# Producing a hyperplane in the vector space of all polynomials

Given $P$, the vector space of all polynomials (over reals), I am supposed to come out with a hyperplane of $P$, and show that it is indeed a hyperplane.

Since a hyperplane is the same as a flat, I gave myself such example $p+M=\{p+q: q \text{ is in } M \}$ where $M$ is a subspace of dimension $n-1$, so $M$ is spanned by basis elements say $(1,x,x^2,...,x^n-2)$. I am now supposed to show that indeed it is a hyperplane. I am thinking of showing that the vectors $1-x^{n-1}, x-x^{n-1},\ldots, x^{n-2}-x^{n-1}$ are linearly independent?

Does this solve the question? Or you think I have to do something different?

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What is $n$? The space of all polynomials does not have a maximal degree. –  Henning Makholm Sep 19 '11 at 0:26
I mean vector space of polynomials with dimension $n$ having as basis $(1,x,x^2,...x^n-1)$ and what should be the case if the space is of dimension infinity? –  smanoos Sep 19 '11 at 0:37
@smanos: Did you mean the last polynomial in your set to be $x^{n-1}$ instead of $x^n-1$? For arbitrary dimension, a subspace $W$ has codimension $k$ if the dimension of the quotient space $V/W$ is $k$. –  Arturo Magidin Sep 19 '11 at 1:12
A polynomial in this vector space is the zero element if and only if each $x^i$ coefficient is zero. So for example you could say all polynomials $P(x)$ such that $P(0)=0$ forms a hyperplane because the orthogonal complement is the space of constant polynomials. –  anon Sep 19 '11 at 1:15

Let $V$ be a vector space. The kernel of any nonzero linear form on $V$ is a linear hyperplane. Any linear hyperplane is of this form. Two nonzero linear forms have the same kernel if and only if they are proportional. The set of linear hyperplanes in $V$ is thus canonically isomorphic to the set $P(V^*)$ of proportionality classes of nonzero linear forms on $V$.

If $f$ is a nonzero linear form on $V$, then $f^{-1}(1)$ is an affine nonlinear hyperplane. Any affine nonlinear hyperplane can be uniquely written in this way. The set of affine nonlinear hyperplanes in $V$ is thus canonically isomorphic to the set of nonzero linear forms on $V$.

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I think this one sounds nice. Thank you –  smanoos Sep 19 '11 at 15:14

You have an $n$-dimensional vector space. You have a basis for said space. Hence you have an isomorphism between your space and $\mathbb{R}^n$. I imagine you can find a hyperplane in $\mathbb{R}^n$. So you should do so, and then pull it back to your space via the isomorphism.

The point is: There is only one $n$-dimensional real vector space (up to isomorphism), so you had might as well be looking at $\mathbb{R}^n$, the simplest avatar of it.

Of course you could solve this directly, but if you are getting confused by the fact that your vectors are polynomials, there is no point in not making it easier for yourself.

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