Ok, so I'm asking this in physics because I'm currently working through part of Srednicki's text on QFT, even though it's really a maths question.

In Srednicki's chapter on non-Abelian gauge theory, he introduces the generators of a Lie group. At the moment we're only analysing $SU(N)$, which is defined by $M M^\dagger = 1$ and $\det(M) = 1$ for all $M \in SU(N)$

And the corresponding conditions on the generators of the group are $T = T^\dagger$ and ${\rm Tr}(T) = 0$ for all $T \in \mathfrak{su}(N)$

Then what I don't understand is that Srednicki tells me that we should normalise our generators so that $${\rm Tr}(T^i T^j) = \frac{1}{2}\delta^{ij}.$$

So presumably this arises because our set of $N^2-1$ generators is a basis for the tangent space of $SU(N)$ at the identity, and we choose it to be orthogonal and then need a condition to normalise the lengths of all of the basis vectors?
Why did the condition Srednicki gave do that? And where did we input that the vectors are orthogonal?