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The question asks to find the kernel of $S: P^2 \to \mathbb{R}$ defined by $S(a+bx+cx^2) = a+b+c$.

I know how to find the kernel of a matrix transformation (it's just the null space of the matrix) but I can't conceptualize a transformation from two different types of vector spaces. How would I go about finding a basis ker(S)?

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Check the definition. The kernel consists of all the vectors (in your case: second degree polynomials) that are mapped to the zero vector (in your case: the zero polynomial).

Now since $S \left( a+bx+cx^2 \right) = a+b+c$, you have: $$S \left( a+bx+cx^2 \right) = 0 \iff a+b+c = 0 $$ So a polynomial of the form $a+bx+cx^2$ is the kernel of $S$ when the sum of its coefficients is equal to $0$.

For example:

  • $x^2-3x+1$ is not in the kernel because $1-3+1 \ne 0$; indeed: $$S(x^2-3x+1) = 1-3+1 = -1$$
  • $x^2-3x+2$ is in the kernel because $1-3+2 = 0$; indeed: $$S(x^2-3x+2) = 1-3+2 = 0$$

To find a basis of this kernel, notice that you can solve the condition $a+b+c = 0$ for one of the coefficients, e.g. $\color{purple}{a=-b-c}$. Vectors in the kernel are thus of the form: $$\color{purple}{\underbrace{-b-c}_{a}}+bx+cx^2 = b\left( \color{blue}{-1+x} \right) + c\left( \color{red}{-1+x^2}\right)$$ Notice that you can always write such a vector as a linear combination of the vectors (polynomials) $\color{blue}{-1+x}$ and $\color{red}{-1+x^2}$ so these two clearly span the kernel. Verify that they are linearly independent and hence form a basis for the kernel.

Another way of looking at this kernel: notice that for a polynomial $p(x)=a+bx+cx^2$, you have $p(1)=a+b+c$ so the condition $a+b+c=0$ boils down to $p(1)=0$, i.e. the polynomial has $x=1$ as a root. This allows the kernel to be described as: $$\mbox{Ker}(S) = \left\{ p(x) \in P^2 \;\vert\; p(1) = 0\right\}$$

You can check that the basis vectors we found above indeed have $x=1$ as a root.

Thanks to zipirovich for pointing this out in the comments.

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    $\begingroup$ The condition that the sum of the coefficients of a polynomial $p(x)$ is zero is the same as saying that $p(1)=0$. So the kernel is $\ker(S)=\{p(x)\in P^2\;\colon\; p(1)=0\}$. And then (part of) the reason these two polynomials work as the basis is that they are linearly independent polynomials both having $x=1$ as a root. $\endgroup$ – zipirovich Dec 18 '16 at 22:19
  • $\begingroup$ @zipirovich Thanks for the nice addition; but I have the feeling that user5319366 is also helped with a more general approach rather than (only) this (elegant) formulation in terms of the polynomial having $x=1$ as root. $\endgroup$ – StackTD Dec 18 '16 at 22:23

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