91Ó°ÊÓ

Let us define the sets P and V as follows.

$\begin{array}{l}\mathrm{P}=\left\{\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+{\mathrm{a}}_1\mathrm{x}+...|\mathrm{n}∈{\mathrm{N}}_0\right\}\\ \mathrm{V}=\left\{\left\{{\mathrm{x}}_0,{\mathrm{x}}_1,{\mathrm{X}}_2.{\mathrm{x}}_3,...\right\}|{\mathrm{x}}_{\mathrm{i}}∈{\mathrm{R}}_{\mathrm{i}},\mathrm{i}∈{\mathrm{N}}_0\right\}\end{array}$

Now, let us define the transformation as follows.

$\mathrm{T}:\mathrm{P}→\mathrm{V}$with $\mathrm{T}\left(\mathrm{f}\left(\mathrm{x}\right)\right)=\left\{\mathrm{f}\left(0\right),\mathrm{f}\left(1\right),\mathrm{f}\left(2\right),....\right\}$.

Let$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+...+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_0$ be an arbitrary polynomial from the P.

Now, determine the derivative at point zero as follows.

$$\begin{gathered} f\left(0\right)=a_0 \\ f\left(1\right)=a_0+a_1+a_2+...+a_n \\ f\left(2\right)=a_0+2a_1+...+2^n{a}_n \\ f\left(3\right)=a_0+3a_1+...+3^n{a}_n \\ â‹\circledR \\ f\left(n\right)=a_0+n{a}_1+...+n^n{a}_n \\ â‹\circledR \end{gathered}$$

From the derivative conclude that the transformation can be defined in another way as follows.

role="math" localid="1659932511053" $T\left(f\right(x\left)\right)=\left\{a_0,\underset{k=0}{\overset{n}{∑}}{a}_k,\underset{k=0}{\overset{n}{∑}}{2}^k{a}_k,...\underset{k=0}{\overset{{}_n}{∑}}{n}^k{a}_k,...\right\}$for an arbitrary polynomial as follows.

$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+...+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_0$.

Consider two linear spaces V and W. A function is said to be linear transformation if the following holds.

$$\begin{gathered} \mathbf{T}\mathbf{(}\mathbf{f}\mathbf{+}\mathbf{g}\mathbf{)}\mathbf{=}\mathbf{T}\mathbf{\left(}\mathbf{f}\mathbf{\right)}\mathbf{+}\mathbf{T}\mathbf{\left(}\mathbf{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{T}\mathbf{\left(}\mathbf{kf}\mathbf{\right)}\mathbf{=}\mathbf{kT}\mathbf{\left(}\mathbf{f}\mathbf{\right)} \end{gathered}$$

For all elements f,g of V and k is scalar.

A linear transformation $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{→}\mathbf{W}$is said to be an isomorphism if and only if $\mathit{k}\mathit{e}\mathit{r}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and $\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{W}$or $\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{W}\mathbf{\right)}$.

To check the first condition as follows.

Consider an arbitrary polynomial f(0) and g(0) from and as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{f}\right(\mathrm{x})+\mathrm{g}(\mathrm{x}\left)\right)=\left\{\mathrm{f}\left(0\right)+\mathrm{g}\left(0\right),\left(\mathrm{f}\left(0\right)+\mathrm{g}\left(1\right)\right),...,\left(\mathrm{f}\left(\mathrm{n}\right)+\mathrm{g}\left(\mathrm{n}\right)\right),...\right\} \\ =\left(\mathrm{f}\left(0\right),\mathrm{f}\left(1\right),...,\mathrm{f}\left(\mathrm{n}\right),...\right)+\left(\mathrm{g}\left(0\right),\mathrm{g}\left(1\right),\mathrm{g}\left(\mathrm{n}\right),...\right) \\ \mathrm{T}\left(\mathrm{f}\right(\mathrm{x})+\mathrm{g}(\mathrm{x}\left)\right)=\mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)+\mathrm{T}\left(\mathrm{g}\right(\mathrm{x}\left)\right) \end{gathered}$$

Now, to check the second condition as follows.

Let$\mathrm{Î\pm }$ be an arbitrary scalar, and $f\left(x\right)∈\mathrm{P}$.

role="math" localid="1659931898993" $$\begin{gathered} \mathrm{T}\left(\mathrm{Î\pm ´Ú}\left(\mathrm{x}\right)\right)=\left(\mathrm{Î\pm ´Ú}\left(0\right),\mathrm{Î\pm ´Ú}\left(1\right),...,\mathrm{Î\pm ´Ú}\left(\mathrm{n}\right),...\right) \\ =\mathrm{Î\pm }\left(\mathrm{f}\left(0\right),\mathrm{f}\left(0\right),...,\mathrm{f}\left(\mathrm{n}\right),0,...\right) \\ \mathrm{T}\left(\mathrm{Î\pm ´Ú}\left(\mathrm{x}\right)\right)=\mathrm{Î\pm °Õ}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \end{gathered}$$

Thus, T is a linear transformation.

A linear transformation $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{→}\mathbf{W}$is isomorphism if and only if $\mathbf{ker}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and$\mathit{I}\mathit{m}\mathbf{\left(}\mathit{t}\mathbf{\right)}\mathbf{=}\mathbf{W}$

Now, check if $\ker \left(t\right)=\left\{0\right\}$.

The domain is the space of all polynomials P and hence the zero in the domain will be zero polynomial as follows.

$\begin{array}{l}\mathrm{p}\left(\mathrm{x}\right)=0×{\mathrm{x}}^{\mathrm{n}}+…+0×\mathrm{x}+0\\ \mathrm{p}\left(\mathrm{x}\right)=0\end{array}$

And the codomain is the space of all sequences V and hence the zero will be localid="1659933798650" $\left(0,0,0,…\right)$.

The kernel as follows.

$\ker \left(\mathrm{T}\right)=\left\{\mathrm{x}∈{\mathrm{P}}_2|\mathrm{T}\left(\mathrm{x}\right)=\left(0,0,0,...\right)\right\}$

Let$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+...+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_0$be an arbitrary polynomial from

The next equation as follows.

$$\begin{gathered} T\left(f\right(x\left)\right)=\left(0,0,0,...\right) \\ \left(f\left(0\right),f\left(1\right),...f^n\left(n\right),0..\right)=(0,0,0,...) \\ \left\{a_0,\underset{k=0}{\overset{n}{∑}}{a}_k,\underset{k=0}{\overset{n}{∑}}{2}^k{a}_k,...\underset{k=0}{\overset{n}{∑}}{n}^k{a}_k,...\right\}=(0,0,0,...) \end{gathered}$$

The last equation of the transformation is different.

Now, the two vectors are equal only and only if their coordinates are the same as follows.

$\begin{array}{l}(\mathrm{x},\mathrm{y})=(\mathrm{z},\mathrm{y})\\ ⇔\mathrm{x}=\mathrm{z}âˆ\P Ä\mathrm{y}=\mathrm{t}\end{array}$

Hence it must be as follows.

$\underset{\mathrm{k}=0}{\overset{\mathrm{n}}{∑}}{\mathrm{n}}^{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}}=0,\left(âˆ\P Ä\mathrm{n}∈{\mathrm{N}}_0\right)\left(âˆ\P Ä\mathrm{k}∈\left(0,1,2,...,\mathrm{n}\right)\right)$.

The conclusion is that it form a system of equations as follows.

$$\begin{gathered} {\mathrm{a}}_0=0 \\ {\mathrm{a}}_0+{\mathrm{a}}_1+...+{\mathrm{a}}_{\mathrm{n}}=0 \\ {\mathrm{a}}_0+2{\mathrm{a}}_1+...+2^n{\mathrm{a}}_n=0 \\ â‹\circledR \\ {\mathrm{a}}_0+n{\mathrm{a}}_1+...+{\mathrm{n}}^{\mathrm{n}}{\mathrm{a}}_{\mathrm{n}}=0 \\ â‹\circledR \end{gathered}$$

The solution of the last system must be only trivial and hence as follows.

${\mathrm{a}}_{\mathrm{k}}=0,\left(âˆ\P Ä\mathrm{k}∈\left(0,1,2,...,\mathrm{n}\right)\right)$.

This implies that as follows.

$\begin{array}{l}\ker \left(\mathrm{T}\right)=\left\{0·{\mathrm{x}}^{\mathrm{n}}+…+0·\mathrm{x}+0\right\}\\ \ker \left(\mathrm{T}\right)=\left\{0\right\}\end{array}$.

A linear transformation $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{→}\mathbf{W}$is isomorphism if and only if $\mathbf{ker}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and$\mathbf{Im}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{W}$

Now, let us define the image of the transformation.

Let$\mathrm{f}\left(\mathrm{x}\right)={\mathrm{a}}_{\mathrm{n}}{\mathrm{x}}^{\mathrm{n}}+...+{\mathrm{a}}_1\mathrm{x}+{\mathrm{a}}_0$ be an arbitrary polynomial from the P.

Then $\mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\left\{{\mathrm{a}}_0,\underset{\mathrm{k}=0}{\overset{\mathrm{n}}{∑}}{\mathrm{a}}_{\mathrm{k}},\underset{\mathrm{k}=0}{\overset{\mathrm{n}}{∑}}{2}^{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}},...\underset{\mathrm{k}=0}{\overset{{}_{\mathrm{n}}}{∑}}{\mathrm{n}}^{\mathrm{k}}{\mathrm{a}}_{\mathrm{k}},...\right\}$.

From the last it seems that one element are in image of the transformation.

Let$(\mathrm{k},0,0,…),k∈\mathrm{R}$ is an element from the space and there is no polynomial f(x) and P so thatrole="math" localid="1659932426790" $T(\mathrm{f}\left(\mathrm{x}\right))=(\mathrm{k},0,0,…)$ holds, which implies that as follows.

$\mathrm{lm}\left(\mathrm{T}\right)≠\mathrm{V}$.

Since, $\mathrm{lm}\left(\mathrm{T}\right)≠\mathrm{V}$.

Therefore, T is not an isomorphism.

Thus, T is a linear transformation also $\ker \left(\mathrm{T}\right)=\left\{0\right\}$,$\mathrm{lm}\left(\mathrm{T}\right)≠\mathrm{V}$ and T is not an isomorphism.