Degrees of Freedom of a Unitary Matrix Proof My book states that a 3$\times$3 unitary matrix has 9 degrees of freedom, but I have trouble seeing why that is true. So I want to try and show that an $n\times n$ matrix has $n^2$ degrees of freedom. 
Base case $n=1$: This is just a matrix with one entry which has length 1. It is clear that any entry with length 1 will do. So I have $1 = 1^2$ degree of freedom
$n \implies n+1$: Suppose we have an $n \times n$ unitary matrix and it has $n^2$ degrees of freedom. I want to now show that a $(n+1) \times (n+1)$ unitary matrix has $(n+1)^2$ degrees of freedom. If I add a $(n+1)$th row and column to the $n \times n$ unitary matrix, I have added $(n+1) + n = 2n + 1$ entries. For every entry $a_{ij}$ I have added, I need to ensure that each $a_{ij}^2 = 0$ in order for the matrix to be unitary. Since there are numerous ways to choose $a_{ij}$ such that $a_{ij}^2$ = 0, I have added $2n+1$ degrees of freedom. So the total degrees of freedom is $n^2 + 2n + 1 = (n+1)^2$.
Can I get some feedback on this proof? Thanks in advance!
Note It seems like my approach above is at a dead end.
 A: copper.hat has already given an answer in the comments. This is the best answer I can think of. But there are also alternative "pieces of evidence" in support of $n^2$ degrees of freedom. Maybe they will be useful too.
We know that the identity matrix $I$ is unitary. Let us look at unitary matrices that are really close to $I$, i.e. at matrices $A=I+\Delta$, where all the elements of $\Delta$ are very small. Since $A$ is unitary, we have
$$
    I = A^H A = (I+\Delta)^H (I + \Delta) = I + \Delta + \Delta^H + \Delta^H \Delta,
$$
and so
$$
    0 = \Delta + \Delta^H + \Delta^H \Delta \approx \Delta + \Delta^H.
$$
In the last approximate equality I used the fact that elements in $\Delta^H \Delta$ are way smaller than elements in $\Delta$ and $\Delta^H$. So, $\Delta$ is approximately skew-Hermitian. And it is quite easy to see that skew-Hermitian matrices have $n^2$ degrees of freedom, because you can choose any complex numbers above the main diagonal (this gives you $2\cdot\frac{n^2-n}{2}$ degrees of freedom) and you can choose any imaginary numbers on the main diagonal (this is $n$ more degrees of freedom). This isn't rigorous, especially in that part when we use the word "approximate". But we saw that there is some sort of a connection between unitary and skew-Hermitian matrices.
This connection is deeper than it seems at first sight. For instance, it can also be demonstrated using the matrix exponential. It turns out that if matrix $X$ is skew-Hermitian, then $e^X$ is unitary. Even more, any unitary matrix is in fact a matrix exponential of some skew-Hermitian matrix. So it's no big surprise that their numbers of degrees of freedom are the same.
Finally, if you'd like to find out how all this stuff can be formulated rigorously, then differential geometry can give you the answers. One of the central notions in differential geometry is that of a smooth manifold. copper.hat's comment can be turned into a rigorous proof of the fact that $U(n)$, the set of all unitary $n \times n$ matrices, is an $n^2$-dimensional submanifold of the Euclidean space $\mathbb{R}^{2n^2}$.
