Question

Consider the vector space P₃

_5.1. Is span {1+x,x+x²,x²+x³,x³+1}=P₃ Motivate your answer

5.2. Let D P₃→ P₂ be the differentiation operator D(a₀+a₁x+a₂x²+a₃x³)=a1+2a₂x+3a₃x²

(i) Find the matrix representation of D relative to the basis {1,x,x²,x³} using the coefficient ordering a0+a₁x+a₂x²+a₃x³→$\begin{bmatrix} a0 \\ a1\\ a2 \\ a3 \end{bmatrix}$

(ii) Find the kernel and range of D

Answer 1

PART 5.1

Relative to basis $(1,x,x²,x³)$

$(1+x)=(1,1,0,0),(x+x²)=(0,1,1,0)$

$(x²+x³)=(0,0,1,1),(x³+1)=(1,0,0,1)$

Matrix P=$\begin{pmatrix} 1& 0&0&1\\ 1& 1&0&0\\0&1&1&0\\0&0&1&1 \end{pmatrix}$

rref of P=$\begin{pmatrix} 1&0&0&1\\ 0&1&0&-1\\0&0&1&1\\0&0&0&0 \end{pmatrix}$

The rank of the matrix is 3, therefore it span P₃

PART 5.2 (i)

$D(1)=0,D(x)=1,D(x²)=2x$

$D(x³)=3x²$

Relative to the basis

$0=(0,0,0,0), 1=(1,0,0,0)$

$2x=(0,2,0,0)$

$D=\begin{pmatrix} 0 & 1&0&0 \\ 0&0&2&0\\0&0&0&3\\0&0&0&0 \end{pmatrix}$

Part 5.2 (ii)

rref $D=\begin{pmatrix} 0&1&0&0\\ 0&0&1&0\\0&0&0&1\\0&0&0&0 \end{pmatrix}$

Range of D is 3

$\begin{bmatrix} 0&1&0&0\\ 0&0&1&0\\0&0&0&1\\0&0&0&0 \end{bmatrix}\begin{bmatrix} x_1\\ x_2\\x_3\\x_4 \end{bmatrix}=\begin{bmatrix} 0\\ 0\\0\\0 \end{bmatrix}$

$x_2=0\quad x_3=0\quad x_4=0$

Taking $x_1=t\>where\>t\in\R$

Kernel $D=[{(t,0,0,0)\>:\>t\in\R}]$