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

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

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

A linear transformation$\mathbf{T}\mathbf{:}\mathbf{V}\mathbf{鈫�}\mathbf{W}$ 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 $\mathbf{dim}\mathbf{\left(}\mathbf{V}\mathbf{\right)}\mathbf{=}\mathbf{dim}\mathbf{\left(}\mathbf{W}\mathbf{\right)}$.

The given transformation as follows.

$\mathrm{T}\left({\mathrm{x}}_{0,}{\mathrm{x}}_1,{\mathrm{x}}_2,{\mathrm{x}}_3,...\right)=\left({\mathrm{x}}_{0,}{\mathrm{x}}_2,{\mathrm{x}}_4,...\right)$, from V to V.

By using the definition of linear transformation as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{A}+\mathrm{B}\right)=\mathrm{T}\left(\mathrm{A}\right)+\mathrm{T}\left(\mathrm{B}\right) \\ \mathrm{T}\left(\mathrm{kA}\right)=\mathrm{kT}\left(\mathrm{A}\right) \end{gathered}$$

Now, to check the first condition as follows.

Consider two infinite sequences V from as follows.

$x=\left(x_0.x_1,...\right)$and$y=\left(y_0.y_1,...\right)$be an infinite sequences from V.

Now, simplify as follows.

$$\begin{gathered} \mathrm{T}(\mathrm{x}+\mathrm{y})=\mathrm{T}\left({\mathrm{x}}_0+{\mathrm{y}}_0,{\mathrm{x}}_1+{\mathrm{y}}_1+...\right) \\ =\left({\mathrm{x}}_0+{\mathrm{y}}_0,{\mathrm{x}}_2+{\mathrm{y}}_2+...\right) \\ =\left({\mathrm{x}}_0+{\mathrm{x}}_2,...\right)+\left({\mathrm{y}}_0+{\mathrm{y}}_2,...\right) \\ \mathrm{T}(\mathrm{x}+\mathrm{y})=\mathrm{T}\left(\mathrm{x}\right)+\mathrm{T}\left(\mathrm{y}\right) \end{gathered}$$

Simplify for the second condition as follows.

$$\begin{gathered} \mathrm{T}\left(\mathrm{kx}\right)=\mathrm{T}\left({\mathrm{kx}}_0,{\mathrm{kx}}_1,{\mathrm{kx}}_2+...\right) \\ =\left({\mathrm{kx}}_0,{\mathrm{kx}}_2,{\mathrm{kx}}_4+...\right) \\ =\mathrm{k}\left({\mathrm{x}}_0,{\mathrm{x}}_2,{\mathrm{x}}_4\right) \\ \mathrm{T}\left(\mathrm{kx}\right)=\mathrm{kT}\left(\mathrm{x}\right) \end{gathered}$$

Thus, T is a linear transformation.

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

Now, it easy to show that$\mathrm{T}\left(0,1,0,1,...\right)=\left(0,0,0,...\right)$ and $\mathrm{T}\left(\left(0,2,0,2,...\right)\right)=\left(0,0,0,...\right)$.

That is the inverse ofrole="math" localid="1659416754005" $\left(0,0,0,鈥�\right)$ does not have unique vector to map to.

This means that T is not an isomorphism.

Thus, T is a linear transformation and not an isomorphism.