If u and v are in U, then Au= Bu and Av= Bv. So A(u+v)= Au+ Av= Bu+ Bv= B(u+v). Similarly for A(au).

First, the correct spelling is "independ[b]e[b]nt". Now, if [X1, X1 + X2, X1 +X2 + X3,....,X1 +X2.....XK] were not independent the there would exist numbers a1, a2, ..., aK, not all 0, such that a1X1+ a2(X1+X2)+ a3(X1+ X2+ X3)+ ...+ aK(X1+ X2+ ...+XK)= 0. rewrite that to collect "like" Xs and show that would imply that [X1, X2, ..., X] are not independent.2. If [X1, X2, X3,......XK] is independant, show that [X1, X1 + X2, X1 +X2 + X3,....,X1 +X2.....XK] is also independant.

Same as 2.3.If [Y, X1, X2, X3,.....XK] is independant show that [Y + X1, Y + X2, Y + X3,....Y+XK] is also independant

Actually, since W is NOT a "proper" subset of itself, I would say this is not true!4. Let U and W denote subspace , and assume that U is a proper subset of W. If dimU = n-1, show that either W = U or W =

Same comment as for (4).5. Let U and W denote the subspaces of , and assume that U is a proper subset of W. If dim W = 1, show that either U = {0} or U = W

Show that it is a linear transformation by showing that T(u+v)= T(u)+6. Let C1, C2,....CN be fixed columns in and define T: ----> by in . Show that T is a linear transformation and find the matrix of T

T(v) and T(au)= aT(u). Consider the matrix having C1, C2, ..., CN as its columns, of course.