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I am currently reading some books on finite Elements and I have a question regarding one step in the proof of one lemma:

Let $P$ be a polynomial of degree $d \geq 1$ that vanishes on a hyperplane L, i.e. the hyperplane $\{x: L(x)=0\}$. Then we can write $P=LQ$, where $Q$ is a polynomial of degree $(d-1)$.

Here L is a non-degenerate linear function. They proceed as follows: Since affine transformations map polynomials of degree $d$ onto polynomials of degree $d$, make an affine change of coordinates such that: $$ L(\hat{x},x_n)=x_n$$ and $L(\hat{x},x_n)=0$ is the $\hat{x}$-axis. Here $\hat{x}=(x_1,...,x_{n-1})$

The rest of the proof is pretty understandable and not hard to figure out yourself, however I have a hard time really understanding this affine transformation of $L$, since I have not really worked with it before. I really want to know how such transformation of a linear function $L$ is done especially in an easy way to understand. For example, in $\mathbb{R}$ for a linear function $f(x)=ax+b$, would the transformation: $$\hat{x}=(x-b)/a$$ be the correct transformation, such that $f(\hat{x})=x$ or am I wrong? If correct, the my question is how would such transformation look like in higher dimension?

Any suggestions, explanation or advices where I can find more information to understand it is greatly appreciated. The book is: The Mathematical Theory of Finite Element Methods" by S. Brenner and L. Scott.

babemcnuggets
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1 Answers1

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I was reading The Mathematical Theory of Finite Element Methods and came across this question. I do think the line: "Make an affine change of coordinates such that $L(\hat{x},x_n) = x_n$ and the hyperplane $L(\hat{x},x_n) = x_n$ is the $\hat{x}$-axis." is very confusing and not understandable to me.

But I assume that as we are not concerned with geometry here we can simply adopt the following definition(an affine transformation is just a linear transformation with translation):

What is the difference between linear and affine function?

And to put this into context, if we have a hyperplane in $\mathbb{R}^3$, say, $x+2y+z = 1$, I would make the affine transformation $z' = x + 2y + z -1 $, so that we would have a new polynomial $P(x,y,z')$ instead of $P(x,y,z)$

Then you can just follow the proof on the book.

Dinoman
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