# Unique linear transformation carrying projective hyperplane to projective hyperplane and point to point - a concise proof

Let $$\mathbb P^n$$ denote the projective $$n$$-space over an algebraically close field $$k$$, i.e. $$\mathbb P^n$$ is given by $$(\mathbb A^{n+1}\setminus \{0\})/ \sim$$ where $$\mathbb A^{n+1}$$ is the affine $$(n+1)$$-space and $$\sim$$ is the equivalence relation identifying points which are scalar multiples of one another, that is for two points $$(a_0, \cdots , a_n)$$ and $$(b_0, \cdots , b_n)$$ in $$\mathbb A^{n+1}$$, $$(a_0, \cdots , a_n) \sim (b_0, \cdots , b_n) \iff \exists \hspace{1mm} \lambda \in k^\times \text{ s.t. } b_j = \lambda a_j \text{ for all }1 \leq j \leq n$$ By a hyperplane in $$\mathbb P^n$$, I shall mean the zero set of some linear homogenous polynomial $$f \in k[x_0, \cdots , x_n]$$, that is some polynomial of the form $$f(x_0, \cdots , x_n) := \sum_{j=0}^n a_j x_j$$ where $$(a_0, \cdots , a_n) \in \mathbb P^n$$.

I have seen the following result get used in a few contexts before, and although I can see intuitively why it must be true, I have been unable to find a rigorous argument justifying the same:

Fact(?) Let $$H$$ be a hyperplane and $$P$$ any point in $$\mathbb P^n \setminus H$$. Then there exists a linear transformation $$A \in \text{GL}_{n+1}(k)$$ such that $$A(H)$$ is the hyperplane $$\{(x_0, \cdots , x_n) : x_0=0\}$$ and $$A(P) = (1, 0, \cdots , 0)$$.

I am looking for a complete and concise proof of this result, which is clean if possible. I believe that one possible argument could rest on the following observations:

1. $$H$$ is uniquely determined by any $$n$$ points on it. So we now pick $$n+1$$ points $$P_1, \cdots , P_n$$ on $$H$$.
2. There exists a linear transformation sending $$P$$ to $$(1, 0, \cdots, 0)$$ and $$P_j$$ to $$(0, \cdots , 0 , 1, 0, \cdots, 0) \in \mathbb P^n$$ ($$0$$ in the $$j$$-th slot, here the $$n+1$$ slots are being called the $$0$$-th, $$1$$-st, ..., $$n$$-th slot slots) for each $$1 \leq j \leq n$$.

I have however been unable to make these clean and rigorous (I keep getting involved with too many linear equations) and am starting to doubt the accuracy of my intuition. I would really appreciate a complete argument for the above "Fact(?)" or a reference containing the same and if possible, suggestions on how to make my idea work.

Edit (Some Progress): Thanks to Roland's comment, I think I have made some progress:

Let $$H$$ be given by the equation $$\sum_{j=0}^n a_j x_j = 0$$. Then in $$\mathbb A^{n+1}$$, $$H$$ remains the same (nevertheless I will call it $$H_0$$ when viewed as a subset of $$\mathbb A^{n+1}$$) while $$P := (p_0, \cdots , p_n)$$ becomes the line $$L_0 := \{(p_0 t, \cdots , p_n t) : t \in k\}$$. I should first show that there is a matrix $$A \in \text{GL}_{n+1}(k)$$ such that $$A(H_0) = H_1$$ and $$A(L_0)=L_1$$, where $$H_1 := \{(0, x_1, \cdots , x_n) : x_j \in k\} \subset \mathbb A^{n+1}$$ and $$L_1$$ is the line $$\{(t, 0, \cdots , 0) : t \in k\} \subset A^{n+1}$$.

So now I can pick $$n$$ linearly independent points $$A_j \in H_0$$ ($$1 \leq j \leq n$$), which is possible since $$H_0$$ is an $$n$$-dimensional subspace of $$\mathbb A^{n+1}$$ and I get a linear transformation $$A \in \text{GL}_{n+1}(k)$$ which sends $$A_j$$ to $$(0, \cdots , 0 , 1, 0, \cdots 0)$$ (with $$1$$ in the $$j$$-th slot) for each $$1 \leq j \leq n$$. Thus $$A$$ sends $$H_0$$ to $$H_1$$. I still have to send $$A(L_0)$$ to $$L_1$$ so I need a linear transformation $$T \in \text{GL}_{n+1}(k)$$ which sends $$A(L_0)$$ (which is also a line through the origin) to $$L_1$$ and leaves $$H_1$$ invariant (as a set).

Finally, we let $$T \in \text{GL}_{n+1}(k)$$ be the linear transformation that sends $$(p_0, \cdots , p_n) \in \mathbb A^{n+1}$$ to $$(1, 0, \cdots , 0)$$ and fixes some basis of $$H_1$$ pointwise.

Upon getting this last linear transformation $$T$$, we note that $$TA \in \text{GL}_{n+1}(k)$$ sends $$H_0$$ to $$H_1$$ and $$L_0$$ to $$L_1$$ in $$\mathbb A^{n+1}$$. Therefore $$TA$$ should also do the required job, namely, send $$H$$ to $$\{(0, x_1, \cdots , x_n)\} \subset \mathbb P^n$$ and $$P$$ to $$(1, 0, \cdots 0)$$, thus completing the proof and making "Fact(?)"$a fact. My only follow-up question: Is this argument correct or are there are any gaps? • You have the right idea, but instead of working in$\mathbb{P}^n$, you can translate it to a problem in$\mathbb{A}^{n+1}$which is easier as it is a problem of linear algebra.$H$corresponds to a$n$-dimensional linear subspace of$\mathbb{A}^{n+1}$and$P$corresponds to a$1$-dimensional linear subspace not contained in$H$. Pick bases associated to$H$and$P$, and from linear algebra you know that you can construct a linear map from its action on a basis. This gives immediately the$A$you are looking for without writting any equations. Nov 20, 2020 at 8:14 • @Roland Thanks for your quick comment, can you check out the edit? Nov 20, 2020 at 9:01 • Sure, but I don't get why you don't do both transformations at the same time.$P=(p_0,...,p_n)$is obviously a basis of$L_0$and$(P,A_1,A_2,...,A_n)$a basis of$\mathbb{A}^{n+1}$. Send$P$to$(1,0,0,...,0)$,$A_i$to$(0,...,1,...,0)\$ and you are done. Nov 20, 2020 at 9:14
• Oh, .. yeah... you are right. I guess the "sending plane to plane and line to line" formulation threw me off for a bit. Nov 20, 2020 at 9:15

A small note: $$GL_{n+1}(k)$$ acts differently for linear systems than for points. If $$H$$ is a hyperplane with equation $$a^T \times x=0$$, then the equation of $$AH$$ is $$(aA^{-1})^T \times x=0$$.
Now, your problem is the following: given nonzero vectors $$a$$ (row) and $$x$$ (column) with $$ax \neq 0$$, find an invertible matrix $$A$$ such that $$aA^{-1}=(1,0,\ldots,0)$$, $$Ax=(r,0,\ldots,0)$$ with $$r \neq 0$$.
Find a basis $$(a_2,\ldots,a_{n+1})$$ of the $$n$$-dimensional space of the row vectors orthogonal to $$x$$. Take $$A_1$$ the matrix the rows of which are $$(a,a_2,\ldots,a_{n+1})$$. Then by definition $$aA_1^{-1}=(1,0,\ldots,0)$$, and $$x_1=A_1x$$ has zero entries at indices $$2 \leq i \leq n+1$$, and is nonzero. So we are done.