91影视

Consider the function$T\left(f\left(t\right)\right)=f\left(t\right)+f\text{'}\text{'}\left(t\right)$from$C^{鈭�}$to$C^{鈭�}$.

A function Dis called a linear transformation onif the function Dsatisfies the following property鈥檚.

1. $\mathbf{D}\left(\mathbf{x}\mathbf{+}\mathbf{y}\right)\mathbf{=}\mathbf{D}\left(\mathbf{x}\right)\mathbf{+}\mathbf{D}\left(\mathbf{y}\right)$for all $\mathbf{x}\mathbf{,}\mathbf{y}鈭�\mathrm{鈩�}$.
2. $\mathbf{D}\left(\mathbf{伪虫}\right)\mathbf{=}\mathbf{伪顿}\left(\mathbf{x}\right)$for all constent.$\mathbf{伪}鈭�\mathrm{鈩�}$

An invertible linear transformation is called isomorphism or dimension of domain and co-domain is not same then the function is not isomorphism.

Assume $f,g鈭�P$then$T\left(f\left(t\right)\right)=f\left(t\right)+f\text{'}\text{'}\left(t\right)$and$T\left(g\left(t\right)\right)=g\left(t\right)+g\text{'}\text{'}\left(t\right)$

Substitute the value $f\left(t\right)+f\text{'}\text{'}\left(t\right)$for $T\left(f\left(t\right)\right)$and $g\left(t\right)+g\text{'}\text{'}\left(t\right)$for $T\left(g\left(t\right)\right)$in $T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$as follows.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)=f\left(t\right)+f\text{'}\text{'}\left(t\right)+g\left(t\right)+g\text{'}\text{'}\left(t\right)$

Now, simplify $T\left(\left\{f+g\right\}\left(t\right)\right)$as follows.

$\begin{array}{rcl}T\left(\left\{f+g\right\}\left(t\right)\right)& =& \left\{f+g\right\}\left(t\right)+{\left\{f+g\right\}}^{\text{'}\text{'}}\left(t\right)\\ & =& f\left(t\right)+g\left(t\right)+f\text{'}\text{'}\left(t\right)+g\text{'}\text{'}\left(t\right)\\ & =& f\left(t\right)+f\text{'}\text{'}\left(t\right)+g\left(t\right)+g\text{'}\text{'}\left(t\right)\\ T\left(\left\{f+g\right\}\left(t\right)\right)& =& T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)\\ & & \end{array}$

Assume $f鈭�P$and $伪鈭�\mathrm{鈩�}$then$T\left(伪f\left(t\right)\right)=\left(伪f\right)\left(t\right)+{\left(伪f\right)}^{\text{'}\text{'}}\left(t\right)$ .

Simplify the equation $T\left(伪f\left(t\right)\right)=\left(伪f\right)\left(t\right)+{\left(伪f\right)}^{\text{'}\text{'}}\left(t\right)$as follows.

$\begin{array}{rcl}T\left(伪f\left(t\right)\right)& =& \left(伪f\right)\left(t\right)+{\left(伪f\right)}^{\text{'}\text{'}}\left(t\right)\\ & =& 伪\left\{f\left(t\right)\right\}+伪\left\{f\text{'}\text{'}\left(t\right)\right\}\\ & =& 伪\left\{f\left(t\right)+f\text{'}\text{'}\left(t\right)\right\}\\ T\left(伪f\left(t\right)\right)& =& 伪T\left(f\left(t\right)\right)\\ & & \end{array}$

As$T\left(\left\{f+g\right\}\left(t\right)\right)=T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)$ and$T\left(伪f\left(t\right)\right)=伪T\left(f\left(t\right)\right)$ , by the definition of linear transformation T is linear.

Differentiate the equation $f\left(t\right)=\cos \left(t\right)$both side with respect to t as follows.

$$\begin{gathered} f\left(t\right)=\cos \left(t\right) \\ f\text{'}\left(t\right)=-\sin \left(t\right) \end{gathered}$$

Again differentiate the equation $f\text{'}\left(t\right)=-\sin \left(t\right)$both side with respect to t as follows.

$$\begin{gathered} f\text{'}\left(t\right)=-\sin \left(t\right) \\ f\text{'}\text{'}\left(t\right)=-\cos \left(t\right) \end{gathered}$$

Substitute the values $\cos \left(t\right)$for $f\left(t\right)$and $-\cos \left(t\right)$for $f\text{'}\text{'}\left(t\right)$in the equation$T\left(f\left(t\right)\right)=f\left(t\right)+f\text{'}\text{'}\left(t\right)$as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=f\text{'}\left(t\right)+f\text{'}\text{'}\left(t\right) \\ T\left(f\left(t\right)\right)=\cos \left(t\right)-\cos \left(t\right) \\ T\left(f\left(t\right)\right)=0 \end{gathered}$$

As $T\left(f\left(t\right)\right)=0$but $f\left(t\right)鈮�0$means dimension of $Ker\left(T\right)$is not zero, by the theorem the transformation T is not an isomorphism.

Hence, the statement is false.