We are tasked with finding the kernel of the given linear map $s$ and its dimension.

Problem Details:

The map is defined as: $s: P_2 \to \mathbb{R}^2$ where $P_2$ is the space of polynomials of degree at most 1 (this is implied as $a + bx$ has no higher-degree terms). For a polynomial $p(x) = a + bx$, the map sends it to: $s(a + bx) = (a - b, a + b).$

Step 1: Kernel of $s$

The kernel of $s$, denoted $\ker(s)$, consists of all polynomials $p(x) \in P_2$ such that $s(p(x)) = 0$. In this case, we require: $s(a + bx) = (a - b, a + b) = (0, 0).$ This gives the following system of equations:

1. $a - b = 0$,
2. $a + b = 0$.

Solving the System:

From the first equation $a - b = 0$, we have: $a = b.$ Substituting $a = b$ into the second equation $a + b = 0$, we get: $a + a = 0 \implies 2a = 0 \implies a = 0.$ Thus, $b = 0$ as well.

The only solution is $a = 0$ and $b = 0$. Therefore, the kernel consists solely of the zero polynomial: $\ker(s) = \{ 0 \}.$

Step 2: Dimension of the Kernel

The kernel is trivial, containing only the zero polynomial. Thus, the dimension of the kernel is: $\dim(\ker(s)) = 0.$

Final Answer:

1. The kernel of $s$ is $\ker(s) = \{ 0 \}$.
2. The dimension of the kernel is $\dim(\ker(s)) = 0$.

Would you like me to further explain this process or discuss the rank-nullity theorem in this context?

Related Questions:

1. What is the rank of $s$, and how does it relate to the dimension of $P_2$?
2. What is the image of $s$, and how can we compute its dimension?
3. How can we generalize this mapping to higher-degree polynomial spaces $P_n$?
4. If $\ker(s)$ is trivial, does that imply $s$ is injective?
5. How can we visualize this map geometrically in $\mathbb{R}^2$?

Tip:

Always check the kernel and image dimensions using the Rank-Nullity Theorem: $\dim(\text{Domain}) = \dim(\ker(s)) + \dim(\text{Image}(s)).$