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Consider two linear spaces Vand W. A function Tfrom Vto Wis called linear transformation if

1. $\mathbf{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{T}\left(f\right)\mathbf{+}\mathbf{T}\left(g\right)$
2. $\mathbf{T}\left(\mathrm{kf}\right)\mathbf{=}\mathbf{kT}\left(f\right)$

for all elements f and g of Vand for all scalars.

Ifis finite dimensional, then

$\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{rank}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{+}\mathbf{nullity}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{imT}\mathbf{\right)}\mathbf{+}\mathbf{dim}\mathbf{\left(}\mathbf{kerT}\mathbf{\right)}$

An invertible linear transformation Tis called an isomorphism.

Consider the linear transformation$T:Z_n鈫�P_{n-1}$ given by $T\left(f\left(t\right))=f\text{'}(t\right)$.

Since,$f\left(t\right),g\left(t\right)鈭�Z_n$

$$\begin{gathered} T\left(f\left(t\right)+g\left(t\right)\right)=T\left(\left(f+g\right)t\right) \\ =\left(f+g\right)\text{'}t=f\text{'}\left(t\right)+g\text{'}\left(t\right) \\ =T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right) \end{gathered}$$

Also, consider a scalar c, then

$$\begin{gathered} T\left(cf\left(t\right)\right)=T\left(\left(cf\right)t\right) \\ =\left(cf\right)\text{'}t=cf\text{'}\left(t\right) \\ =cT\left(f\left(t\right)\right) \end{gathered}$$

Therefore, T is a linear transformation.

Consider any non-zero polynomial,

$f\left(t\right)=a_0+a_{1}t+...+a_{n-1}{t}^{n-1}鈭�P_{n-1}$

Then,

$$\begin{gathered} T\left(f\left(t\right)\right)=f\text{'}\left(t\right) \\ =a_1+2a_2+...+n{a}_n{t}^{n-1} \end{gathered}$$

Since, it non-zero there exists$0鈮�i鈮�n-1$ such that $a_i鈮�0$.

Thus,$T\left(f\left(t\right)\right)$ is also a non-zero polynomial in $P_{n-1}$.

Hence, $\ker \left(T\right)=\left\{0\right\}$.

Therefore, $\dim \left({\mathrm{Z}}_{\mathrm{n}}\right)=\mathrm{n}=\dim \left({\mathrm{p}}_{\mathrm{n}-1}\right)$.

Thus, T is an isomorphism.