$T$:$P_1$ → $R^2; T[p(x)]=[p(0),p(1)]$
I need help, not sure how to start. Thanks in advance!
$P$=Polynomials
Apparently this theorem is meant to help: If $V$ and $W$ have the same dimension $n$, a linear transformation $T:V→W$ is an isomorphism if it is either one-to-one or onto.
Can you verify that $T$ is a linear transformation? This is routine and I will assume that you have verified this. A linear transformation is one-to-one if and only of $Tp=0$ implies $p=0$. So consider $p(x)=ax+b$ in $P_1$. Suppose $Tp=0$. Then $(p(0),p(1))$ is the zero vector in $\mathbb R^{2}$ which means $p(0)=0$ and $p(1)=0$. But $p(x)=ax+b$ so we get $0+b=0$ and $a+b=0$. These two equations imply $a=b=0$ which means $p$ is the zero polynomial or the zero element of $P_1$. We have proved that $T$ is one-to-one. Since $\mathbb R^{2}$ has dimension 2 we only have to prove that $P_1$ also has dimension 2. For this verify that $p_1(x)=1$ and $p_2(x)=x$ form a basis for $P_1$.