Suppose a linear map $T: \mathbf{R}^2 \to \mathbf{R}^2 $ and let the subring of $End_{\mathbf{R} }(\mathbf{R}^2)$ generated by $T$ and $R$, i.e. $$\mathbf{R}[T] = \{c_0+c_1T + ... +c_nT^n : c_0, ... c_n \in \mathbf{R}\} $$
Define $\phi: \mathbf{R}[x] \to \mathbf{R} [T] $ by $$\phi(c_0 + c_1x + ... +c_nx^n)=c_0 + c_1T + ... + c_nT^n$$ for all $c_0+c_1x+... + c_nx^n \in \mathbf{R}[x]$ where $\mathbf{R}[x]$ is the polynomial ring in $x$. Then $\phi$ is clearly a ring homomorphism and surjective.
My question is how to show that $\phi$ is not injective
Some ideas: So $\phi$ is injective iff $ker(\phi) = \{0\}$. However, $T = T_A, \text{say matrix } A = $$\begin{bmatrix} 1 & 2 \\ 0 & 3 \end{bmatrix} $$ \implies A^2-4A+3I=0 \implies {T_A}^2 -4T_A+3I=0$ $$ \mathbb R[x] \to End_{\mathbf{R} }(\mathbf{R}^2)$$
$$x^2 - 4x+3 \to T^2-4T+3I=0$$ Please help or direct me. Due to the lack of my abstract algebra background, I have a hard time how to show that $\phi$ is not one-to-one.