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 i s a v e c t o r i a l p l a n e . h i s b a s e i s

B = (\vec{i}; \vec{j}) . f i s a n e n d o m o r p h i s m d e f i n e d

b y f (\vec{i}) = - \frac{\sqrt{2}}{2} \vec{i} + \frac{\sqrt{2}}{2} \vec{j} a n d f (\vec{j}) = \frac{\sqrt{2}}{2} \vec{i} - \frac{\sqrt{2}}{2} \vec{j}

1) S h o w t h a t k e r f i s a v e c t o r i a l s t r a i g h

l i n e a n d h i s b a s e i s {\vec{e}}_{1} = \sqrt{2} \vec{i} + \sqrt{2} \vec{j}

2) s h o w t h a t G, t h e s e t o f v e c t o r s \vec{u}

\in E s u c h a s f (\vec{u}) = \sqrt{2} \vec{u} i s a v e c t o r i a l s t r a i g h

l i n e a n d h i s B a s e i s {\vec{e}}_{2} = \vec{i} + \vec{j}

3) D e t e r m i n e t h e m a t r i x A^{'} o f f i n

B^{'} i f B^{'} = ({\vec{e}}_{1}; {\vec{e}}_{2}) .