Question Number 192239 by mathdave last updated on 12/May/23
Commented by mathdave last updated on 12/May/23
$${can}\:{someone}\:{pls}\:{help}\:{me}\:{out}\:{it}\:{very}\:{urgent} \\ $$
Answered by aleks041103 last updated on 13/May/23
$${Choose}\:{basis}\:\left\{{v}_{\mathrm{1}} ,…,{v}_{{n}} \right\}\:{for}\:{V}\:{and}\: \\ $$$${basis}\:\left\{{u}_{\mathrm{1}} ,…,{u}_{{n}} \right\}\:{for}\:{U} \\ $$$${then}\:{we}\:{can}\:{associate}\:{a}\:{matrix}\:{M}\in{M}_{{n}} \left(\mathbb{R}\right)\:{with} \\ $$$${the}\:{linear}\:{operator}\:{F}:{V}\rightarrow{U}\:{as} \\ $$$${F}\:\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{a}_{{i}} {v}_{{i}} \right)=\underset{{j}=\mathrm{1}} {\overset{{n}} {\sum}}\left(\underset{{k}=\mathrm{1}} {\overset{{n}} {\sum}}{M}_{{jk}} {a}_{{k}} \right){u}_{{j}} \\ $$$${F}\:−\:{nonsing}.\Leftrightarrow{det}\left({M}\right)\neq\mathrm{0}\Leftrightarrow\exists{M}^{−\mathrm{1}} \Leftrightarrow \\ $$$$\Leftrightarrow\exists{F}\:^{−\mathrm{1}} :{U}\rightarrow{V}\Leftrightarrow{F}\:{is}\:{bijection}\:{from}\:{V}\:\:{to}\:{U} \\ $$$$\Leftrightarrow{F}\:−\:{isomorhism}\:\:\:\Box. \\ $$$$ \\ $$