Consider the linear transformation $\ltdefn{R}{P_2}{P_2}$ defined by \begin{equation*} \lt{R}{a+bx+cx^2}= (15a+8b-4c)+(-12a-6b+3c)x+(24a+14b-7c)x^2 \end{equation*} and the vectors \begin{align*} \vect{w}_1 &=1-x+x^2 & \vect{w}_2 &=x+2x^2 & \vect{w}_3 &=1+4x^2 & \end{align*}
Then compute \begin{align*} \lt{R}{\vect{w}_1} &= \lt{R}{1-x+x^2} = 3-3x+3x^2 =3\vect{w}_1\\ \lt{R}{\vect{w}_2} &= \lt{R}{x+2x^2} = 0+0x+0x^2 =0\vect{w}_2\\ \lt{R}{\vect{w}_3} &= \lt{R}{1+4x^2} = -1-4x^2 =(-1)\vect{w}_3\\ \end{align*}
So $\vect{w}_1$, $\vect{w}_2$, $\vect{w}_3$ are eigenvectors of $R$ with eigenvalues (respectively) $\lambda_1=3$, $\lambda_2=0$, $\lambda_3=-1$. Notice how the eigenvalue $\lambda_2=0$ indicates that the eigenvector $\vect{w}_2$ is a nontrivial element of the kernel of $R$, and therefore $R$ is not injective (Exercise CB.T15).