I hope you liked the title.
I have a question that is as follows:
Consider the linear transformation $T: P_3(\mathbb{R}) \to P_3(\mathbb{R})$ given by $$T(f(x))=f(0)+f'(x)+f''(x)$$ Where the prime symbol denotes differentiation
a) Determine the matrix representation of $T$ with respect to the standard ordered basis of $P_3(\mathbb{R})$
Note: Standard basis $\{1,x,x^2,x^3\}$
Is this correct?
$ \begin{pmatrix} 0&0&0&0\\1&0&0&0\\2&2&0&0\\0&6&3&0 \end{pmatrix}$
??
$T(1)=1=1.1+0.x+0.x^2+0.x^3$
$T(x)=1$
$T(x^2)=0+2x+2=2.1+2.x+0.x^2+0.x^3$
$T(x^3)=0+3x^2+6x=0.1+6.x+3.x^2+0.x^3$
I don't see how you got that matrix when the matrix should be $$\begin{pmatrix} 1&1&2&0\\0&0&2&6\\0&0&0&3\\0&0&0&0 \end{pmatrix}$$
If $f(x) \in Ker(T)$, then $T(f(x))=0$ which would give $f(0)+f'(x)+f''(x)=0$ . Now since any $f(x) \in P_3$ will look like $a_0+a_1x+a_2x^2+a_3x^3$, we will have $(a_0)+(a_1+2a_2x+3a_3x^2)+(2a_2+6a_3x)=0$ which gives $(a_0+a_1+2a_2)+x(2a_2+6a_3)+x^2(3a_3)=0$. Now this is $0$ for all $x$. Hence $3a_3=0$, $2a_2+6a_3=0$ and $a_0+a_1+2a_2=0$. Solvin these together will give you $a_0=-a_1$. Hence $f(x) =a_0(1-x)$.
Note you can also find these equations by solving for $X=(a_0,a_1,a_2,a_3)$ such that $AX=0$, where $A$ is your basis matrix