91影视

Consider a linear transformation T from $P_3$to${\mathrm{鈩�}}^{2脳2}$.

A function Dis called a linear transformation on Rif 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.

Theorem: Consider a linear transformationT defined from $\mathbf{T}\mathbf{:}\mathbf{V}鈫�\mathbf{W}$then the transformation Tis an isomorphism if and only if$\mathbf{dim}\left(\mathbf{Ker}\left(\mathbf{T}\right)\right)\mathbf{=}\mathbf{0}$where

$\mathbf{Ker}\left(\mathbf{T}\right)\mathbf{=}\left\{\mathbf{f}\left(\mathbf{x}\right)鈭�\mathbf{P}\mathbf{:}\mathbf{T}\left\{\mathbf{f}\left(\mathbf{x}\right)\right\}\mathbf{=}\mathbf{0}\right\}$and.

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

Consider $f鈭�P_3$then $T\left(f\left(t\right)\right)=\left[\begin{array}{cc}a-b& c\\ c& d\end{array}\right]$where$f\left(t\right)=a+bt+c{t}^2+d{t}^3$.

Assume, the polynomials$f\left(t\right)=a_1+b_{1}t+c_1{t}^2+d_1{t}^3$and $g\left(t\right)=a_2+b_{2}t+c_2{t}^2+d_2{t}^3$.

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

$\begin{array}{rcl}T\left(\left\{f+g\right\}\left(t\right)\right)& =& T\left(\left(a_1+a_2\right)+\left(b_1+b_2\right)t+\left(c_1+c_2\right)t^2+\left(d_1+d_2\right)t^3\right)\\ & =& \left[\begin{array}{cc}\left(a_1+a_2\right)-\left(b_1+b_2\right)& \left(c_1+c_2\right)\\ \left(c_1+c_2\right)& \left(d_1+d_2\right)\end{array}\right]\\ & =& \left[\begin{array}{cc}{a}_1-b_1& c_1\\ c_1& d_1\end{array}\right]+\left[\begin{array}{cc}{a}_2-b_2& c_2\\ c_2& d_2\end{array}\right]\\ & =& T\left(f\left(t\right)\right)+T\left(g\left(t\right)\right)\end{array}$

Assume$f鈭�P$ and $伪鈭�\mathrm{鈩�}$thenrole="math" localid="1660386123620" $T\left(伪f\left(t\right)\right)=\left[\begin{array}{cc}伪a-伪b& 伪c\\ 伪c& 伪d\end{array}\right]$

Simplify the equation $T\left(伪f\left(t\right)\right)=\left[\begin{array}{cc}伪a-伪b& 伪c\\ 伪c& 伪d\end{array}\right]$as follows.

$\begin{array}{rcl}T\left(伪f\left(t\right)\right)& =& \left[\begin{array}{cc}伪a-伪b& 伪c\\ 伪c& 伪d\end{array}\right]\\ & =& \left[\begin{array}{cc}伪\left(a-b\right)& c\\ c& d\end{array}\right]\\ & =& 伪\left[\begin{array}{cc}a-b& c\\ c& d\end{array}\right]\\ & =& 伪T\left(f\left(t\right)\right)\\ & & \end{array}$

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.

Consider a function$f\left(t\right)鈭�P_3$ such that$f\left(t\right)=1+t$.

Simplify the equation$\left(f\left(t\right)\right)=\left[\begin{array}{cc}a-b& c\\ c& d\end{array}\right]$if $f\left(t\right)=1+t$ as follows.

$$\begin{gathered} T\left(f\left(t\right)\right)=\left[\begin{array}{cc}a-b& c\\ c& d\end{array}\right] \\ T\left(f\left(t\right)\right)=\left[\begin{array}{cc}1-1& 0\\ 0& 0\end{array}\right] \\ T\left(f\left(t\right)\right)=\left[\begin{array}{cc}0& 0\\ 0& 0\end{array}\right] \end{gathered}$$

As $T\left(f\left(t\right)\right)=0$but$f\left(t\right)鈮�0$implies $\dim \left(kernal\left(T\right)\right)鈮�\left\{0\right\}$.

Therefore, the linear transformation T is not an isomorphism.

Hence, the statement is false.