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The following question is from Golan's linear algebra book. I have posted a solution in the answers.

Problem: Let $F$ ba field and let $V$ be a vector subspace of $F[x]$ consisting of all polynomials of degree at most 2. Let $\alpha:V\rightarrow F[x]$ be a linear transformation satisfying

$\alpha(1)=x$

$\alpha(x+1)=x^5+x^3$

$\alpha(x^2+x+1)=x^4-x^2+1$.

Determine $\alpha(x^2-x)$.

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2 Answers 2

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Your calculation looks fine. Equivalently, but with rather less work, note that $$x^2-x=(x^2+x+1)-2(x+1)+1,$$ so we can compute $\alpha(x^2-x)$ directly by linearity, without computing $\alpha$ at the most common basis elements.

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The idea here is to determine the action on each of the basis elements $1$, $x$, and $x^2$.. By linearity we see

$\alpha(x)=\alpha((x+1)-1)=\alpha(x+1)-\alpha(1)=x^5+x^3-x$

$\alpha(x^2)=\alpha((x^2+x+1)-(x+1))=\alpha(x^2+x+1)-\alpha(x+1)=-x^5+x^4-x^3-x^2+1$

Using linearity again we see

$\alpha(x^2-x)=\alpha(x^2)-\alpha(x)=-x^5+x^4-x^3-x^2+1-x^5-x^3+x=-2x^5+x^4-2x^3-x^2+x+1$

and if the field is of characteristic 2, the $x^5$ and $x^3$ terms disappear.

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Looks good to me. Another method: $x^2-x=(x^2+x+1)-2(x+1)+1$, so $\alpha(x^2-x)=\alpha(x^2+x+1)-2\alpha(x+1)+\alpha(1)$, etc. –  Gerry Myerson Jun 1 '12 at 23:47
    
The original question does not mention that the field is of characteristic 2, does it? –  lhf Jun 2 '12 at 3:14
    
It does not specify the characteristic, so I thought I would account for all possibilities. –  Potato Jun 2 '12 at 3:20

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