UCSD ECE269 Handout #5 Prof. Young-Han Kim Wednesday, January 31, Solutions to Homework Set #2 (Prepared by TA Alankrita Bhatt)

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1 UCSD ECE69 Handout #5 Prof. Young-Han Kim Wednesday, January 3, 08 Solutions to Homework Set # (Prepared by TA Alankrita Bhatt). Zero nullspace. Let A F m n. Prove that the following statements are equivalent. (a) N(A) {0}. (b) A T is onto (i.e., R(A T ) F n ). (c) Columns of A are independent. (d) A is tall (i.e., n m) and full-rank (i.e., rank(a) min(m,n) n). Solution: We will show the chain of equivalences (a) (b) (c) (d) (a). (a) (b): By the rank nullity theorem, we have dim(n(a)) + rank(a) n, which implies rank(a) n (since dim(n(a)) 0). Since rank(a) rank(a T ), we then have rank(a T ) n. Since rank is equivalent to the dimension of the column space, the dimension of the column space of A T is n. Because each column vector in A T is of length n, this means that R(A T ) F n. (b) (c): Since A T is onto, rank(a T ) dim(r(a T )) n. Because rank(a) rank(a T ) n, the dim(r(a)) n. Note now that A has n column vectors and for them to span a space of dimension n, all of these column vectors have to be independent. (c) (d): If the columns of A are independent, since each column vector is of length m, there cannot be more than m of them (since more than m vectors of length m necessarily need to be dependent). Thus n m. Since n independent vectors span a space of dimension n, we know that dim(r(a)) n rank(a) n min(m,n). (d) (a): By the rank nullity theorem, rank(a)+dim(n(a)) n. Since rank(a) n, we have dim(n(a)) 0, which implies that N(A) {0}.. Rank of AA T. Let A F m n. (a) Suppose that F R. Prove that rank(aa T ) rank(a) or provide a counterexample. (b) Suppose that F F. Repeat part (a). (c) Suppose that F C. Repeat part (a). (d) Suppose that F C. Prove that rank(aa H ) rank(a) or provide a counterexample. Solution: (a) If AA T x 0, then x T AA T x (A T x) T (A T x) A T x 0, which implies that A T x 0. Thus, for every x N(AA T ), x N(A T ), or equivalently, N(AA T ) N(A T ). Conversely, if A T x 0, then AA T x 0, which implies that N(A T ) N(AA T ). Hence,

2 N(AA T ) N(A T ) and by the rank nullity theorem, rank(a) rank(a T ) m dim(n(a T )) m dim(n(aa T )) rank(aa T ). (b) Consider [ ]. 0 0 In F,AA T 0. Thus, rank(a) but rank(aa T ) 0. (c) Consider [ ] i. 0 0 Again, AA T 0. Thus, rank(a) but rank(aa T ) 0. (d) We can show that rank(a) rank(aa H ) using the same proof as part (a), with A T replaced by A H. Indeed, (A H x) H (A H x) 0 A H x 0. Note, however, that this proof does not work for parts (b) and (c) since in F or C, (A T x) T (A T x) 0 A T x Oddtown. Recall Oddtown with n people forming clubs according to the following rules: Each club has an odd number of members. Each pair of clubs share an even number of members. In class, we discuss that n singleton clubs, namely, {},{},...,{n} are compliant. We now form more interesting clubs. (a) For n 4, form 4 clubs, each with more than one member. (b) For n 6, form 6 clubs, not all of them of equal sizes. Solution: We give an n n matrix A, the rows of which represent clubs and the columns of which represent the people. A ij is if person j is a member of club i, and 0 otherwise. (a) is a valid 4 4 matrix

3 (b) is a valid 6 6 matrix Rank of a sum. Let A,B F m n. Show that rank(a+b) rank(a)+rank(b). Solution: Let r A and r B denote the rank of A and B respectively. Consider a basis (u,u,,u ra ) that spans R(A) and another basis (v,v,,v rb ) that spans R(B). We will show that the set of vectors (u,u,,u ra,v,v,,v rb ) spans R(A+B). Consider the column space of A+B. If the columns of A are denoted by (a,a, a n ), and those of B are denoted by (b,b, b n ), the columns of A+B are denoted by (a +b,a + b,,a n +b n ). Thus, any linear combination of (a +b,a +b,,a n +b n ) can be written as a linear combination of the n vectors (a,b,a,b,,a n,b n ). Since (u,u,,u ra ) is a basis for R(A) and (v,v,,v rb ) a basis for R(B), span(a,b,a,b,,a n,b n ) span(u,u,,u ra,v,v,,v rb ). Since the vectors (u,u,,u ra,v,v,,v rb ) span the column space of (A+B), it immediately follows that rank(a+b) dim(r(a+b)) r A +r B rank(a)+rank(b). 5. Rank of a product. Let A R 6 4 has rank and B R 4 5 has rank 3. (a) Find the smallest possible value r min of rank(ab). Find specific A and B such that rank(ab) r min. (b) Find the largest possible value r max of rank(ab). Find specific A and B such that rank(ab) r max. Solution: (a) We first prove that nullity(ab) nullity(a) + nullity(b) () for any pair of matrices A R m n and B R n k. To show this, we decompose N(AB) by N(B) and its orthogonal complement N(B) R(B T ) as Then, N(AB) {z R k : ABz 0} {z R k : Bz 0}+{z R(B T ) : Bz N(A)} N(B)+V. nullity(ab) dim(n(ab)) dim(n(b))+dim(v) nullity(b)+dim(v), 3

4 and it suffices to show that dim(v) dim(n(a)) nullity(a). To upperbounddim(v), suppose that z,...,z k form a basis for V. Then Bz,...,Bz k must be independent; otherwise, α Bz + +α k Bz k B(α z + +α k z k ) 0 implies that z α z + +α k z k 0 and z N(B), which is a contradiction to the assumption that z R(B T ) N(B). But at the same time, Bz,...,Bz k N(A) and thus k dim(n(a)). Therefore, dim(v) dim(n(a)). By the rank nullity theorem, () implies or equivalently, n rank(ab) (k rank(a))+(n rank(b)), rank(ab) rank(a)+rank(b) k, where k is the number of rows of B. Thus, specializing to our problem, we have This lower bound is tight, as shown by rank(ab) and B of ranks and 3, respectively, and AB, which has rank. (b) Recall that rank(ab) min(rank(a), rank(b)). This upper bound is tight, as shown by and B , and AB, 4

5 which has rank. 6. Bilinearity. Let V be a vector space over R and, be an inner product. Show that, is linear in the second argument, namely, x,αy +βz α x,y +β x,z. Solution: First of all, since V is over R, we have symmetry of inner product (instead of conjugate symmetry), i.e. a,b b,a. Then, x,αy +βz αy +βz,x α y,x +β z,x α x,y +β x,z Where the first equality follows by linearity in the first argument, and the second equality follows by symmetry. 7. Inner product of polynomials. Let P 3 be the vector space of all polynomials of degree 3 with real coefficients, that is, Let K : P 3 P 3 R be defined as P 3 {α 0 +α x+α x +α 3 x 3 : α 0,α,α,α 3 R}. K(p,q) p(x)q(x)dx. (a) Show that K(, ) represents an inner product for P 3. (b) Find an orthogonal basis for P 3 using Gram Schmidt orthogonalization. Solution: (a) We will show that the three properties that an inner product is required to satisfy hold for K(, ). Linearity in the first argument. K(αp +βp,q) α (αp (x)+βp (x))q(x)dx (αp (x)q(x)+βp (x)q(x))d(x) αp (x)q(x)dx+ p (x)q(x)dx+β αk(p,q)+βk(p,q). βp (x)q(x)dx p (x)q(x)dx 5

6 Conjugate symmetry. K(q,p) K(p,q). q(x)p(x)dx q(x)p(x)dx p(x)q(x)dx p(x)q(x)dx Positive definiteness. Notethat p(x) 0 p P 3,x [,]. Therefore, K(p,p) p(x)p(x)dx p(x) dx 0. If we have K(p,p) p(x) dx 0, since p(x) is a continuous function we necessarily need p(x) to be identically 0. (b) First of all, recall that,x,x,x 3 form a basis for P 3 (cf. HW, Problem 8(a)). We will use this basis to construct an orthonormal basis using Gram Schmidt orthogonalization. p 0 (x), p 0 (x) ( p (x) x p (x) x x p (x) x ( ) x dx 3 x p (x) x /3 x /3 x, ) ( 3 3 x3 dx x 45 8 ( ) 45 p 3 (x) x 3 8 x5 dx ( x 3 ) dx x x, p 3 (x) x3 (3/5)x x 3 (3/5)x ( x ) (x 3 35 ) 8 x. x ) dx x 3, ( x ) ( ) x4 dx x 6