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Find a vector space complement to $\mathbb{R}(1 + x + x^2) + \mathbb{R}(x - x^2 - x^3)$ in $\mathbb{R}[x]\le_3$. (This is referring to the set of polynomials with degree less than or equal to 3.)

I am not sure I completely understand the process of finding vector space complements and it would be good to have one example of how to do it! Thanks for your help.

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What is $\,\Bbb R(1+x+x^2)\,$? The ideal generated by the polynomial $\,1+x+x^2\,$? Or the scalar multiples of this polynomial? Or something else? –  DonAntonio Sep 27 '12 at 3:26
    
Ah I apologise, it refers to scalar multiples of the polynomial, from the real field. Thanks for the clarification. –  Peter Sep 27 '12 at 4:49
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1 Answer

It's easy to show that the polynomial 1 is not in the given vector space (that is, it's easy to show that there are no real numbers $a,b$ such that $a(1+x+x^2)+b(x-x^2-x^3)$ is identically 1 --- can you do this?), and it's just as easy to show that $x^3$ is not in the space spanned by $1,1+x+x^2,x-x^2-x^3$ so you can take as your complement the space spanned by 1 and $x^3$ (in your notation, ${\bf R}(1)+{\bf R}(x^3)$).

A more systematic approach would be to form the matrix $$\pmatrix{1&1&1&0\cr0&1&-1&-1\cr}$$ whose rows come in what I hope is the obvious way from your given polynomials $1+x+x^2$ and $x-x^2-x^3$, and then find a basis for the nullspace of this matrix. Have you learned how to do that? You can bring the matrix to reduced row-echelon form, $$\pmatrix{1&0&2&1\cr0&1&-1&-1\cr}$$ and the read out a basis for the nullspace as $\{{(-2,1,1,0),(-1,1,0,1)\}}$ and then interpret these vectors as the polynomials $-2+x+x^2,-1+x+x^3$. Then ${\bf R}(-2+x+x^2)+{\bf R}(-1+x+x^3)$ is your complement.

This is not the same as the first answer I gave, but that's OK --- a given vector space has lots of complements.

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