# Need help finding the kernel of a linear transformation $P^2 \to \mathbb{R}$

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.

• 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. – zipirovich Dec 18 '16 at 22:19
• @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. – StackTD Dec 18 '16 at 22:23