Null space and kernel of matrix representation

Question

Let $P_3(\mathbb{C})$ be the complex vector space of complex polynomials of degree $2$ or less. Let $\alpha,\beta\in\mathbb{C}, \alpha\neq\beta$. Consider the function $L:P_3(\mathbb{C}) \mapsto \mathbb{C}^2$ given by

$$L(p)=\begin{bmatrix} p(\alpha) \\ p(\beta)\\ \end{bmatrix}, \text{ for } p\in P_3(\mathbb{C})$$

For the basis $v=(1,X,X^2)$ for $P_3(\mathbb{C})$ and the standard basis $E = (e_1,e_2)$ for $\mathbb{C}^2$. Find the matrix representation $_E[L]_v$ and determine the null space $N(_E[L]_v)$ and find a basis for the ker(L).

I have found the matrix representation:

$$_E[L]_v = [L(v)]_E = [L(1)]_E\ [L(X)]_E\ [L(X^2)]_E = \begin{bmatrix} 1\quad \alpha \quad \alpha^2 \\ 1\quad \beta \quad \beta^2 \end{bmatrix}$$

By using ERO we can reduce the matrix to: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$

I am uncertain how to find the null space $N(_E[L]_v)$ and a basis for the kernel.

Answer 1

You're doing good and the matrix is exactly what you found. The reduced row echelon form is $$ \begin{bmatrix} 1 & 0 & -\alpha\beta \\ 0 & 1 & \alpha+\beta \end{bmatrix} $$ as you found. Now you can determine a basis for the null space of the matrix as generated by $$ \begin{bmatrix} \alpha\beta \\ -(\alpha+\beta) \\ 1 \end{bmatrix} $$ and this is the coordinate vector of a polynomial generating the kernel, which is thus $$ q(X)=\alpha\beta-(\alpha+\beta)X+X^2 $$

As a check: this polynomial $q$ has $\alpha$ and $\beta$ as roots and so $L(q)=0$. The kernel has dimension $1$ by the rank nullity theorem.

Answer 2

It is probably better to do that via polynomials.

Suppose $p \in P_3(\mathbb{C})$ is such that $p(\alpha) = 0 = p(\beta)$. Then $p$ is divisible by both $x - \alpha$ and $x - \beta$.

Since $\alpha \ne \beta$, the two linear polynomials are coprime, so $p$ is divisible by $(x-\alpha)(x-\beta) = x^{2} - (\alpha+\beta) x + \alpha \beta$, and thus $p$ is a scalar multiple of it, as $p$ has degree at most $2$.

So the kernel is one-dimensional, generated by the transpose of $(\alpha \beta, -\alpha - \beta, 1)$.

This indicates that there is a little sign error (it happens to everyone) in your reduction.