I would like to show that $U(n)$, the unitary group of degree $n$, is a properly embedded Lie subgroup of $GL(n, \mathbb C)$ of dimension $n^2$, and find the matrices that are in the tangent space at the identity.
To show the first part, I think I can argue it is a kernel $\Phi^{-1}(I_n)$, where $\Phi(A) = \overline {A^T}A$. However, I am not sure how to show its dimension, and how to identify matrices that are in the tangent space at the identity.
To figure out what the tangent space to $U(n)$ at the identity, use the trajectories of curves definition of the tangent space. Let the tangent space to the identity of $U(n)$ be denoted $\mathfrak{u}(n)$. Let $A\in \mathfrak{u}(n)$ and let $\gamma:(-\varepsilon,\varepsilon)\to U(n)$ so that $\gamma(0)=I_n$ and $\gamma'(0)=A$. By definition, as $\gamma(t)\gamma(t)^\dagger=I_n$, and so differentiating at $t=0$ we get $$ 0=\frac{d}{dt}\bigg|_{t=0}\gamma(t)\gamma(t)^\dagger=\gamma'(0)\gamma(0)^\dagger+\gamma(0)\gamma'(0)^\dagger=A+A^\dagger. $$ You can use this idea to argue that $\mathfrak{u}(n)$ consists of the set of $n\times n$ skew-Hermitian matrices. I.e. $$ \mathfrak{u}(n)=\{A\in \mathfrak{gl}(n,\Bbb{C}):A=-A^{\dagger}\}. $$ Next, you need to come up with a basis for this space and hence count its dimension. For $n=2$, a basis over $\Bbb{R}$ is given by $$ \begin{bmatrix} i&0\\ 0&0 \end{bmatrix}, \:\: \begin{bmatrix} 0&0\\ 0&i \end{bmatrix},\:\: \begin{bmatrix} 0&1\\ -1&0 \end{bmatrix},\:\: \begin{bmatrix} 0&i\\ i&0 \end{bmatrix}. $$ I leave the generalization of this to you.