= Solution
Over $\mathbb R$ or $\mathbb C$, a <Lie algebra> is a <vector space> with a <bilinear map> $[\ ,\ ]$ which is alternating and obeys the <Jacobi identity>:
$$
[X,X]=0,\qquad [X,[Y,Z]]+[Y,[Z,X]]+[Z,[X,Y]]=0.
$$
Bilinearity and alternation imply $[X,Y]=-[Y,X]$.
For a <Matrix Lie group>, identify the <tangent space> $\mathfrak g=T_I G$ with derivatives $\gamma'(0)$ of smooth curves through $I$. Product curves show that the sum of two such derivatives is again tangent, and reparametrization supplies scalar multiples. In particular this is a real <vector space>, even when the matrices have complex entries. For $X,Y\in\mathfrak g$, choose a curve $\gamma(s)$ with $\gamma'(0)=X$. Conjugating a curve with derivative $Y$ shows that
$$
\gamma(s)Y\gamma(s)^{-1}\in\mathfrak g\qquad\text{for every sufficiently small }s.
$$
Differentiate this curve in the finite-dimensional <vector space> $\mathfrak g$. The result is
$$
\boxed{[X,Y]=XY-YX\in\mathfrak g.}
$$
The matrix <commutator> is bilinear and alternating, and expanding the six terms proves its <Jacobi identity>. Thus this construction gives the <Lie algebra of a matrix Lie group>, with the appropriate bracket, using actual group curves rather than an assumed commutator closure.
For the <unitary group>, differentiating $\gamma(t)^\dagger\gamma(t)=I$ gives $X^\dagger+X=0$. Conversely, if $X^\dagger=-X$, then $e^{tX}$ is unitary and is a curve with derivative $X$. Consequently
$$
\boxed{\mathfrak u(N)=\{X:X^\dagger=-X\},\qquad\dim_{\mathbb R}\mathfrak u(N)=N^2.}
$$
This is the <unitary Lie algebra>. The diagonal entries are purely imaginary, contributing $N$ real parameters, and each upper off-diagonal entry contributes two real parameters. A <matrix-unit basis of the unitary Lie algebra> is
$$
iE^{(i,i)}\quad(1\leq i\leq N),\qquad E^{(i,j)}-E^{(j,i)},\quad i(E^{(i,j)}+E^{(j,i)})\quad(1\leq i<j\leq N).
$$
These $N+2\binom N2=N^2$ anti-Hermitian <matrix units> and their combinations are linearly independent over $\mathbb R$ and span every allowed entry.
The symplectic stabilizer is a <subgroup>: the identity preserves $J$, and if $MJM^T=J$ and $NJN^T=J$, then
$$
(MN)J(MN)^T=M(NJN^T)M^T=J.
$$
Multiplying $MJM^T=J$ on the left by $M^{-1}$ and on the right by $M^{-T}$ gives $M^{-1}JM^{-T}=J$. Products and inverses remain unitary. This is the <compact symplectic group>, often denoted $USp(2n)$ or $Sp(n)$.
Differentiating the stabilizer equation gives $XJ+JX^T=0$. For $X=\begin{pmatrix}A&B\\C&D\end{pmatrix}$, this says
$$
B^T=B,\qquad C^T=C,\qquad D=-A^T.
$$
Together with $X^\dagger=-X$, these are equivalently
$$
\boxed{X=\begin{pmatrix}A&B\\-\overline B&\overline A\end{pmatrix},\qquad A^\dagger=-A,\quad B^T=B.}
$$
These conditions are also sufficient: $e^{tX}$ is unitary, and
$$
\frac{d}{dt}\bigl(e^{tX}Je^{tX^T}\bigr)=e^{tX}(XJ+JX^T)e^{tX^T}=0.
$$
Hence it stays in the subgroup. They characterize its <compact symplectic Lie algebra> without adding any trace condition. In fact the trace automatically vanishes. The free anti-Hermitian block $A$ has $n^2$ real parameters, and the complex <symmetric matrix> $B$ has $n(n+1)$ real parameters. Thus
$$
\boxed{\dim_{\mathbb R}\mathfrak h=n^2+n(n+1)=n(2n+1).}
$$
Here is a complete <matrix-unit basis of the compact symplectic Lie algebra>. The generators supplying $A$ are
$$
H_i=i(E^{(i,i)}-E^{(i+n,i+n)})\quad(1\leq i\leq n),
$$
$$
P_{ij}=E^{(i,j)}-E^{(j,i)}+E^{(i+n,j+n)}-E^{(j+n,i+n)},
$$
$$
Q_{ij}=i\bigl(E^{(i,j)}+E^{(j,i)}-E^{(i+n,j+n)}-E^{(j+n,i+n)}\bigr)\quad(i<j).
$$
The two families supplying the real and imaginary parts of $B$ are, for $1\leq i\leq j\leq n$,
$$
U_{ij}=\frac{E^{(i,j+n)}+E^{(j,i+n)}-E^{(i+n,j)}-E^{(j+n,i)}}{1+\delta_{ij}},
$$
$$
V_{ij}=\frac{i\bigl(E^{(i,j+n)}+E^{(j,i+n)}+E^{(i+n,j)}+E^{(j+n,i)}\bigr)}{1+\delta_{ij}}.
$$
The denominator merely avoids double-counting diagonal entries. Each displayed generator obeys both defining tangent conditions. The $H,P,Q$ generators form a real basis of the allowed $A$ blocks, and the $U,V$ generators form a real basis of the complex symmetric $B$ blocks. Thus they are independent and their total number is $n(2n+1)$. \b[They give all generators required by the real compact algebra, including the case $n=1$.]
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