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# Kernel and range of a linear function

## Problem

Recall that $\mathcal{P}_d$ is the vector space of all polynomials of degree at most $d$. Let $f:\mathcal{P}_2\rightarrow\mathcal{P}_2$ be a linear function defined by $p\mapsto x\cdot p'$, where $p'$ is the derivative of $p$. Find $\ker(f)$ and $\range(f)$.

## Thought

As always, we should start with the definition. Recall that

$$
\begin{aligned}
\ker(f) &= \{p\in\mathcal{P}_3: f(p) = 0\} \text{ and } \\
\range(f) &= \{f(p): p\in\mathcal{P}_3\}.
\end{aligned}
$$

For example, $1$ is an element in $\ker(f)$ since $f(1) = x\cdot (1)' = 0$, and $f(x^2) = x\cdot (x^2)' = x\cdot (2x) = 2x^2$ is an element in $\range(f)$.

We may rely on the matrix representation to find the kernel and the range.

## Sample answer

Let $\alpha = \{1, x, x^2\}$ be a basis of $\mathcal{P}_2$. By direct computation, we have

$$
\begin{aligned}
f(1) &= 0 + 0x + 0x^2, \\
f(x) &= 0 + 1x + 0x^2, \\
f(x^2) &= 0 + 0x + 2x^2.
\end{aligned}
$$

Therefore,

$$
A = [f]_\alpha^\alpha = \begin{bmatrix}
| & | & | \\
[f(1)]_\alpha & [f(x)]_\alpha & [f(x^2)]_\alpha \\
| & | & | \\
\end{bmatrix} = \begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 2
\end{bmatrix}.
$$

Using the techniques in [How to find a basis of the kernel?](https://hackmd.io/@jephianlin/SyPyuWrXT) and [How to find a basis of the column space?](https://hackmd.io/@jephianlin/HJs0C-rQ6), we know that

$$
\begin{aligned}
\ker(A) &= \left\{\begin{bmatrix} x \\ 0 \\ 0 \end{bmatrix}: x\in\mathbb{R}\right\} = \vspan(\{\be_1\}) \text{ and} \\
\Col(A) &= \left\{\begin{bmatrix} 0 \\ y \\ z \end{bmatrix}: y,z\in\mathbb{R}\right\} = \vspan(\{\be_2, \be_3\}).
\end{aligned}
$$

Since the mapping $p\mapsto [p]_\alpha$ is sending $1\mapsto\be_1$, $x\mapsto\be_2$, $x^2\mapsto\be_3$. We may translate the previous equalities into

$$
\begin{aligned}
\ker(f) &= \vspan(\{1\}) \text{ and} \\
\range(f) &= \vspan(\{x, x^2\}).
\end{aligned}
$$

*This note can be found at Course website > Learning resources.*