Linear subvarieties of $\Bbb{P}^n$ and change of coordinates I am looking at problem 4.11 of Fulton's Algebraic Curves. He asks to show that if $V = V(H_1,\ldots, H_r)$ where $H_i's$ are all linear forms of degree $1$, then there is a projective change of coordinates $T$ of $\Bbb{P}^n$ such that 
$$T^{-1}V = V(X_{m+2},\ldots,X_{n+1}).$$
The number $m$ is then called the dimension of $V$. At first the problem seemed straightforward to me: we have $H_i = \sum_{j=1}^{n+1} a_{ij}X_j$ and considering the $r \times (n+1)$ matrix $A = (a_{ij})$, we know that there are invertible $r \times r$ and $(n+1) \times (n+1)$ matrices $Q,P$ such that 
$$A':= QAP^{-1}$$
is a matrix with a $(\operatorname{rank} A) \times (\operatorname{rank} A) $ copy of the identity as a lower right block and zeros everywhere else. 


My problem is: The thing is this operation of doing $QAP^{-1}$ is not a projective change of coordinates. For it to be one, it seems I am only allowed to right multiply and not left multiply. I am getting confused with something fundamental here? 


 A: What you are trying to do is to find bases of $k^r,k^{n+1}$ such that the linear map
$$
k^{n+1} \rightarrow k^r
$$
given by the $F_1,...,F_r$ "looks nice". But what Fulton wants you to do is to move the zero set of the $F_1,...,F_r$ inside $\Bbb{P}^n$. Do you see the difference?
More explicitly, a projective change of coordinates of $\Bbb{P}^n$ is by definition an automorphism of $\Bbb{A}^{n+1}$, given by linear forms. So indeed, your $QAP^{-1}$ is not a projective change of coordinates, since it is a linear map from $k^{n+1}$ to $k^r$ (I use the vector space notation, to make the distinction clear).
What you can do to solve the problem is, instead of changing bases, just change variables.
Given $V=V(F_1)$, with $F_1=\sum_{i=1}^{n+1} a_{1i}X_i$, we can assume w.l.o.g that $a_{1n+1} \neq 0$. Define a polynomial map $k[X_1,...,X_{n+1}] \rightarrow k[X_1,...,X_{n+1}]$ by leaving $X_1,...,X_n$ unchanged and mapping
$$
X_{n+1} \mapsto \frac{1}{a_{1n+1}} \left(X_{n+1}-\sum_{i=1}^n a_{1i}X_i \right)
$$
The associated automorphism $T_1: \Bbb{A}^{n+1} \rightarrow \Bbb{A}^{n+1}$ satisfies $T_1^{-1}(V(F_1))=V(X_{n+1})$.
Now use induction on $r$ to finish the proof.
