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$T: P_2 \to P_2$ defined by $T(p(x)) = p(x) + p(-x)$. Find basis for $\ker T$

Here is my solution: $$p(x) = ax^2 + bx + c$$ $$p(x) + p(-x) = 0 \to 2ax^2+2c =0$$ So, a = c = 0.

So, basis of $\ker T$ is $x$.

The thing that I'm not sure is: because $2ax^2 + 2c = 0$ with all x, so $a = c = 0$. Can I say that?

Thanks :)

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depends on the characteristic of the field, it is obviously not true if $\operatorname{char}(K)=2$. – Julian Kuelshammer Nov 9 '12 at 14:35
Can you tell me clearer please :) – hqt Nov 9 '12 at 14:37
expanding to an answer – Julian Kuelshammer Nov 9 '12 at 14:37
up vote 2 down vote accepted

Just realized that my comment is already the solution:

If $\operatorname{char}(K)=2$. Then $2=0$ and the kernel of $p(x)\mapsto p(x)+p(-x)=2p(x)=0$ for all $p$, so that $\ker T=P_2$.

If $\operatorname{char}(K)\neq 2$, then $2$ is invertible in $K$, so that you can divide the equation $2ax^2+2c=0$ by $2$ to get $a\cdot x^2+c\cdot 1=0$. Now $\{1,x^2\}$ is linearly independent in $P_2$ (by definition). So that $a=0=c$ by the definition of linear independence.

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