91影视

Consider the function T(f (t)) = f (4t - 3 ) from $P_2\mathrm{to}{P}_2$.

A function D is called a linear transformation on localid="1659425827307" $\mathbf{鈩�}$if the function Dsatisfies the following property鈥檚.

$$\begin{gathered} \mathbf{\left(}\mathbf{a}\mathbf{\right)}\mathbf{}\mathbf{D}\mathbf{}\mathbf{(}\mathbf{x}\mathbf{+}\mathbf{y}\mathbf{)}\mathbf{}\mathbf{=}\mathbf{D}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{+}\mathbf{D}\mathbf{\left(}\mathbf{y}\mathbf{\right)}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}\mathbf{x}\mathbf{,}\mathbf{y}\mathbf{}\mathbf{鈭�}\mathbf{鈩�} \\ \mathbf{\left(}\mathbf{b}\mathbf{\right)}\mathbf{}\mathbf{D}\mathbf{}\mathbf{\left(}\mathbf{伪虫}\mathbf{\right)}\mathbf{}\mathbf{=}\mathbf{伪顿}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{}\mathbf{for}\mathbf{}\mathbf{all}\mathbf{}\mathbf{constant}\mathbf{}\mathbf{伪}\mathbf{鈭�}\mathbf{鈩�} \end{gathered}$$

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

Assume $f,g鈭�P_2\mathrm{then}T(f(t\left)\right)=f(4t-3)\mathrm{and}T\left(g\right(t\left)\right)=g(4t-3)$.

Substitute the value f ( 4t - 3 ) for T ( f (t) )and g ( 4t - 3 ) for T ( f (t) ) in T ( f (t) ) +T ( f (t) ) as follows.

T( f (t) ) +T ( f (t) ) =f (4t - 3) + g (4t - 3)

Now, simplifyT ( { f + g } (t))as follows.

T({ f + g } (t)) = { f + g } ( 4t - 3)

T ( { f + g } (t)) = f ( 4t - 3) +g ( 4t - 3)

T ( { f + g } (t)) = T ( f(t) + T( g (t))

$\mathrm{Assume}f鈭�P_2\mathrm{and}伪鈭�\mathrm{鈩�}\mathrm{then}\mathit{T}\left(\right(伪f\left)\right)=\left(伪f\right)(4t-3).$

Simplify the equation $\mathit{T}\left(\right(伪f)(t\left)\right)=\left(伪f\right)(4t-3).$as follows.

$$\begin{gathered} \mathit{T}\left(\right(伪f)(t\left)\right)=\left(伪f\right)(4t-3) \\ \mathit{T}\left(\right(伪f\left)\right(t\left)\right)=伪f(4t-3) \\ \mathit{T}\left(\right(伪f\left)\right(t\left)\right)=伪T\left(f\right(t\left)\right). \end{gathered}$$

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

As $\left\{1,t,t^2,...\right\}$is the basis element of P and the matrix corresponding to basis is upper triangular matrix.

Each row and column of upper triangular matrix is linearly independent.

Therefore, the rank (T) is n.

The Kernel (T) is defined as follows.

dim(T) = dim (Kernal (T)) +rank (T)

n= dim ( Kernel (T)) + 0

dim (Kernel (T)) = 0

Therefore, the dimension of is Kernel (T).

Theorem: Consider a linear transformation T defined from $T:V鈫�W$then the transformation T is an isomorphism if and only ifKer (T) = 0 where .

$ker\left(T\right)=\left\{f\left(x\right)鈭�P:T\left\{f\left(x\right)\right\}=0\right\}$

As the dimension of Kernel (T) is 0, by the theorem the function T is an isomorphism.

Hence, the statement is True.