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 $T\left(f\left(x\right)\right)=\left\{f\left(0\right),f\text{'}\left(0\right),f"\left(0\right),....\right\}$.

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 P.

Now, determine the derivative at point zero as follows.

localid="1659926956866" $$\begin{gathered} \mathrm{f}\left(0\right)={\mathrm{a}}_0 \\ \mathrm{f}\text{'}\left(0\right)={\mathrm{a}}_1 \\ \mathrm{f}"\left(0\right)=2{\mathrm{a}}_2 \\ {\mathrm{f}}^{\left(3\right)}\left(0\right)=6{\mathrm{a}}^3 \\ â‹\circledR \\ {\mathrm{f}}^{\left(\mathrm{n}\right)}\left(0\right)=\mathrm{n}!{\mathrm{a}}_{\mathrm{n}} \\ {\mathrm{f}}^{\left(\mathrm{n}+1\right)}\left(0\right)=0 \end{gathered}$$

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

$\mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\left\{{\mathrm{a}}_0,{\mathrm{a}}_1,2{\mathrm{a}}_2,...\mathrm{n}!{\mathrm{a}}_{\mathrm{n}},0,0,...\right\}$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 T 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$\mathbf{ker}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{\left\{}\mathbf{0}\mathbf{\right\}}$and$\mathbf{im}\mathbf{\left(}\mathbf{T}\mathbf{\right)}\mathbf{=}\mathbf{W}$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(x) and g(x) from and as follows.

$$\begin{gathered} T\left(f\right(x)+g(x\left)\right)=\left\{f\left(0\right)+g\left(0\right),\left(f\left(0\right)+g\left(0\right)\right)\text{'},...,{\left(f\left(0\right)+g\left(0\right)\right)}^{n-1},0,0...\right\} \\ =\left(f\left(0\right)+g\left(0\right),f\text{'}\left(0\right)+g\text{'}\left(0\right),...,f^n\left(0\right)+g^n\left(0\right),0...\right) \\ =\left(f\left(0\right),f\text{'}\left(0\right),...,f^n\left(0\right),0,...\right)+\left(g\left(0\right),g\text{'}\left(0\right),...,g^n\left(0\right),0,...\right) \\ T\left(f\right(x)+g(x\left)\right)=T\left(f\right(x\left)\right)+T\left(g\right(x\left)\right) \end{gathered}$$

Now, to check the second condition as follows.

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

$$\begin{gathered} \mathrm{T}\left(\mathrm{Î\pm ´Ú}\left(\mathrm{x}\right)\right)=\left(\mathrm{Î\pm ´Ú}\left(0\right),\mathrm{Î\pm ´Ú}\text{'}\left(0\right),...,{\mathrm{Î\pm ´Ú}}^{\mathrm{n}}\left(0\right),0,...\right) \\ =\mathrm{Î\pm }\left(\mathrm{f}\left(0\right),\mathrm{f}\text{'}\left(0\right),...,{\mathrm{f}}^{\mathrm{n}}\left(0\right),0,...\right) \\ \mathrm{T}\left(\mathrm{Î\pm ´Ú}\left(\mathrm{x}\right)\right)=\mathrm{Î\pm °Õ}\left(\mathrm{f}\left(\mathrm{x}\right)\right) \end{gathered}$$

Thus, T is a linear transformation.

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

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.

localid="1659935142710" $\begin{array}{l}\mathrm{p}\left(\mathrm{x}\right)=0×{\mathrm{x}}^{\mathrm{n}}+…+0×\mathrm{x}+0\\ \mathrm{p}\left(\mathrm{x}\right)=0\end{array}$

And the codomain is the space of all sequences V and hence the zero will be $\left(0.0.0,…\right)$.

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 P

The next equation as follows.

$$\begin{gathered} T\left(f\right(x\left)\right)=\left(0,0,0,...\right) \\ \left(f\left(0\right),f\text{'}\left(0\right),...f^n\left(0\right),0..\right)=(0,0,0,...) \\ \left(a_0,a_1,2a_2,...,n!a_n,0,0,...\right)=(0,0,0,...) \end{gathered}$$

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.

localid="1659933297901" $\begin{array}{l}(\mathrm{x},\mathrm{y})=(\mathrm{z},\mathrm{y})\\ ⇔\mathrm{x}=\mathrm{z}âˆ\P Ä\mathrm{y}=\mathrm{t}\end{array}$

Hence it must be as follows.

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

This implies that as follows.

localid="1659929108435" $\begin{array}{l}\ker \left(\mathrm{T}\right)=\left\{0·{\mathrm{x}}^{\mathrm{n}}+…+0·\mathrm{x}+0\right\}\\ \ker \left(\mathrm{T}\right)=\left\{0\right\}\end{array}$.

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 $\mathrm{T}\left(\mathrm{f}\right(\mathrm{x}\left)\right)=\left({\mathrm{a}}_0,{\mathrm{a}}_1,2{\mathrm{a}}_2,...,\mathrm{n}!{\mathrm{a}}_{\mathrm{n}},0,0,...\right)$.

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

The reason for this is simple.

Since, it has the polynomial and the transformation can be defined by the derivatives.

Suppose that a polynomial of degree n, and all the derivatives greater than will be equal to zero.

So, only finally many elements will be non-zero numbers, the rest of them is equal to zero.

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.