Answer
See below
Work Step by Step
Assume $x=(x_1,x_2,x_3)\\
y=(y_1,y_2,y_3)\\
c\in R$
$T(f+g)=T((f+g)(0),(f+g)(1))\\
=T(f(0)+g(0),f(1)+g(1))\\
=(f(0),g(0)+(f(1),g(1))\\
=T(f)+T(g)$
$T(cf)=T(c(f)(0),(cf)(1))\\
=(cf(0),cf(1))\\
=c(f(0),f(1))\\
=cT(f)$
Hence, $T$ is a linear transformation
Obtain: $Ker(T)=\{f \in C[0,1]:T (f)=0\}\\
=\{f \in C[0,1]: T(f)=0\}\\
=\{f \in C[0,1]: (f(0),f(1))=(0,0)$
Assume $f(x)=x^2-x\\
\rightarrow f(0)=f(1)=0\\
\rightarrow T(f)=(0,0)\\
\rightarrow f \in Ker(T)$
then $Ker(T)=\ne \{0\}$
$T$ is not one-to-one
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^3\\
\dim [Ker(T)]+2=\infty\\
\dim [Ker(T)]=\infty$
$Ker(T)$ is an infinite space.
$Ker(T)=\{f \in C[0,1]:f(0)=f(1)=0\}$
