Degree $1$ birational map of $\mathbb{P}_k^n$ I wonder why the degree $1$ birational maps of $\mathbb{P}_k^n$ are the automorphisms of $\mathbb{P}_k^n$? In particular, why are they defined everywhere on $\mathbb{P}_k^n$?
I know each degree $1$ birational maps of $\mathbb{P}_k^n$ is of the form $\phi:=[f_0:...:f_n]$, where $f_i$ is a homogeneous linear polynomial. I know $\phi$ is defined everywhere iff the associated matrix (coefficients of $f_i$) of $\phi$ is corresponding to a matrix in $\text{PGL}_{n+1}(k)$, i.e. $\text{Aut}(\mathbb{P}_k^n)=\text{PGL}_{n+1}(k)$. But how to see the set of degree $1$ birational maps of $\mathbb{P}_k^n$ is $\text{Aut}(\mathbb{P}_k^n)$?
Also why does an automorphism of $\mathbb{P}^n$ have to be of degree $1$?

The key to this is the matrix of linear forms. If the matrix is not full rank, then the image of the map associated to this matrix of linear forms is supported inside some closed linear subvariety, which implies it is not birational. So birationality is equivalent to the matrix being full rank, which is equivalent to it being defined everywhere: the points where the map associated to the matrix isn't defined is exactly the projectivization of the kernel of this matrix.
