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)?


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.

  • 3
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.