Exercise 2.4.23

The transformation is linear since

where $m$ is a integer large enough such that $\sigma (k)=\tau (k)=0$ for all $k>m$. It would be injective by following argument. Since $T(\sigma )={\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n\sigma (i)x^i=0$ means $\sigma (i)=0$ for all integer $i\le n$, with the help of the choice of $n$ we can conclude that $\sigma =0$. On the other hand, it would also be surjective since for all polynomial ${\sum <!--\; FUNCTION\; APPLICATION\; -->}_{i=0}^n{a}_i{x}^i$ we may let $\sigma (i)=a_i$ and thus $T$ will map $\sigma$ to the polynomial.