Maximal tori of $SO(n)$ and $Sp(n)$ I know that the maximal torus of the group $GL_n (\mathbb{C})$ is the set of all diagonal matrices
$$
\left(
\begin{matrix}
x_1 & 0 & \cdots & 0 \\
0 & x_2 & \cdots & 0 \\
0 & 0 & \ddots & 0 \\
0 & 0 & \cdots & x_n
\end{matrix}
\right)
$$
But what are the maximal tori of $SO(n)$ and $Sp(n)$? What about the maximal split tori?
 A: What a maximal torus looks like depends on your choice of bilinear form.  For example, in the split group $G = \operatorname{SO}_{2n}(\mathbb C)$, we can choose the symmetric, nondegenerate bilinear form $B: \mathbb C^{2n} \times \mathbb C^{2n} \rightarrow \mathbb C$ given by
$$B(x_1, ... , x_{2n}, y_1, ... , y_{2n}) = x_1 y_{2n} + x_2y_{2n-1} + \cdots + x_{2n}y_1$$
By definition, $G$ is the group of invertible linear operators $T$ on $\mathbb C^{2n}$ satisfying $B(Tv,Tw) = B(v,w)$ for all $v, w \in \mathbb C^{2n}$.
In coordinates, if  $J_0$ is matrix with $1$s on the antidiagonal, and zeroes elsewhere, and 
$$J = \begin{pmatrix} & J_0 \\ J_0 &  \end{pmatrix}$$
then 
$$G = \{ g \in \operatorname{GL}_{2n}(\mathbb C) : \space ^tg Jg = J \}$$
Up to isomorphism, this does not depend on the choice of nondegenerate, symmetric bilinear form $B$ (equivalently, it does not depend on the choice of invertible symmetric matrix $J$).  If you prefer another matrix $J'$ in place of $J$, you will define another special orthogonal group $G'$, and there will exist a $g \in \operatorname{GL}_{2n}(\mathbb C)$ such that $gGg^{-1} = G'$.
To find a maximal torus of $G$, we let $D$ be the group of diagonal invertible matrices in $\operatorname{GL}_{2n}(\mathbb C)$.  It is easy to see that 
$$T := D \cap G = \{ \begin{pmatrix} t_1 \\ & \ddots \\ & & t_n \\ & & & t_n^{-1} \\ & & & & \ddots \\ & & & & & & t_1^{-1} \end{pmatrix} \}$$
It is clear that $T$ is a torus in $G$.  I claim it is moreover a maximal torus in $G$.  If not, then the general theory of linear algebraic groups tells you that $T$ is contained in a maximal torus $T_1$ of $G$.  Every element of $T_1$ must commute with every element of $T$.  But there are no elements in $\operatorname{GL}_{2n}(\mathbb C)$ which commute with every element of $T$, except for diagonal matrices.  Thus $T_1$ must consist of diagonal matrices, which forces $T_1$ to equal $T$.
