(1). start with a basis of W. extend this to a basis of V.

so you should have something like:

B = {w_{1},....,w_{k},v_{1},...,v_{n-k}}, where dim(W) = k, and dim(V) = n.

so what you want to prove is dim(V/W) = n-k. well, {v_{1},...,v_{n-k}} is a set of n-k vectors, so if:

{v_{1}+W,....,v_{n-k}+W} is a basis for V/W, we're good to go.

to prove linear independence, you need to show that if there are c_{1},...,c_{n-k}with:

c_{1}(v_{1}+W) +...+ c_{n-k}(v_{n-k}+ W) = W

then c_{1}=...= c_{n-k}= 0.

show that the above means that c_{1}v_{1}+...+ c_{n-k}v_{n-k}is in W.

now assume that not all c_{j}= 0, but we have c_{1}v_{1}+...+ c_{n-k}v_{n-k}in W.

then (since the w_{i}form a basis for W) we have:

c_{1}v_{1}+...+ c_{n-k}v_{n-k}= b_{1}w_{1}+...+ b_{k}w_{k},

so -b_{1}w_{1}-...- b_{k}w_{k}+ c_{1}v_{1}+...+ c_{n-k}v_{n-k}= 0.

why does this contradict the linear independence of B?

to show that the {v_{1}+W,....,v_{n-k}+W} span V/W, pick any u+W in V/W.

write u = a_{1}w_{1}+...+ a_{k}w_{k}+ a_{k+1}v_{1}+...+a_{n}v_{n-k}

(which we can do since B is a basis for V).

note that if u = w + v, where w is in W, then u + W = (w + v) + W = (v + w) + W = (v + W) + (w + W) = v + W + W = v + W

(W = 0 + W is the identity of V/W, and w + W = W if and only if w - 0 = w is in W).

(2). these sorts of exercises are tedious, and involve working through what rank(f) means. rank(f) = dim(f(U)) if f:U→U (f(U) is also called im(f)).

so if rank(f) = k, for example, this means we have {u_{1},...,u_{k}} with B = {f(u_{1}),...,f(u_{k})} a basis for f(U).

and if rank(g) = m, we have {v_{1},...,v_{m}} with B' = {g(v_{1}),...,g(v_{m})} a basis for g(U).

show that B U B' spans (f+g)(U). this means that some linearly independent subset C of B U B' is a basis for (f+g)(U).

hence rank(f+g) = dim((f+g)(U)) = |C| ≤ |B U B'| ≤ |B| + |B'| = dim(f(U)) + dim(g(U)) = rank(f) + rank(g).

i don't want to spoil all your fun, so try working through the rest and let us know where you get stuck.