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

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

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

A linear transformation role="math" localid="1659755138732" $\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$is said to be an isomorphism if and only if $\mathbf{ker}\left(T\right)\mathbf{=}\left\{0\right\}\mathbf{}\mathbf{and}\mathbf{}\mathbf{im}\left(T\right)\mathbf{=}\mathbf{W}\mathbf{}\mathbf{or}\mathbf{}\mathbf{dim}\left(V\right)\mathbf{=}\mathbf{dim}\left(W\right)$.

Consider that all the real numbers ${\mathrm{c}}_0,{\mathrm{c}}_1,...,{\mathrm{c}}_{\mathrm{n}}$are distinct

Since, any non-zero polynomial of degree has at most n roots.

Therefore, there exist no non-zero polynomial of degree has at most n roots such that role="math" localid="1659755299180" $\mathrm{f}\left({\mathrm{c}}_0\right)=\mathrm{f}\left({\mathrm{c}}_1\right)=...f\left({\mathrm{c}}_{\mathrm{n}}\right)=0$.

Then, for all non-zero polynomial of degree has at most n as follows.

$\mathrm{T}\left(\mathrm{f}\left(\mathrm{t}\right)\right)鈮�\left[\begin{array}{c}0\\ 0\\ 鈰�\\ 0\end{array}\right]$

Thus, the kernel as follows:

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

And the dimension as follows.

$$\begin{gathered} \dim \left({\mathrm{P}}_{\mathrm{n}}\right)=\dim \left({\mathrm{R}}^{\mathrm{n}+1}\right) \\ =\mathrm{n}+1 \end{gathered}$$

Thus, $\mathrm{T}\left(\mathrm{f}\left(\mathrm{t}\right)\right)=\left[\begin{array}{c}\mathrm{f}\left({\mathrm{c}}_0\right)\\ \mathrm{f}\left({\mathrm{c}}_1\right)\\ 鈰�\\ \mathrm{f}\left({\mathrm{c}}_{\mathrm{n}}\right)\end{array}\right]$is an isomorphism from ${\mathrm{P}}_{\mathrm{n}}\mathrm{to}{\mathrm{R}}^{\mathrm{n}+1}$for distinct real numbers ${\mathrm{c}}_0,{\mathrm{c}}_1,...,{\mathrm{c}}_{\mathrm{n}}$.