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Boris Novikov
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Fix $a\in V, a\ne 0$ and for $L\in {\rm Hom}(V,V)$ denote $L_a:x\to L(x)+a$. SetLet $$G=\{L_a|L\in {\rm Hom}(V,V) \}$$$G$ to be the set of all $L_a (L\in {\rm Hom}(V,V)$ and $$\lambda (L_a,M_a)(x) =L_a(x)+M_a(x)-a=(L+M)(x)+a.$$

Fix $a\in V, a\ne 0$ and for $L\in {\rm Hom}(V,V)$ denote $L_a:x\to L(x)+a$. Set $$G=\{L_a|L\in {\rm Hom}(V,V) \}$$ and $$\lambda (L_a,M_a)(x) =L_a(x)+M_a(x)-a=(L+M)(x)+a.$$

Fix $a\in V, a\ne 0$ and for $L\in {\rm Hom}(V,V)$ denote $L_a:x\to L(x)+a$. Let $G$ to be the set of all $L_a (L\in {\rm Hom}(V,V)$ and $$\lambda (L_a,M_a)(x) =L_a(x)+M_a(x)-a=(L+M)(x)+a.$$

Source Link
Boris Novikov
  • 3.1k
  • 1
  • 16
  • 17

Fix $a\in V, a\ne 0$ and for $L\in {\rm Hom}(V,V)$ denote $L_a:x\to L(x)+a$. Set $$G=\{L_a|L\in {\rm Hom}(V,V) \}$$ and $$\lambda (L_a,M_a)(x) =L_a(x)+M_a(x)-a=(L+M)(x)+a.$$