91影视

Consider two linear spaces V and W. A transformation T is said to be a linear transformation if it satisfies the properties,

$$\begin{gathered} \mathit{T}\left(f+g\right)\mathbf{=}\mathit{T}\mathbf{\left(}\mathit{f}\mathbf{\right)}\mathbf{+}\mathit{T}\mathbf{\left(}\mathit{g}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathit{T}\left(kv\right)\mathbf{=}\mathit{k}\mathit{T}\mathbf{\left(}\mathit{v}\mathbf{\right)} \end{gathered}$$

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

An invertible linear transformation is called an isomorphism.

Consider the transformation $T\left(f\left(t\right)\right)=f\left(2t\right)-f\left(t\right)$, from $P_2$to $P_2$.

Check whether the transformationsatisfies the below two conditions or not.

role="math" localid="1659847299713" $$\begin{gathered} 1.T(\mathrm{f}+\mathrm{g})=T\left(f\right)+T\left(g\right) \\ 2.T\left(kf\right)=kT\left(v\right) \end{gathered}$$

Verify the first condition.

Let $f\left(t\right)$ and $g\left(t\right)$ be two polynomial functions from. Then,$T\left(f\left(t\right)\right)=f\left(2t\right)-f\left(t\right),T\left(g\left(t\right)\right)=g\left(2t\right)-g\left(t\right)$

Find $T\left(f\left(t\right)+g\left(t\right)\right)$.

$$\begin{gathered} T\left(f\left(t\right)+g\left(t\right)\right)=T\left(\left(f+g\right)\left(t\right)\right) \\ =\left(f+g\right)\left(2t\right)-(f+g)\left(t\right) \\ =f\left(2t\right)+g\left(2t\right)-f\left(t\right)-g\left(t\right) \\ =(f\left(2t\right)-f(t\left)\right)+\left(g\left(2t\right)-g\left(t\right)\right) \end{gathered}$$

$T\left(f\left(t\right)+g\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$

It is clear that, the first condition$T\left(f+g\right)=T\left(f\right)+T\left(g\right)$ is satisfied.

Verify the second condition.

Let k be an arbitrary scalar, and$f\left(t\right)鈭�P_2$ as follows.

$$\begin{gathered} T\left(\left(kf\right)\left(t\right)\right)=\left(kf\right)\left(2t\right)-\left(kf\right)\left(t\right) \\ =k\left(f\right(2t\left)\right)-k\left(f\left(t\right)\right) \\ =k\left(f\right(2t\left)\right)-f\left(t\right) \\ =kT\left(f\left(t\right)\right) \end{gathered}$$

It is clear that, the second condition$T\left(kf\right)=kT\left(f\right)$ is also satisfied.

Thus, T is a linear transformation.

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

Now, check whether$\ker \left(T\right)=\left\{0\right\}$.

According to the definition of the kernel of a transformation,

$ker\left(T\right)=\left\{f鈭�P_{2}T\left(f\left(t\right)\right)=0\right\}$.

Consider a polynomial function $f\left(t\right)鈭�P_2$as$f\left(t\right)=a+bt+c{t}^2$

Then,

$$\begin{gathered} T\left(f\left(t\right)\right)=0 \\ f\left(2t\right)-f\left(t\right)=0 \\ a+b\left(2t\right)+c{\left(2t\right)}^2-\left(a+bt+c{t}^2\right)=0 \\ a+2bt+4c{t}^2-a-bt-c{t}^2=0+0t+0t^2 \end{gathered}$$

$bt+3c{t}^2=0t+0t^2$

Comparing both sides, it can be concluded that role="math" localid="1659848345313" $b=0,c=0$

Here, the variable is a free variable. This means, the kernel of the transformation T will be $ker\left(T\right)=\left\{a+0t+0t^2|a鈭�\mathrm{鈩�}\right\}$.

Clearly, $\ker \left(T\right)鈮�\left\{0\right\}$.

It can be concluded that, one of the property of the isomorphism definition is not satisfied.

Therefore, the transformation T is not an isomorphism.

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