Question #197139

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.




1
Expert's answer
2021-05-24T15:21:18-0400

1.Let {v1,v2,...,vmv_1,v_2,...,v_m } is linearly independent in V and wW.w\in W.


Suppose (v1w,v2w,...,vmw)(v_1-w,v_2-w,...,v_m-w) is Linearly dependent. Then there esists scalars a1,a2,...,ama_1,a_2,...,a_m , not all zero such that


 a1(v1w),a2(v2w)..,am(vmw)a_1(v_1-w),a_2(v_2-w)..,a_m(v_m-w)


Rearranging the equation-


a1v1+...+amvm=(a1+...+am)wa_1v_1+...+a_mv_m=(a_1+...+a_m)w


If a1+...+ama_1+...+a_m were zero, then the equation above would contradict the linear independence of {v1,v2,...,vmv_1,v_2,...,v_m }. Thus a1+....+am0.a_1+....+a_m\neq 0.


Thus divide both sides of the equation by (a1+...+am)(a_1+...+a_m) showing that wspan(v1,v2,..,vm)w\in span(v_1,v_2,..,v_m)


dim span(v1w,v2w,...,vmw)m1.\Rightarrow \text{dim span}(v_1-w,v_2-w,...,v_m-w)\ge m-1.


2.Suppose u1,u2,...,umu_1,u_2,...,u_m are finite-dimensional subspaces of V.


Thus, Each uju_j has a finite basis.


Concatenate these lists to get a spanning list of length dim(u1)++dim(um) for u1++um.dim(u_1) + · · · + dim(u_m) \text{ for } u_1 + · · · + u_m.


This shows that u1++umu_1+· · ·+u_m is finite dimensional and since any spanning list can be reduced


to a basis then dim(u1++um)dim(u1)++dim(um).dim(u_1 + · · · + u_m) ≤ dim(u_1) + · · · + dim(u_m).


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