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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.

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  • $\begingroup$ You’re almost there. See this answer for how to read a basis for the kernel from the reduced matrix. $\endgroup$ – amd Apr 3 '18 at 20:01
  • $\begingroup$ So it is possible to write the RREF matrix: $\begin{bmatrix} 1 \quad 0 \quad - \alpha\beta \\ 0 \quad 1 \quad \alpha + \beta \\ \end{bmatrix},$ as the following: $x_1 = \alpha\beta$, $x_2 = -\alpha-\beta$, $x_3 = x_3$ as $x_3$ is a free variable we can put in 1, so $x_3=1$ This way we have that $ L_v= (\alpha\beta, -\alpha-\beta, 1)^T$ Which means that the basis for the kernel is equal to $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix}$? $\endgroup$ – Simbörg Apr 3 '18 at 21:15
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    $\begingroup$ Your reasoning is a bit off. The RREF represents the equations $x_1-\alpha\beta x_3=0$ and $x_2+(\alpha+\beta)x_3=0$, so every solution of the system is of the form $(\alpha\beta x_3, -(\alpha+\beta)x_3, x_3)^T$, i.e., a multiple of $(\alpha\beta, -\alpha-\beta,1)^T$. $\endgroup$ – amd Apr 3 '18 at 21:45
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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.

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  • $\begingroup$ I am not sure I understand your argument for the kernel, but I will try to see if I understand it correctly. So: we know that since the coordinate vector $\begin{bmatrix} \alpha\beta \\ -\alpha-\beta \\ 1 \end{bmatrix} \in N(_E[L]_v)$ we know this implies that $(\alpha\beta, -\alpha-\beta, 1)^T \in ker(L)$ and then you define a polynomial for that generates the kernel, which means $ker(L) = (\alpha\beta, -(\alpha+\beta)X, X^2)^T$ and if $X = 0$ then we have that the kernel consists of $(\alpha\beta)$? $\endgroup$ – Simbörg Apr 4 '18 at 9:27
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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.

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  • $\begingroup$ Your approach makes sense, it is just that the method I have in my textbook is that: The null space of a matrix A, N(A), is equal to the null space of the RREF H, N(H). So I believe that I have to reduce the matrix representation and find the null space with this method (and yes there was a slight error in my computation, it should be correct now) $\endgroup$ – Simbörg Apr 3 '18 at 18:18
  • $\begingroup$ @Simbörg, you should have been taught that once you have transformed your matrix in the block form $[I \mid A]$, where $I$ is an appropriate identity matrix, then the space of solutions of the associated homogeneous system (i.e. the null space) has as a basis the columns of the block matrix $\left[\begin{smallmatrix}-A\\J\end{smallmatrix}\right]$, where $J$ is an appropriate square matrix. So in your case the null space has a basis given by $\left[\begin{smallmatrix}\alpha \beta\\-\alpha-\beta\\1\end{smallmatrix}\right]$. $\endgroup$ – Andreas Caranti Apr 4 '18 at 8:28

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