Answer to Question #232037 in Linear Algebra for Amuj

Question

Given a transformation  defined as. Find  and , range of                                                                

Answer 1

$Let L: R^{3} \rightarrow R^{2} \ be\ defined \ by:\\ L\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right)=\left(\begin{array}{l} x_{1}+x_{2} \\ x_{2}+x_{3} \end{array}\right)\\ \therefore{ker} L= \begin{aligned} &\operatorname{ker} L= \\ &\left\{\left(\begin{array}{l} x_{1} \\ x_{2} \\ x_{3} \end{array}\right) \mid\left(\begin{array}{l} x_{1}+x_{2} \\ x_{2}+x_{3} \end{array}\right)=\left(\begin{array}{l} 0 \\ 0 \end{array}\right)\right\} \end{aligned} \\ The \ homogeneous \ system \ coefficient \ matrix: \left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right) \text { has r.r.e.f. }\left(\begin{array}{ccc} 1 & 0 & -1 \\ 0 & 1 & 1 \end{array}\right) x_{3} \ is \ arbitrary, \\ x_{1}=x_{3}, x_{2}=-x_{3} \\ \operatorname{ker} L=\operatorname{span}\{\underbrace{(1,-1,1)}\} \\ \\\\\\ \quad basis \ for \ ker L. \\ \\\operatorname{ker} L \neq\{(0,0,0)\}\\ \begin{aligned} &\text { range } L=\left\{\left(\begin{array}{l} x_{1}+x_{2} \\ x_{2}+x_{3} \end{array}\right) \mid \text { for all } x_{1}, x_{2}, x_{3}\right\} \\ &=\left\{x_{1}\left(\begin{array}{l} 1 \\ 0 \end{array}\right)+x_{2}\left(\begin{array}{l} 1 \\ 1 \end{array}\right)+x_{3}\left(\begin{array}{l} 0 \\ 1 \end{array}\right) \mid\right. \text { for } \\ &\text { all } \left.x_{1}, x_{2}, x_{3}\right\} \end{aligned}\\ Find \ a \ basis \ for \ range \ L= span\{(1,0),(1,1),(0,1)\} \\ \begin{aligned} &\left(\begin{array}{lll} 1 & 1 & 0 \\ 0 & 1 & 1 \end{array}\right) \text { has r.r.e.f. }\left(\begin{array}{lll} 1 & 0 & -1 \\ 0 & 1 & 1 \end{array}\right) \\ &\Rightarrow \text { range } L=\text { span }\{\underbrace{(1,0),(1,1)}\} \end{aligned}\\ range L=R^{2} \\\Rightarrow L \ is \ onto.$