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)?
 A: 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.
