How to find $\ker(T)$

Question

Let $T : P_3 \to \mathbb{R}^3$ be defined by $T(p(x)) = \begin{bmatrix}p(1)\\p(2)\\p(3)\end{bmatrix}$. Show that $T$ is a linear transformation, and find $\ker(T)$. $P_3$ := {$p(x) = a_0 + a_1x + a_2x^2 + a_3x^3 | a_1, a_2, a_3, a_4 \in \mathbb{R}$}

I know that $T(v+w) = T(v) + T(w)$ and $cT(v) = T(cv)$ to prove it's a linear transformation. Am I just substituting $v$ and $w$ in the $p(x)$ equations?

And I'm really stuck on how to solve for the kernel. I know that it is like the nullspace but it there a different way to solve for it since it is for a linear transformation? I'm slightly confused on what I'm trying to manipulate.

Answer 1

Hint: The kernel is made of all polynomials in $P_3$ that have $1$, $2$, and $3$ as roots. For example $(x-1)(x-2)(x-3)$ is an element in the kernel.

Answer 2

To check for the LT let consider

• $T(p(x)+q(x)) = \begin{bmatrix}p(1)+q(1)\\p(2)+q(2)\\p(3)+q(3)\end{bmatrix}=?$
• $T(c\cdot p(x)) = \begin{bmatrix}c\cdot p(1)\\c\cdot p(2)\\c\cdot p(3)\end{bmatrix}=?$

For the second part, you can proceed exactly for the ordinary vector space, let consider that each polynomial in the form $p(x) = a_0 + a_1x + a_2x^2 + a_3x^3$ in the basis $\cal B=\{1,x,x^2,x^3\}$ is represented by its components $(a_0,a_1,a_2,a_3)$.

Answer 3

As @ir7 suggests, if $p \in \ker T$ then

$$\pmatrix{0 \\ 0 \\ 0} = T(p) = \pmatrix{p(1) \\ p(2) \\ p(3)} \implies p(1) = p(2) = p(3) = 0 \implies (x-1)(x-2)(x-3) \text{ divides } p$$

Therefore, $$\ker T = \operatorname{span}\{(x-1)(x-2)(x-3)\} = \{\alpha (x-1)(x-2)(x-3) : \alpha \in \mathbb{R}\}$$

Answer 4

$$T(p(x)) = \begin{bmatrix}p(1)\\p(2)\\p(3)\end{bmatrix} =\begin{bmatrix}0\\0\\0\end{bmatrix} $$

$$p(1)=p(2)=p(3)=0$$

$$ p(x)=a(x-1)(x-2)(x-3)$$

That is $$ Ker(T)=\{p(x): p(x)=a(x-1)(x-2)(x-3)\}$$