textrm{GL}_n (k)$ ($k$ algebraically closed as usual), and $\sigma:G \rightarrow G$ the automorphism given by $x \mapsto sxs^{-1}$.  Let $A = k[G]$, and $\sigma^{\ast}: A \rightarrow A$ the $k$-algebra homomorphism $\sigma^{\ast}(f)(x) = f(sxs^{-1})$.  I'm trying to understand why $\sigma^{\ast}$ is semisimple, in the sense that if $W$ is any finite dimensional $k$-vector subspace of $A$ which is stable under $\sigma^{\ast}$, then the restriction of $\sigma^{\ast}$ to $W$ defines a diagonalizable linear transformation.
We have that $s(a_{ij})s^{-1} = (\lambda_i \lambda_j^{-1}a_{ij})$ for any $(a_{ij}) \in G$.  Therefore, $\sigma^{\ast}$ has the effect on monomials $$\prod\limits_{i,j} X_{ij}^{n_{ij}} \mapsto \prod\limits_{i,j} [\lambda_i \lambda_j^{-1}]^{n_{ij}} X_{ij}^{n_{ij}}$$ Also, $\sigma^{\ast}$ stabilizes the determinant polynomial: $$\sigma^{\ast}(\textrm{Det}) = \sum\limits_{\tau \in S_n} \textrm{sgn}(\tau) \sigma^{\ast}\prod\limits_{i=1}^n X_{i, \tau(i)}$$ with $\sigma^{\ast}\prod\limits_{i=1}^n X_{i, \tau(i)} = \prod\limits_{i=1}^n \lambda_i \lambda_{\tau(i)}^{-1} X_{i, \tau(i)} = \prod\limits_i X_{i, \tau(i)}$.  We can identify $k[G]$ with the localization of the polynomial ring in $X_{ij}$ at the determinant polynomial, and it follows that $\sigma^{\ast}$ acts on $k[G]$ by scaling the expressions $\frac{\prod\limits_{i,j} X_{ij}^{n_{ij}}}{\textrm{Det}^k}$, which span $k[G]$ as a vector space over $k$.
I'm trying to put all this information together to conclude that every $\sigma^{\ast}$-stable finite dimensional subspace of $k[G]$ has a basis consisting of $\sigma^{\ast}$-eigenvectors.  Do you think this is a viable approach?
 A: Actually I think we are done by the above argument.  If $V$ is a (non-necessarily finite dimensional) vector space, $\phi: V \rightarrow V$ is a vector space isomorphism, and $V$ is a union of finite dimensional $\phi$-stable subspaces $W_i$ such that $\phi|W_i$ is semisimple, then $\phi$ is semisimple.  
This is follows on account of two facts: (i) for any finite collection $S$ of the $i$, the restriction of $\phi$ to $\sum\limits_{i \in S} W_i$ is a semisimple linear endomorphism of $\sum\limits_{i \in S} W_i$, and (ii): if $W \subseteq W'$ are finite dimensional $\phi$-invariant subspaces, and $\phi|W'$ is semisimple, then so is $\phi|W$.  
It follows that $\phi$ is semisimple, because if $W$ is a finite dimensional $\phi$-invariant subspace, then $W$ is contained in a finite sum of the $W_i$.  
Now let $V = k[G]$, and $\phi = \sigma^{\ast}$.  We have shown that there is a basis $v_i$ for $V$, and scalars $0 \neq c_i \in k$, such that $\phi(v_i) = c_iv_i$ for all $i$.  Here the $v_i$ are monomials divided by some power of the determinant.  Hence the restriction of $\phi$ to $kv_i$ is a $\phi$-stable semisimple linear transformation.  One can take as the $W_i$ all possible finite spans of the $v_i$.  By fact (i), all these $W_i$ are $\phi$-invariant.
