HW#3, solutions to selected problems

3) (a) Suppose c_nx^n+x_{n-1}x^{n-1}+...+c_0=0 for all real numbers x. We need to show that c_n=...=c_0=0. If p is a degree n polynomial in x and a is a real number such that p(a)=0, then p equals x-a times a degree n-1 polynomial. (Proof: we can divide polynomials to get p=(x-a)q+r where q has degree n-1 and the remainder r is a number. Substituting x=a into this equation, we find that r=0.) Taking a_1,...,a_{n+1} to be any n+1 distinct real numbers, we can apply the above fact repeatedly to deduce that c_nx^n+...+c_0 is divisible by (x-a_1)(x-a_2)...(x-a_{n+1}). This implies that c_n=...=c_0=0, because a nonzero polynomial of degree n or less cannot be divisible by a polynomial of degree n+1 (because when you multiply two nonzero polynomials, the degree adds).

(b) By part (a), V contains a subspace of dimension n+1. By a theorem proved in class, dim(V) >= n+1. Since n can be any integer, it follows that dim(V)=infinity.

4) Let v_1,...,v_k be a basis for V, and let w_1,...,w_l be a basis for W. I claim that v_1,...,v_k,w_1,...,w_l are linearly independent. To see this, suppose a_1v_1+...+a_kv_k+b_1w_1+...+b_lw_l=0. We need to show that a_1=...=a_k=b_1=...=b_l=0. To simplify the notation, let v=a_1v_1+...+a_kv_k and w=b_1w_1+...+b_lw_l. Then v+w=0, so v=-w. Now v is in V and -w is in W, so v and w are both in the intersection of V and W. But this intersection was assumed to be {0}. So v=w=0. Since v_1,...,v_k are independent, v=0 implies a_1=...=a_k=0. Since w_1,...,w_l are independent, w=0 implies b_1=...=b_l=0.

So we have k+l linearly independent vectors in R^n. Therefore k+l is less than or equal to n. Since k=dim(V) and l=dim(W), this proves what we wanted.

5) This problem is more advanced than the others. Let's see how to solve it.

(a) implies (b): assume that (a) holds. How do we find v and W fulfilling condition (b)? Notice that if such v and W have been found, then we must have v in S, because v-v=0 must be in the subspace W. So, not knowing any better, let's try taking v to be an arbitrary vector in S. We know that this exists since S is assumed nonempty. We need to find a subspace W such that x is in S iff x-v is in W. The latter condition is clearly equivalent to saying that x is in W iff x+v is in S. (Right?) So let's simply define W to be the set of x in V such that x+v is in S. We just need to check that W is a subspace.

Certainly W is nonempty, because by definition 0 is in W, because 0+v=v is in S. To prove closure under addition, let x and y be in W. We need to show that x+y is in W. Let's apply the definition of W. We know that x+v and y+v are in S, and we want to show that x+y+v is in S. Let's try to use property (a). We have x+y+v=(2x+v)/2+(2y+v)/2. If we just knew that 2x+v and 2y+v are in S, we would be done by property (a). But all we know is that x+v and y+v are in S. Hmm... wait a minute! If we know that W is closed under scalar multiplication, then 2x and 2y are in W, so 2x+v and 2y+v are in S, and then we get closure under addition. OK, so we just have to check closure under scalar multiplication. Suppose x is in W, so that x+v is in S, and suppose c is a scalar. We need to show that cx is in W, i.e. that cx+v is in S. We cleverly write cx+v=c(x+v)+(1-c)v. Since x+v and v are in S, it follows from (a) that cx+v is in S. Cool!

(b) implies (a): suppose (b) holds. We need to prove (a). Certainly S is nonempty, because v in S, since v-v=0 must be in the subspace W. Now suppose x,y are in S and c is a scalar. We need to show that cx+(1-c)y is in S. By (b), this is equivalent to showing that cx+(1-c)y-v is in W. We cleverly write cx+(1-c)y-v=c(x-v)+(1-c)(y-v). Since x and y are in S, x-v and y-v are in W, so c(x-v)+(1-c)(y-v) is in W since W is a subspace. QED.

Polished proof that (a) implies (b): suppose (a) holds. Choose any v in S; this must exist since S is nonempty. Let W be the set of x in V such that x+v is in S. Clearly x is in S iff x-v is in W. So we just need to show that W is a subspace of V. W is nonempty because 0 is in W because v is in S. To prove closure under scalar multiplication, suppose x is in W, i.e. x+v is in S, and let c be a scalar. Then cx+v=c(x+v)+(1-c)v is in S by (a), so cx is in W. To prove closure under addition, suppose x and y are in W. By the above, 2x and 2y are in W, so 2x+v and 2y+v are in S. By (a), x+y+v=(2x+v)/2+(2y+v)/2 is in S, so x+y is in W.

6) Suppose c_1v_1+...+c_nv_n=0. Taking the inner product of the left side with itself, we get c_1^2||v_1||^2+...+c_n^2||v_n||^2=0. The cross terms involving inner products of v_i with v_j are zero by assumption. Now ||v_1||,...,||v_n|| are positive because v_1,...,v_n are assumed nonzero. So everything in this equation is nonnegative. So every term must be zero, so c_1=...=c_n=0.