Answer
See below
Work Step by Step
Define $T:R^2 \rightarrow P_1(R)$ by $T(a,b)=ax+b$
Assume $(a,b);(c,d) \in R^2\\
\alpha \in R$
Obtain: $T((a,b),(c,d))=T(a+c,b+d)\\
=(a+c)x+(b+d)\\
=ax+cx+b+d\\
=T(a,b)++T(c,d)$
then $T(\alpha (a,b))=T(\alpha a, \alpha b)\\
=\alpha ax+\alpha b\\
=\alpha(ax+b)\\
=\alpha T(a,b)$
Hence, $T$ is a linear transformation.
Obtain: $Ker(T)=\{x \in R^2:T (x)=0\}\\
=\{(a,b) \in R^2: T(a,b)=0\}\\
=\{ (a,b)\in R^2: ax+b=0\}\\
\rightarrow a=b=0$
then $Ker(T)=\{(0,0)\}=\{0\}$
$T$ is one-to-one.
Apply Rank Nullity Theorem:
$\dim [Ker(T)]+\dim [Rng(T)]=\dim R^2\\
0+\dim [Rng(T)]=2\\
\dim [Rng(T)]=2 $
$Rng(T) \subset P_1(R)$ and $\dim P_1(R)=2=\dim [Rng(T)]$
$T$ is onto (2)
From $(1),(2)$, $T$ is an isomorphism.
