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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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    the last element should be $x^{n-1}$ under the comments.2011-09-19

2 Answers 2

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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 you2011-09-19
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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.