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E-is-a-vectorial-plane-his-base-is-B-i-j-f-is-an-endomorphism-defined-by-f-i-2-2-i-2-2-j-and-f-j-2-2-i-2-2-j-1-Show-that-ker-f-is-a-vectorial-straigh-li




Question Number 97380 by mathocean1 last updated on 07/Jun/20
E is a vectorial plane. his base is   B=(i^→ ;j^→ ). f is an endomorphism defined  by f(i^→ )=−((√2)/2)i^→ +((√2)/2)j^→  and f(j^→ )=((√2)/2)i^→ −((√2)/2)j^→   1)Show that ker f is a vectorial straigh  line and his base is e_1 ^→ =(√2)i^→ +(√2)j^→   2)show that G, the set of vectors u^→    ∈ E such as f(u^→ )=(√2)u^→  is a vectorial straigh  line and his Base is e_(2  ) ^→ =i^→ +j^→   3) Determine the matrix A′ of f in  B′ if B′=(e_1 ^→ ;e_2 ^→ ).
$${E}\:{is}\:{a}\:{vectorial}\:{plane}.\:{his}\:{base}\:{is}\: \\ $$$${B}=\left(\overset{\rightarrow} {{i}};\overset{\rightarrow} {{j}}\right).\:{f}\:{is}\:{an}\:{endomorphism}\:{defined} \\ $$$${by}\:{f}\left(\overset{\rightarrow} {{i}}\right)=−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{i}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{j}}\:{and}\:{f}\left(\overset{\rightarrow} {{j}}\right)=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{i}}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\overset{\rightarrow} {{j}} \\ $$$$\left.\mathrm{1}\right){Show}\:{that}\:{ker}\:{f}\:{is}\:{a}\:{vectorial}\:{straigh} \\ $$$${line}\:{and}\:{his}\:{base}\:{is}\:\overset{\rightarrow} {{e}}_{\mathrm{1}} =\sqrt{\mathrm{2}}\overset{\rightarrow} {{i}}+\sqrt{\mathrm{2}}\overset{\rightarrow} {{j}} \\ $$$$\left.\mathrm{2}\right){show}\:{that}\:{G},\:{the}\:{set}\:{of}\:{vectors}\:\overset{\rightarrow} {{u}} \\ $$$$\:\in\:{E}\:{such}\:{as}\:{f}\left(\overset{\rightarrow} {{u}}\right)=\sqrt{\mathrm{2}}\overset{\rightarrow} {{u}}\:{is}\:{a}\:{vectorial}\:{straigh} \\ $$$${line}\:{and}\:{his}\:{Base}\:{is}\:\overset{\rightarrow} {{e}}_{\mathrm{2}\:\:} =\overset{\rightarrow} {{i}}+\overset{\rightarrow} {{j}} \\ $$$$\left.\mathrm{3}\right)\:{Determine}\:{the}\:{matrix}\:{A}'\:{of}\:{f}\:{in} \\ $$$${B}'\:{if}\:{B}'=\left(\overset{\rightarrow} {{e}}_{\mathrm{1}} ;\overset{\rightarrow} {{e}}_{\mathrm{2}} \right). \\ $$

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