Question

dimension of subspaces

Answer 1

if \[v _{1},...,v _{n}\] are generators of S and \[u _{1},...,u _{n}\] are generators of T, why \[\dim(L(v _{1}+u _{1},...,v _{n}+u _{n})) \le \dim(L(v _{1},...,v _{n},u _{1},...,u _{n}))\]?

Answer 2

What does L mean?

Answer 3

lineal

Answer 4

Set of all linear combinations ?

Answer 5

yes

Answer 6

Okay. It suffices to show that
\[\large L(u_1+v_1,u_2+v_2...,u_n+v_n)\subseteq L(v_1,v_2...,v_n,u_1,u_2, ... u_n)\]

Answer 7

Because clearly, if one space is a subspace of the other, the dimension of the subspace may not exceed the dimension of the space which contains it.

Answer 8

ok, thanks :)