Linear Functional Operations in Vector Spaces and Fields

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daveb
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A linear functional is a function g:V to F where V is a vector space over a field F such that if u and v are elements of V and a is an element of F, then g(u+v) = g(u) + g(v) and
g(au) = ag(u)
Let G be the space of all linear functionals on V. Then if [tex]\oplus_{1}[/tex] and [tex]\otimes_{1}[/tex] are (repectively) addition and scalar multiplication in V, [tex]\oplus_{2}[/tex] and [tex]\otimes_{2}[/tex] are (repectively) addition and multiplication in F, and [tex]\oplus_{3}[/tex] and [tex]\otimes_{3}[/tex] are (repectively) addition and scalar multiplication in G, am I correct in saying that the correct way is to write
g(u[tex]\oplus_{1}[/tex]v) = g(u) [tex]\oplus_{3}[/tex] g(v) and
g(a[tex]\otimes_{1}[/tex]u) = a[tex]\otimes_{3}[/tex]g(u)?

(God I hope the latex worked out..it's my first time)
 
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daveb said:
A linear functional is a function g:V to F where V is a vector space over a field F such that if u and v are elements of V and a is an element of F, then g(u+v) = g(u) + g(v) and
g(au) = ag(u)
Let G be the space of all linear functionals on V. Then if [tex]\oplus_{1}[/tex] and [tex]\otimes_{1}[/tex] are (repectively) addition and scalar multiplication in V, [tex]\oplus_{2}[/tex] and [tex]\otimes_{2}[/tex] are (repectively) addition and multiplication in F, and [tex]\oplus_{3}[/tex] and [tex]\otimes_{3}[/tex] are (repectively) addition and scalar multiplication in G, am I correct in saying that the correct way is to write
g(u[tex]\oplus_{1}[/tex]v) = g(u) [tex]\oplus_{3}[/tex] g(v) and
g(a[tex]\otimes_{1}[/tex]u) = a[tex]\otimes_{3}[/tex]g(u)?

(God I hope the latex worked out..it's my first time)
No, this is not correct. g(u) and g(v) are elements of F. Therefore you should write:
g(u[tex]\oplus_{1}[/tex]v) = g(u) [tex]\oplus_{2}[/tex] g(v)
and
g(a[tex]\otimes_{1}[/tex]u) = a[tex]\otimes_{2}[/tex]g(u)

Are the following what you are looking for (where g, h[tex]\in[/tex]G)?
(g[tex]\oplus_{3}[/tex]h)(u) = g(u) [tex]\oplus_{2}[/tex] h(u)
and
(a[tex]\otimes_{3}[/tex]g)(u) = a[tex]\otimes_{2}[/tex]g(u)
 
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Right! Forgot about that.