1. Give the dimension of each of the following vector spaces over R.

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1 Answers to Exercise Set II.2. Drills 1. Give the dimension of each of the following vector spaces over R. (a) dimr 4 = 4, (b) dimp 5 = 6, (c) dimm 3,4 = 12, (d) dimf X = 4, (X = {1,2,3,4}) 2. Thedimensionof Cwhenit isconsideredasavectorspaceoverris2. Thedimension of C n considered as a vector space over R is 2n. 3. True or False: (a) False (b) True (c) True (d) True (e) False; dim M should be 51 (f) True (g) True (h) True (i) False 4. Answer questions (a) From dim(h +K)+dimH K = dimh +dimk, we have 5+2 = 4+dimK and hence dimk = 3. (b) We have dimp 10 = 11 and dim(m + N) = dimm + dimn dim(m N) = = 10. Hence is M +N P 10. (c) The zero operator (on any vector space) is the only linear operator of rank zero. (d) According to the form of vectors in the kernel, we have dimkert = 3 Thus the rank of T is dimp 4 dimkert = 5 3 = Find the rank and the nullity (a) T 1 : F 2 F 3 sending (x 1,x 2 ) to (0,x 1,x 2 ): rank = 2, nullity = 0. (b) T 2 : F 3 F 3 sending (x 1,x 2,x 3 ) to (0,x 1,x 2 ): rank = 2, nullity = 1. (c) S ω : R 3 R 3 given by S ω (x) = ω x, where ω 0: rank = 2, nullity = 1. (d) D : P n P n sending p(x) P n to its derivative p (x): rank = n, nullity = 1. (e) M : P 3 P 5, sending p(x) to x 2 p(x): rank = 4, nullity = 0. (f) E : P 2 P 2, sending p(x) to 1 (p(x)+p( x)): rank = 2, nullity = 1. 2 (g) Q : P 2 P 2, sending p(x) to 1 2 (p(x) p( x)): rank = 1, nullity = 2.

2 (h) δ : M 2,2 M 2,2, given by δ(x) = AX XA. The operator δ sends [ ] x11 x 12 x 21 x 22 to Hence we have: rank = 2, nullity = 2. [ ][ ] [ ][ ] [ ] 1 0 x11 x 12 x11 x x12 =. 0 2 x 21 x 22 x 21 x x 21 0 (i) Φ : M 2,2 M 2,2 given by Φ(X) = BXC: Φ sends [ ] x11 x 12 x 21 x 22 to Hence we have: rank = 1, nullity = 3. [ ][ ][ ] [ ] 1 0 x11 x x11 x = x 21 x x 11 x True or false (a) True (b) False: take T = D on P 2, sending p(x) to its derivative p (x) (c) True: from T 2 (V) T(V) and dimt 2 (V) = dimt(v) we have T 2 (V) = T(V). (d) True: from kert kert 2 and dimkert = dimv rank T = dimv rank T 2 = dimkert 2, we have kert = kert 2. (e) False: letting S = I and T = I, we have rank (S +T) = 0, but rank(s)+rank(t) rank(st) = rank(i)+rank( I) rank( I) = rank(i) = n. (f) True: from S 2 = O we get S(V) kers and hence dims(v) dimkers. 7. We are given matrix A and its reduced row echelon form B (a) Basis for the range of M A : (2,1,0,1), (4i,2i,i,2i), (7i,3i,5i,4i). (b) Basisforker(M A ): (4i 8, 2 5i,i,1,0,0), (8i 7, 4 2i,0,2,1,0), (0,0,0,0,0,1). (c) (1,3i,6i, 1+2i,3,0) = (1,2i,3i,1,1,0)+i(0,1,5,2,2i,0) 2i(0,0,1, i,2i,0). (d) rank = 3, nullity = Find bases for kernel and range (a) Basis for kernel: {(i,1,0), (1, 1,1)}. Basis for range: {(i,1)}. (b) Basis for kernel: {1+x, x+x 3 }. Basis for range: {1, x}.

3 [ ] x11 x (c) Φ : M 2,2 M 2,2 sends 12 x 21 x 22 to [ ][ ] [ ][ ] [ ] 1 3 x11 x 12 x11 x x22 3x = 11 4x x 21 x 22 x 21 x x 21 0 Basis for kerφ: {[ ] 1 0, 0 1 Basis for the range of T: {[ ] 0 1, 0 0 [ ]} [ ]} A polynomial p(x) of degree 4 satisfying p(x+1) p(x) = x(x+1)(x+2): p(x) = 1 4 x(x+1)(x+2)(x+3). We can check this as follows: 4(p(x+1) p(x)) = (x+1)(x+2)(x+3)(x+4) x(x+1)(x+2)(x+3) = (x+1)(x+2)(x+3)[(x+4) x] = 4(x+1)(x+2)(x+3). Hence n k= 0 k(k +1)(k +2) = n(n+1)(n+2)(n+3). 4 Exercises 1. Suppose that b 1, b 2,..., b n is a basis of V. Thus each vector v can be written in a unique way as a linear combination of b 1, b 2,..., b n, say v = a 1 b 1 +a 2 b 2 + +a n b n ( ) To show that b 1, b 2,..., b n are linear independent, we suppose a 1 b 1 +a 2 b 2 + +a n b n = 0. The last identity is a way to express 0 as a linear combination of b 1, b 2,..., b n. Another expression is 0b 1 +0b b n = 0.

4 Since such an expression is unique, necessarily we have a 1 = 0, a 2 = 0,..., a n = 0. The identity ( ) above shows that b 1, b 2,..., b n is spanning. Next, assume that b 1, b 2,..., b n are linearly independent and spanning. Here spanning means that every vector v can be written as as linear combination of b 1, b 2,..., b n such as ( ) above. To check that the linear combination ( ) is unique, suppose we also have v = a 1b 1 +a 2b 2 + +a nb n. Subtract this identity from ( ), we obtain 0 = (a 1 a 1 )b 1 +(a 2 a 2 )b 2 + +(a n a n )b n. From the linear independence of b 1, b 2,..., b n we see that a k a k = 0 and hence a k = a k (1 k n). 2. Let M be the subspace spanned by w 1,w 2,...,w 50. Then dimm 50. On the other hand, since M contains 50 linearly independent vectors v 1,v 2,...,v 50, we have dimm 50. ThusdimM = 50. Nowthe 50 vectors w 1,w 2,...,w 50 spansasubspace M of dimension 50 and hence they form a basis of that subspace. In particular, they are linearly independent. 3. Take any nonzero vector e 1 and let e 2 = Te 1. Then Te 2 = TTe 1 = T 2 e 1 = e 1. Now we check that e 1, e 2 are linearly independent. Suppose a 1 e 1 +a 2 b 2 = 0. ( ) Then a 1 Te 1 +a 2 Te 2 = 0 and hence a 1 e 2 a 2 e 1 = 0. Together with ( ), we obtain a 1 (a 1 e 1 +a 2 b 2 ) a 2 (a 1 e 2 a 2 e 1 ) = 0, or (a a 2 2)e 1 = 0. Since e 1 is nonzero, we must have a a 2 2 = 0. Since the vector space is assumed to be real, a 1 and a 2 are real numbers and hence a 2 1+a 2 2 = 0 implies a 1 = a 2 = 0. So e 1, e 2 are linearly independent vectors in the 2 dimensional space and [ hence] they form a basis B. From Te 1 = e 2 and Te 2 = e 1, we see that 0 1 [T] B = Clearly, T and T 2 has the same range and hence they have the same rank, namely r. The domain of T 2 has the dimension dim(v V) = 2n. So the nullity of T 2 is 2n r.

5 5. Note that x kert (2) T (2) x (Tx,Tx) = (0,0) Tx = 0 x kert. So we have kert (2) = kert. Hence rank T (2) = dimv dimkert (2) = dimv dimkert = rank T. 6. We have dimt(v)+dimkert = dimv = 3. From T 2 = O we see that T(V) kert and hence dimt(v) dimkert. Thus 3 = dimt(v)+dimkert dimt(v)+dimt(v) = 2dimT(V). Thus dimt(v) 3/2. Since dimt(v) is an integer, we must have dimt(v) 1. This shows that the rank of T is at most (a) The inequality rank(s +T) rank(s)+ rank(t) follows from two obvious facts: (S + T)(V) S(V) + T(V) and dim(m + N) dimm + dimn (which is a result of dim(m +N) + dim(m N) = dimm + dimn). (b) Since (ST)(V) S(V), we have dim(st)(v) dims(v), or rank ST rank S. Next, from the obvious fact kert ker(st) we have dimkert dimker(st) and hence dim(st)(v) = dimv dimker(st) dimv dimkert = dimt(v), This shows rank(st) rank(t). 8. From S = (ST)S we see that rank(st) rank(s). On the other hand, we have rank(st) rank(s). Hence rank(st) = rank(s). Similarly, using S = S(TS), we can show rank(st) = rank(t). 9. Consider the induced linear transformation M A : F n F m defined by M A x = Ax. Then M A is a rank one linear mapping and hence its range is spanned by a single vector in F m, say a = [a 1 a 2 a m ]. It is known that the columns of A span the range of M A. So all columns of A are scalar multiples of a, say ab 1, ab 2,..., ab n. Thus a 1 b 1 a 1 b 2 a 1 b n a 2 b 1 a 2 b 2 a 2 b n A = [ab 1 ab 2 ab n ] =. a m b 1 a m b 2 a m b n 10. Consider the linear mapping T: P 4 C 5 given by T(p(x)) = (p(0), p (0), p (0), p(1), p (1)).

6 We need to prove that the range of T is C 5. Since dimp 4 = dimc 5 = 5, it is enough to verify that kert = {0}. Suppose p(x) b 0 + b 1 x + b 2 x 2 + b 3 x 3 + b 4 x 4 kert. From p(0) = 0, we get b 0 = 0. From p (0) = 0, we get b 1 = 0. From p (0) = 0, we get b 2 = 0. Thus we have p(x) = b 3 x 3 + b 4 x 4. Now p(1) = 0 gives b 3 +b 4 = 0 and p (1) = 0 gives 3b 3 + 4b 4 = 0. From the last two identities we obtain b 3 = b 4 = 0. Hence kert = {0}. 11. (a) We can write q(x) = cx n +p(x), where c is a scalar and p(x) P n 1. Then (x+a)q(x+1) xq(x) = (x+a)(c(x+1) n ) x(cx n )+((x+a)p(x+1) xp(x)) = cx((x+a) n x n )+ac(x+1) n +(x+a)p(x+1) xp(x) Clearly ac(x+1) n, (x+a)p(x+1), xp(x) P n. Now cx((x+a) n x n ) = cx((x+a) x)((x+1) n 1 +(x+1) n 2 x+ +(x+1)x n 2 +x n 1 ) = acx((x+1) n 1 +(x+1) n 2 x+ +(x+1)x n 2 +x n 1 ) P n. Hence (x+a)q(x+1) xq(x) P n. (b) Consider the linear operator T : P n P n defined by T(q(x)) = (x+a)q(x+1) xq(x). It is enough to check that kert = {0}. Let q(x) kert. Then (x+a)q(x+1) xq(x) = 0, or (x+a)q(x+1) = xq(x). ( ) Letting x = 0, we have aq(1) = 0. Since a 0, we must have q(1) = 0. Letting x = 1 in ( ), we have (1+a)q(2) = 0. Since a > 0, we must have q(2) = 0. Suppose that k is any positive integer with q(k) = 0. Then letting x = k in ( ), we have (k+a)q(k+1) = q(k) = 0 and consequently q(k+1) = 0. Hence we can use induction to prove that q(k) = 0 for all positive integers. Since a polynomial cannot have infinitely many roots unless it is the zero polynomial, we see that q(x) = (a) For q(x) in P n, we can write q(x) = cx n + p(x), where c is come scalar and p(x) P n 1. Thus, putting r(x) = xp(x+1) (x+1)p(x), we have xq(x+1) (x+1)q(x) = cx(x+1) n c(x+1)x n +r(x) = cx(x+1)((x+1) n 1 x n 1 )+r(x) = cx(x+1)((x+1) n 2 +(x+1) n 3 x+ (x+1)x n 3 +x n 2 )+r(x)

7 which is in P n by inspecting each term in the last expression. (b) Consider the linear operator T on P n defined by T(q(x)) = xq(x+1) (x+1)q(x). It is enough to show that T(P n ) P n. In order to show this, it suffices to check kert {0}. Let q(x) = x. Then q(x) 0 and T(q(x)) = x(x+1) (x+1)x = 0 and hence q(x) kert. 13. (Proof of Theorem 2.3.2) Take a basis W = {w 1, w 2,..., w r } in M N. Extend it to a basis U = {w 1,..., w r, u 1, u 2,..., u s } of M, and a basis V = {w 1,..., w r, v 1, v 2,..., v t } in N. We verify that R = {w 1, w 2,..., w r u 1, u 2,..., u s, v 1, v 2,..., v t } form a basis of M +N. First we check that R is linearly independent. Suppose a 1 w 1 +a 2 w 2 + +a r w r +b 1 u 1 +b 2 u 2 + +b s u s +c 1 v 1 +c 2 v 2 + +c t v t = 0. Then we have h : = a 1 w 1 +a 2 w 2 + +a r w r +b 1 u 1 +b 2 u 2 + +b s u s = c 1 v 1 c 2 v 2 c t v t. ( ) The first expression of h shows h M and the second expression shows h N. Thus h M N. Hence we can write h as a linear combination of vectors in W : h = h 1 w 1 +h 2 w 2 + +h r w r. In view of h = c 1 v 1 c 2 v 2 c t v t, this gives h 1 w 1 +h 2 w 2 + +h r w r +c 1 v 1 +c 2 v 2 + +c t v t = 0. Since vectors in V are linearly independent, we obtain h 1 = h 2 = = h r = c 1 = c 2 = = c t = 0. Thus ( ) becomes a 1 w 1 +a 2 w 2 + +a r w r +b 1 u 1 +b 2 u 2 + +b s u s = 0. Since vectors in U are linearly independent, we have a 1 = a 2 = = a r = b 1 = b 2 = = b s = 0.

8 Hence all scalar coefficients of ( ) are zeros. Next we show that M +N = span R = span (U V ). This follows immediately from M = span U and N = span V. 14. (Hard problem.) Let p(x) be a polynomial of degree n. We prove the linear independence of p(x), p(x+1),..., p(x+n) by induction on the degree of n. First let us notice that the degree of p(x+1) p(x) is exactly n. This can be seenas follows. Write p(x) = cx n +q(x) with q(x) P n 1. Then p(x) = c((x+1) n x n )+(q(x+1) q(x)). Clearly q(x+1) q(x) P n 2. By using the binomial expansion of (x+1) n we see that the degree of (x + 1) n x n is exactly n 1. The assertion is obvious when p(x) is a nonzero constant. Now we assume the validity of the statement in case the degree of p(x) is n 1. Let p(x) be a polynomial of degree n and suppose a 0 p(x)+a 1 p(x+1)+a 2 p(x+2)+ +a n p(x+n) = 0. ( ) The nth derivative of the above identity gives us cn!(a 0 + a 1 + a a n ) = 0, where c is the leading coefficient of p(x). Hence a 0 +a 1 + +a n = 0 ( ) We can rewrite the left hand side of ( ) as a telescoping sum (a 0 +a 1 + +a n )p(x)+(a 1 +a 2 + +a n )(p(x+1) p(x)) +(a 2 + +a n )(p(x+2) p(x+1))+ + +(a n 1 +a n )(p(x+n 1) p(x+n 2))+a n (p(x+n) p(x+n 1)) Hence, in view of ( ), we can rewrite ( ) as b 0 q(x)+b 1 q(x+1)+ +b n 1 q(x+n 1) = 0 where q(x) = p(x + 1) p(x) is a polynomial of degree n 1 and b k = n j= k+ 1 a j. By the induction hypothesis, we know b 0 = b 1 = = b n 1 = 0. Now it is clear that a 0 = a 1 = = a n = 0.