Answer to Question #197139 in Linear Algebra for Kaygee

Question

Suppose V1,V2,VM is linearly independent in V and W€V.Prove that dim span(V1+W1,V2+W,.....VM+W)> or equal to m-1.

2.Suppose U1,U2,.....Um are finite dimensional subspaces of V.Prove that U1+U2+.....Um is finite dimensional and dim(U1+U2+.....Um)<or equal to dim U1+dim U2+.......dim Um.

Answer 1

1.Let {$v_1,v_2,...,v_m$ } is linearly independent in V and $w\in W.$

Suppose $(v_1-w,v_2-w,...,v_m-w)$ is Linearly dependent. Then there esists scalars $a_1,a_2,...,a_m$ , not all zero such that

 $a_1(v_1-w),a_2(v_2-w)..,a_m(v_m-w)$

Rearranging the equation-

$a_1v_1+...+a_mv_m=(a_1+...+a_m)w$

If $a_1+...+a_m$ were zero, then the equation above would contradict the linear independence of {$v_1,v_2,...,v_m$ }. Thus $a_1+....+a_m\neq 0.$

Thus divide both sides of the equation by $(a_1+...+a_m)$ showing that $w\in span(v_1,v_2,..,v_m)$

$\Rightarrow \text{dim span}(v_1-w,v_2-w,...,v_m-w)\ge m-1.$

2.Suppose $u_1,u_2,...,u_m$ are finite-dimensional subspaces of V.

Thus, Each $u_j$ has a finite basis.

Concatenate these lists to get a spanning list of length $dim(u_1) + · · · + dim(u_m) \text{ for } u_1 + · · · + u_m.$

This shows that $u_1+· · ·+u_m$ is finite dimensional and since any spanning list can be reduced

to a basis then $dim(u_1 + · · · + u_m) ≤ dim(u_1) + · · · + dim(u_m).$