Trace of product of $SU(N)$ generators positive definite I am studying the $SU(N)$ group. Let's name $F_a, F_b, \dots$ the $N^2-1$ generators of $SU(N)$ (so Hermitian traceless matrices). There is a theorem that states that there is a set of generators such that:
$$
\operatorname{Tr}(F'_a F'_b)=\frac{1}{2}\delta_{ab}
$$
In order to prove this theorem we need to prove that the matrix $g_{ab}$ defined:
$$
g_{ab}=\operatorname{Tr}(F_a F_b)
$$
is a symmetric, real and positive definite matrix. Although the first two statements are easy to prove, I am having trouble to prove the positiveness.
Any suggestion?
 A: Since the question seems to have gone unanswered, I'll provide a sketch.
We compute:
\begin{align}
\sum_{ab} v_a^* g_{ab} v_b &= \sum_{ab} v_a^* \text{Tr}(F_a F_b) v_b
\\&= \sum_{ab} v_a^* \left( \sum_c(F_a F_b)_{cc} \right) v_b
\\&= \sum_{ab} v_a^* \left( \sum_c \sum_d (F_a)_{cd} (F_b)_{dc} \right) v_b
\\&= \sum_{ab} v_a^*\left(\sum_c \sum_d  (F_a)_{dc}^* (F_b)_{dc} \right) v_b \tag{By Hermiticity of the $F_b$}
\\&= \sum_{dc} \left(\sum_{a} (F_a)_{dc} v_a \right)^* \left( \sum_b (F_b)_{dc} v_b\right) \tag{Rearranging}
\\&=\sum_{dc} \left(\sum_{a} (F_a)_{dc} v_a \right)^* \left( \sum_a (F_a)_{dc} v_a\right) \tag{Changing a summation index}
\\&= \sum_{dc} \left|\sum_{a} (F_a)_{dc} v_a \right|^2 
\\&\geq 0
\end{align}
Note that this only shows that $g$ is positive semidefinite. The proof of definiteness escapes me, but all you have to do is show that $g$ is invertible. Somehow we must have to use the fact that the $F_a$ are a basis for the Hermitian traceless matrices. Having a basis and invertibility go hand in hand. The proof will be something like "Suppose $gv = 0$, then... $A$ (some matrix to be determined) is Hermitian and traceless, so $A$ is a linear combination of the $\{F_a\}$... Since the $\{F_a\}$ are linearly independent, $v = 0$." I'll fill in the details if I figure them out at work tomorrow.
A: My original answer is a reasonable direct approach, but, as you can see, proving definiteness is quite difficult.
Here's a less direct approach that gets the job done, but leans on a nontrivial theorem for the conclusion. We will prove that $\langle A , B \rangle \equiv \text{Tr}(AB)$ is an inner product on the space of Hermitian traceless matrices.
Bilinearity is immediate from the linearity of matrix multiplication and the linearity of trace.
It is a standard result that $\text{Tr}(AB) = \text{Tr}(BA)$. Note that $\langle A, B \rangle$ is a real inner product because the product of Hermitian matrices is again Hermitian and so has real trace (since the trace is the sum of the eigenvalues and Hermitian matrices have real eigenvalues).
Now to show that $\langle \cdot , \cdot \rangle$ is positive definite. We have
\begin{align}
\langle A , A \rangle &= \text{Tr}(AA)
\\&= \sum_b (AA)_{bb}
\\&= \sum_b \sum_c A_{bc} A_{cb}
\\&= \sum_b \sum_c A_{bc} A_{bc}^* \tag{$A$ is Hermitian}
\\&= \sum_b \sum_c |A_{bc}|^2 \geq 0
\end{align}
This shows that $\langle \cdot , \cdot \rangle$ is positive. To see that it is definite, suppose $\langle A , A \rangle = 0$. Then the calculation above shows $\sum_{bc} |A_{bc}|^2 = 0$. But this can only happen if $A_{bc} = 0$ for every $b$ and $c$, i.e. if $A = 0$.
Thus, we have established that $\langle A, B \rangle \equiv \text{Tr}(AB)$ is an inner product on the space of Hermitian traceless matrices. But now $g$ is just the matrix representation of that inner product in the $\{F_a\}$ basis. Since $g$ is the matrix of an inner product, it is a symmetric, real, and positive definite matrix.
In fact, more is true. Since $g$ is the matrix of a real inner product, Sylvester's law of inertia (this is the promised nontrival theorem, although it is a very fundamental result) guarantees that there is a real invertible matrix $S$ such that $SgS^T = I$ In other words, you can write each basis element $F_a'$ as a real linear combination of the $\{F_a\}$.
I think this is the way the problem should be solved. Since we needed Sylvester's law of inertia to conclude, I doubt there's going to be a simpler way to get there. You'll either quote the law of inertia or end up proving it yourself.
