This proof is significantly more involved than the others we have done in this c

Question

This proof is significantly more involved than the others we have done in this class. As such, I will break it down into several steps. Show that if S_1, S_2 are subspaces of R^n, then dim(S_1 + S_2) = dim(S_1) + dim(S_2) - dim(S_1 S_2). Here is the set up. Let {u_1, ... U_k} be a basis for S_1 S_2. Extend it to a basis {u_1, ... U_k, w_1, ...w_r} for S_1, and let V = span{v_1}, . Also, extend it to a basis {u_1, ... u_k, w_1, ...w_r} for S_2. (i) Show that V (S_1 intersection S_2) = {0}. (ii) Show that V intersection S_2 = V (S_1 S_2). (iii) Show that S_1 + S_2 = V + S_2. (iv) What does Problem 2 say about dim(V + S_2)? Now verify the formula above.

Explanation / Answer

(i) Given that V is span of {v1,....,vm} and S1 intersection S2 is span of {u1,...,uk}. As {u1,...uk,v1,...vm} is a bases each of v1,....vm,u1,...uk are linearly independent. Intersection of spans of two lineaely independent vectors is {0} from definition of linear independence. Hence intersection of V with intersection of S1 and S2 is {0}.

(ii) Using (i) we can see that V = S1 S2 union {0}. So V intersection S2 is just {0}. Hence done.

(iii) V + S2 = S1 S2 union {0} + S2 = S1 + S2.

(iv) I do not know anything about Problem 2. But we can see that (V+S2) is S1+S2 and dimension is equal to dim(S1) + dim(S2) - dim(S1 intersection S2) = r+k+m . And this we can verify using the set theory that S1+S2 is span of {u1,...,uk,v1,....vm,w1,...,wr} from definition of +.

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