91Ó°ÊÓ

Consider two linear spacesand. A functionfromtois called linear transformation if

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

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

If Vis finite dimensional, then

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

An invertible linear transformation Tis called an isomorphism.

Consider the linear transformation$T:P_{n-1}→Z_n$given by $T\left(f\left(t\right)\right)={∫}_0^{t}f\left(x\right)dx$.

Consider any non-zero polynomial,

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

Then,

n-1

$$\begin{gathered} T\left(f\left(t\right)\right)={∫}_0^t\left(a_{0}t+a_{1}t+...+a_{n-1}{t}^{n-1}\right)dt \\ ={\left[a_{0}t+a_1\frac{t^2}{2}+...+a_{n-1}\frac{t^n}{n}\right]}_0^t \end{gathered}$$

On applying the limits,

$T\left(f\left(t\right)\right)=a_{0}t+a_1\frac{t^2}{2}+...+a_{n-1}\frac{t^n}{n}$

Since, it non-zero there existslocalid="1659421702278" $0≤i≤n-1$such that $a_i≠0$.

Thus,$T\left(f\left(t\right)\right)$is also a non-zero polynomial in $Z_n$.

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

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

Thus, T is an isomorphism.