Question Number 215525 by zetamaths last updated on 09/Jan/25
$$\Phi\::\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}^{\mathrm{2}} \: \\ $$$$\:\:\:\left({x}.;{y}\right)\mid\rightarrow\left(\mathrm{2}{x}+\mathrm{3}{y}:\mathrm{3}{y}\right) \\ $$$${find} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$\Phi\in\mathscr{L}\left(\mathbb{R}^{\mathrm{2}} \right) \\ $$$${find}\:{ker}\left(\Phi\right)\:{and}\:{the}\:{im}\left(\Phi\right) \\ $$$${f}\:{o}\:{f}\:=? \\ $$$$ \\ $$
Answered by MrGaster last updated on 09/Jan/25
$$\left.\mathrm{ker}\left(\Phi\right)=\left\{{x},{y}\right\}\in\:\mathbb{R}^{\mathrm{2}} \mid\Phi\left({x},{y}\right)=\left(\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$\begin{pmatrix}{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{3}}\end{pmatrix}\begin{pmatrix}{{x}}\\{{y}}\end{pmatrix}=\begin{pmatrix}{\mathrm{0}}\\{\mathrm{0}}\end{pmatrix} \\ $$$$\begin{cases}{\mathrm{2}{x}+\mathrm{3}{y}=\mathrm{0}}\\{\mathrm{3}{y}=\mathrm{0}}\end{cases} \\ $$$${y}=\mathrm{0}\Rightarrow\mathrm{2}{x}=\mathrm{0}\Rightarrow{x}=\mathrm{0} \\ $$$$\mathrm{kel}\left(\Phi\right)=\left\{\left(\mathrm{0},\mathrm{0}\right)\right\}\checkmark \\ $$$$\mathrm{im}\left(\Phi\right)=\left\{\Phi\left({x},{y}\right)\mid\left({x},{y}\right)\in\:\mathbb{R}^{\mathrm{2}} \right\} \\ $$$$\mathrm{im}\left(\Phi\right)=\mathrm{sqan}\left(\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix},\begin{pmatrix}{\mathrm{3}}\\{\mathrm{3}}\end{pmatrix}\right) \\ $$$$\mathrm{Reduse}\:\mathrm{to}\:\mathrm{echelon}\:\mathrm{form}: \\ $$$$\begin{pmatrix}{\mathrm{2}}&{\mathrm{3}}\\{\mathrm{0}}&{\mathrm{3}}\end{pmatrix}\sim\begin{pmatrix}{\mathrm{2}}&{\mathrm{0}}\\{\mathrm{0}}&{\mathrm{3}}\end{pmatrix} \\ $$$$\mathrm{im}\left(\Phi\right)=\mathrm{sqan}\left(\left(\begin{pmatrix}{\mathrm{2}}\\{\mathrm{0}}\end{pmatrix},\begin{pmatrix}{\mathrm{0}}\\{\mathrm{3}}\end{pmatrix}\right)\right. \\ $$$$=\mathrm{sqan}\left(\begin{pmatrix}{\mathrm{1}}\\{\mathrm{0}}\end{pmatrix},\begin{pmatrix}{\mathrm{0}}\\{\mathrm{1}}\end{pmatrix}\right) \\ $$$$=\mathbb{R}^{\mathrm{2}} \checkmark \\ $$$$\mathrm{Conclusion}: \\ $$$$\mathrm{ker}\left(\Phi\right)=\left\{\left(\mathrm{0},\mathrm{0}\right)\right\} \\ $$$$\mathrm{im}\left(\Phi\right)=\mathbb{R}^{\mathrm{2}} \\ $$$${f}=\mathrm{2} \\ $$