91影视

Consider the function $T\left(f\right)=\frac{f\left(t+h\right)-f\left(t\right)}{h}$from $P_{2}to{P}_2$.

A function D is called a linear transformation on$\mathbf{鈩浓划鈩浓划鈩�}$if the function D satisfies the following properties.

(a) $\mathit{D}\left(x+y\right)\mathbf{=}\mathit{D}\left(x\right)\mathbf{+}\mathit{D}\left(y\right)\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{x}\mathbf{,}\mathit{y}\mathbf{鈭�}\mathbf{鈩�}$.

(b) $\mathit{D}\left(伪x\right)\mathbf{=}\mathit{伪}\mathit{D}\left(x\right)\mathbf{}\mathit{f}\mathit{o}\mathit{r}\mathbf{}\mathit{a}\mathit{l}\mathit{l}\mathbf{}\mathit{c}\mathit{o}\mathit{n}\mathit{s}\mathit{t}\mathit{a}\mathit{n}\mathit{t}\mathbf{}\mathit{伪}\mathbf{鈭�}\mathbf{鈩�}\mathit{伪}\mathbf{鈭�}\mathbf{鈩�}\mathit{伪}\mathbf{鈭�}\mathbf{鈩�}\mathbf{.}$.

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

Assume $f,g鈭�P_2$then $T\left(f\right)=\frac{f\left(t+h\right)-f\left(t\right)}{h}andT\left(g\right)=\frac{g\left(t+h\right)-g\left(t\right)}{h}$.

Substitute the value$\frac{f\left(t+h\right)-f\left(t\right)}{h}$for$T\left(f\left(t\right)\right)$ and$\frac{g\left(t+h\right)-g\left(t\right)}{h}$ 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.

$T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)=\frac{f\left(t+h\right)-f\left(t\right)}{h}+\frac{g\left(t+h\right)-g\left(t\right)}{h}$

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

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

Assume$f鈭�P_2$ and $伪鈭�\mathrm{鈩�}$then$T\left(\left(伪f\right)\left(t\right)\right)=\frac{\left(伪f\right)\left(t+h\right)-\left(伪f\right)\left(t\right)}{h}$ .

Simplify the equation$T\left(\left(伪f\right)\left(t\right)\right)=\frac{\left(伪f\right)\left(t+h\right)-\left(伪f\right)\left(t\right)}{h}$ as follows.

$$\begin{gathered} T\left(\left(伪f\right)\left(t\right)\right)=\frac{\left(伪f\right)\left(t+h\right)-\left(伪f\right)\left(t\right)}{h} \\ =伪\frac{f\left(t+h\right)-f\left(t\right)}{h} \\ T\left(伪f\left(t\right)\right)=伪T\left(f\left(t\right)\right) \end{gathered}$$

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.

As$\left\{1,t,t^2\right\}$ is the basis element of$P_2$ , the matrix B is defined as follows.

$$\begin{gathered} T\left\{1\right\}=\frac{1-1}{h} \\ =0 \\ T\left\{1\right\}=0.1+0.t+0.t^2 \end{gathered}$$

The image of$T$ at point t is defined as follows.

$$\begin{gathered} T\left\{t\right\}=\frac{t+h-t}{h} \\ =\frac{h}{h} \\ T\left\{t\right\}=1.1+0.t+0.t^2 \end{gathered}$$

The image of T at point$t^2$ is defined as follows.

$$\begin{gathered} T\left\{t^2\right\}=\frac{{\left(t+h\right)}^2-t^2}{h} \\ =\frac{t^2+h^2+2ht-t^2}{h} \\ =h+2t \\ T\left\{t^2\right\}=h.1+2.t+0.t^2 \end{gathered}$$

Therefore, the matrix$B=\left[\begin{array}{ccc}0& 1& h\\ 0& 0& 2\\ 0& 0& 0\end{array}\right]$ .

The kernel(T) is defined as follows.

$Kernel\left(T\right)=\left\{f\left(t\right)=\overline{)0}f鈭�P_2\right\}$

Assume$f\left(t\right)=a+bt+c{t}^2$ , simplify$\frac{f\left(t+h\right)-f\left(t\right)}{h}=0$ as follows.

$$\begin{gathered} \frac{f\left(t+h\right)-f\left(t\right)}{h}=\frac{a+b\left(t+h\right)+c{\left(t+h\right)}^2-a-bt-c{t}^2}{h} \\ =\frac{bh+c\left(t^2+h^2+2ht\right)-c{t}^2}{h} \\ =\frac{bh+c{t}^2+c{h}^2+c2ht-c{t}^2}{h} \\ \frac{f\left(t+h\right)-f\left(t\right)}{h}=\frac{bh+c{h}^2+c2ht}{h} \end{gathered}$$

Further, simplify the equation as follows.

$$\begin{gathered} \frac{f\left(t+h\right)-f\left(t\right)}{h}=\frac{bh+c{h}^2+c2ht}{h} \\ \frac{f\left(t+h\right)-f\left(t\right)}{h}=b+ch+2ct \end{gathered}$$

As$\frac{f\left(t+h\right)-f\left(t\right)}{h}=0$ , simplify the equation$b+ch+2ct=0$ as follows.

$$\begin{gathered} b+ch+2ct=0.1+0.t+0.t^2 \\ b+ch=0 \\ 2c=0 \\ c=0 \end{gathered}$$

Substitute the value 0 for c in$b+ch=0$ as follows.

$$\begin{gathered} b+ch=0 \\ b=0 \end{gathered}$$

As $b=c=0$, the kernel(T) is spanned$\left\{1\right\}$ by implies$dim\left(kernel\left(T\right)\right)=1$ .

The rank(T) is defined as follows.

$$\begin{gathered} dim\left(T\right)=dim\left(kernel\left(T\right)\right)+rank\left(T\right) \\ 3=1+rank\left(T\right) \\ rank\left(T\right)=2 \end{gathered}$$

The rank(T) is defined as follows.

$rank\left(T\right)=\left\{f\left(t\right)=\overline{)0}伪鈭�R\right\}$

As rank(T) is spanned by$\left\{t,t^2\right\}$ implies the rank(T)=2 .

Therefore, the dimension of is and kernel(T) is 2 and the basis of kernel(T) is 1 and the basis of rank(T) is $\left\{t,t^2\right\}$.

Theorem: Consider a linear transformation T defined from $\mathit{T}\mathbf{:}\mathit{V}\mathbf{鈫�}\mathit{W}$then the transformation is an isomorphism if and only if Ker(T) = 0 where $\mathit{k}\mathit{e}\mathit{r}\left(T\right)\mathbf{=}\left\{f\left(x\right)鈭�P:T\left\{f\left(x\right)\right\}=0\right\}$.

As the dimension of kernel(T) is 1 , by the theorem the function T is not an isomorphism.

Hence, the rank of transformation T is 2 , kernel of the function T is 1 and linear transformation is not an isomorphism.