Inscrutable proof in Humphrey's book on Lie algebras and representations This is a question pertaining to Humphrey's Introduction to Lie Algebras and Representation Theory
Is there an explanation of the lemma in §4.3-Cartan's Criterion? I understand the proof given there but I fail to understand how anybody could have ever devised it or had the guts to prove such a strange statement...
Lemma: Let $k$ be an algebraically closed field of characteristic $0$. Let $V$ be a finite dimensional vector space over $k$, and $A\subset B\subset \mathrm{End}(V)$ two subspaces. Let $M$ be the set of endomorphisms $x$ of $V$ such that $[x,B]\subset A$. Suppose $x\in M$ is such that $\forall y\in M, \mathrm{Tr}(xy)=0$. Then, $x$ is nilpotent.
The proof uses the diagonalisable$+$nilpotent decomposition, and goes on to show that all eigenvalues of $x$ are $=0$ by showing that the $\mathbb{Q}$ subspace of $k$ they generate has only the $0$ linear functional.
Added: (t.b.) here's the page from Google books for those without access:

 A: This lemma troubled me until I saw proof of Cartan's criterion for complex case in an appendix of Fulton & Harris book.
Denote $x_s = d(\lambda_1, \ldots, \lambda_n)$ the semisimple part of $x$, and $\bar{x}_s = d(\bar{\lambda}_1, \ldots, \bar{\lambda}_n)$, the matrix you get when you ''complex-conjugate'' $x_s$.
Lemma is then replaced with the following simple observation:
$\operatorname{Tr}(x \bar{x}_s) = |\lambda_1|^2 + \ldots + |\lambda_n|^2 = 0$ implies $x$ nilpotent.
For completeness, proof of Cartan's criterion using this fact:

Let $V$ be a finite-dimensional
  complex vector space, $L$ Lie subalgebra of $\mathfrak{gl} V$.
Let's assume that $\operatorname{Tr}(xy) = 0$, for all $x \in [LL], y \in L$.
  If we show that every $x \in [LL]$ is nilpotent, it will follow (by
  Engel's theorem) that $[LL]$ is nilpotent and that will imply that $L$
  is solvable. We'll do this using the aforementioned simple
  observation. Write $x = \sum_i [y_i, z_i]$.
  $$\operatorname{Tr}(x\bar{x}_s) = \sum_i \operatorname{Tr}([y_i,
 z_i]\bar{x}_s) = \sum_i \operatorname{Tr}(y_i, [z_i, \bar{x}_s])$$
  Now, we can use Lagrange's interpolation to write
  $\operatorname{ad}(\bar{x}_s)$ as a polynomial in
  $\operatorname{ad}(x_s)$ without constant term, so it follows that $[z_i, \bar{x}_s]$ is an
  element of $[L, L]$. Now, our assumption gives us $\operatorname{Tr}(x
 \bar{x}_s) = 0$.

As you can see, more general proof follows this one closely. 
Using the nice properties of complex conjugation, we don't have to check $\operatorname{Tr}(xy) = 0$ for all $y \in M$ to get $x$ nilpotent, we just have to check that for one element in $M$, namely $\bar{x}_s$.
I can imagine Cartan proving first the complex case, then generalizing the proof to the one you read in Humphreys.
A: This doesn't entirely answer your question but the key ingredients are (1) the rationals are nice in that their squares are non-negative (2) you can get from general field elements to rationals using a linear functional f (3) getting a handle on x by way of the eigenvalues of s.
