Answers Exercise Set II.2 2007 PDF

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Answers to Exercise Set II. Drills 1. Give the dimension of each of the following vector spaces over R. (a) dim R4 = 4, (b) dim P5 = 6, (c) dim M3,4 = 12, (d) dim FX = 4, (X = {1, 2, 3, 4}) 2. The dimension of C when it is considered as a vector space over R is 2. The dimension of Cn considered as a...

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    

 1. Give the dimension of each of the following vector spaces over . (a) dim  = 4,

(b) dim P = 6,

(c) dim M, = 12,

(d) dim FX = 4, (X = {1, 2, 3, 4}) 2. The dimension of  when it is considered as a vector space over  is 2. The dimension of n considered as a vector space over  is 2n. 3. True or False: (a) False (f) True

(b) True (g) True

(c) True (h) True

(d) True

(e) False; dim M should be 51

(i) False

4. Answer questions (a) From dim(H + K) + dim H ∩ K = dim H + dim K, we have 5 + 2 = 4 + dim K and hence dim K = 3. (b) We have dim P = 11 and dim(M + N ) = dim M + dim N − dim(M ∩ N ) = 7 + 8 − 5 = 10. Hence is M + N = P  . (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 dim ker T = 3 Thus the rank of T is dim P − dim ker T = 5 − 3 = 2. 5. Find the rank and the nullity (a) T :  →  sending (x , x) to (0, x , x): rank = 2, nullity = 0. (b) T :  →  sending (x , x, x ) to (0, x, x ): rank = 2, nullity = 1. (c) Sω :  →   given by Sω () = ω × , where ω = : rank = 2, nullity = 1. (d) D : Pn → Pn sending p(x) ∈ Pn to its derivative p′ (x): rank = n, nullity = 1. (e) M : P → P, sending p(x) to x p(x): rank = 4, nullity = 0. (f) E : P  → P, sending p(x) to (p(x) + p(−x)): rank = 2, nullity = 1. (g) Q : P  → P, sending p(x) to (p(x) − p(−x)): rank = 1, nullity = 2.

(h) δ : M, → M ,, given by δ(X) = AX − XA. The operator δ sends          x x 1 0 x  x 1 0 x x 0 to − = x x x x x x x 0 2 0 2

 −x  . 0

Hence we have: rank = 2, nullity = 2. (i) Φ : M, → M, given by Φ(X) = BXC: Φ sends        1 0 x  x 1 1 x x x to = 1 0 0 0 x x x x x

 x . x

Hence we have: rank = 1, nullity = 3. 6. True or false (a) True (b) False: take T = D on P , sending p(x) to its derivative p′(x) (c) True: from T (V ) ⊆ T (V ) and dim T (V ) = dim T (V ) we have T  (V ) = T (V ). (d) True: from ker T ⊆ ker T  and dim ker T = dim V − rank T = dim V − rank T  = dim ker T  , we have ker T = ker T . (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 = O we get S(V ) ⊆ ker S and hence dim S(V ) ≤ dim ker S . 7. We are given matrix A and its reduced row echelon form B (a) Basis for the range of MA : (2, 1, 0, 1), (4i, 2i, i, 2i), (7i, 3i, 5i, 4i). (b) Basis for ker(MA ): (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 = 3. 8. 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}. Basis for range: {1, x}.

(c) Φ : M, → M, sends 

1 3 0 5





  x  x − x x

x x 



to

x x



x x

x x

  1 3 0 = 0 5 4x

Basis for ker Φ:



   1 0 0 3 , 0 1 0 4

Basis for the range of T :



   0 1 0 0 , 0 0 1 0

 3x − 3x − 4x . 0

9. A polynomial p(x) of degree 4 satisfying p(x + 1) − p(x) = x(x + 1)(x + 2): p(x) =

1 x(x + 1)(x + 2)(x + 3). 4

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 

k(k + 1)(k + 2) =

n(n + 1)(n + 2)(n + 3) . 4

 1. Suppose that , , . . . ,  n is a basis of V . Thus each vector  can be written in a unique way as a linear combination of  , , . . . , n , say  = a  + a  +    + a nn

( ∗)

To show that  , , . . . ,  n are linear independent, we suppose a  + a  +    + an n = . The last identity is a way to express  as a linear combination of  , , . . . , n . Another expression is 0  + 0 +    + 0 n = .

Since such an expression is unique, necessarily we have a = 0, a = 0, . . . , an = 0. The identity (∗) above shows that ,  , . . . , n is spanning. Next, assume that , , . . . , n are linearly independent and spanning. Here “spanning” means that every vector  can be written as as linear combination of  , , . . . , n such as (∗) above. To check that the linear combination (∗) is unique, suppose we also have  = a ′ + a′  +    + a ′nn . Subtract this identity from (∗), we obtain  = (a ′ − a )  + (a ′ − a)  +    + (a ′n − an )n . From the linear independence of ,  , . . . , n we see that a′k − a k = 0 and hence a′k = ak (1 ≤ k ≤ n). 2. Let M be the subspace spanned by ,  , . . . ,   . Then dim M ≤ 50. On the other hand, since M contains 50 linearly independent vectors ,   , . . . ,  , we have dim M ≥ 50. Thus dim M = 50. Now the 50 vectors ,  , . . . ,  spans a subspace M of dimension 50 and hence they form a basis of that subspace. In particular, they are linearly independent. 3. Take any nonzero vector  and let   = T  . Then T   = T T  = T   = − . Now we check that ,  are linearly independent. Suppose a + a = .

( ∗)

Then aT   + aT   =  and hence a  − a   = . Together with (∗), we obtain a (a  + a) − a  (a  − a ) = , or (a  + a) = . Since  is nonzero, we must have a + a = 0. Since the vector space is assumed to be real, a and a are real numbers and hence a + a = 0 implies a = a = 0. So ,  are linearly independent vectors in the 2–dimensional space and  hence they form a basis B. From T  =   and T  = − , we see that 0 −1 . [T ]B = 1 0 4. Clearly, T and T has the same range and hence they have the same rank, namely r. The domain of T has the dimension dim(V × V ) = 2n. So the nullity of T is 2n − r.

5. Note that  ∈ ker T

⇔ T  ≡ (T , T ) = (, ) ⇔ T  =  ⇔  ∈ ker T .

So we have ker T  = ker T . Hence rank T  = dim V − dim ker T  = dim V − dim ker T = rank T. 6. We have dim T (V ) + dim ker T = dim V = 3. From T  = O we see that T (V ) ⊆ ker T and hence dim T (V ) ≤ dim ker T . Thus 3 = dim T (V ) + dim ker T ≥ dim T (V ) + dim T (V ) = 2 dim T (V ). Thus dim T (V ) ≤ 3/2. Since dim T (V ) is an integer, we must have dim T (V ) ≤ 1. This shows that the rank of T is at most 1. 7. (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 ) ≤ dim M + dim N (which is a result of dim(M + N ) + dim(M ∩ N ) = dim M + dim N ). (b) Since (ST )(V ) ⊆ S(V ), we have dim(ST )(V ) ≤ dim S(V ), or rank ST ≤ rank S. Next, from the obvious fact ker T ⊆ ker(ST ) we have dim ker T ≤ dim ker(ST ) and hence dim(ST )(V ) = dim V − dim ker(ST ) ≤ dim V − dim ker T = dim T (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(T S), we can show rank(ST ) = rank(T ). 9. Consider the induced linear transformation MA : n →  m defined by MA = A. Then M A is a rank one linear mapping and hence its range is spanned by a single vector in m , say  = [a a    a m ] ⊤ . It is known that the columns of A span the range of MA. So all columns of A are scalar multiples of , say b , b , . . . , bn . Thus A = [b

b





a  b  a  b  bn] =  

am b

 ab n ab n   . 

ab  ab 

 

a mb

   am b n

10. Consider the linear mapping T : P →  given by T (p(x)) = (p(0), p ′(0), p′′ (0), p(1), p′(1)).

We need to prove that the range of T is  . Since dim P  = dim   = 5, it is enough to verify that ker T = {0}. Suppose p(x) ≡ b + b x + b x + b x + b  x ∈ ker T . From p(0) = 0, we get b = 0. From p′(0) = 0, we get b  = 0. From p ′′(0) = 0, we get b = 0. Thus we have p(x) = b x + b x. Now p(1) = 0 gives b + b  = 0 and p′ (1) = 0 gives 3b + 4b = 0. From the last two identities we obtain b = b = 0. Hence ker T = {0}. 11. (a) We can write q(x) = cxn + p(x), where c is a scalar and p(x) ∈ Pn− . Then (x + a)q(x + 1) − xq(x) = (x + a)(c(x + 1)n ) − x(cxn ) + ((x + a)p(x + 1) − xp(x)) = cx((x + a)n − xn) + ac(x + 1) n + (x + a)p(x + 1) − xp(x) Clearly ac(x + 1)n , (x + a)p(x + 1), xp(x) ∈ Pn. Now cx((x + a)n − xn ) = cx((x + a) − x)((x + 1)n− + (x + 1) n− x +    + (x + 1)xn− + xn− ) = acx((x + 1) n− + (x + 1)n− x +    + (x + 1)xn− + xn− ) ∈ P n. Hence (x + a)q(x + 1) − xq(x) ∈ P n. (b) Consider the linear operator T : Pn → P n defined by T (q(x)) = (x + a)q(x + 1) − xq(x). It is enough to check that ker T = {0}. Let q(x) ∈ ker T . 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) = 0. 12. (a) For q(x) in Pn, we can write q(x) = cxn + p(x), where c is come scalar and p(x) ∈ P n−. 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)xn + r(x) = cx(x + 1)((x + 1)n− − xn−) + r(x) = cx(x + 1)((x + 1)n− + (x + 1)n−x +    (x + 1)x n− + xn−) + r(x)

which is in Pn by inspecting each term in the last expression. (b) Consider the linear operator T on Pn defined by T (q(x)) = xq(x + 1) − (x + 1)q(x). It is enough to show that T (Pn) = P n. In order to show this, it suffices to check ker T = {0}. Let q(x) = x. Then q(x) = 0 and T (q(x)) = x(x + 1) − (x + 1)x = 0 and hence q(x) ∈ ker T . 13. (Proof of Theorem 2.3.2) Take a basis W

= {, , . . . ,  r } in M ∩ N . Extend it

to a basis U = { , . . . , r , , , . . . , s } of M , and a basis V = {, . . . ,  r ,   , , . . . , t } in N . We verify that R = {,  , . . . ,  r ,  , . . . , s ,  ,  , . . . ,  t} form a basis of M + N . First we check that R is linearly independent. Suppose a  + a +    + a r r + b  + b +    + b s s + c  + c  +    + ct  t = . Then we have  : = a + a   +    + ar  r + b  + b  +    + bs s = −c − c   −    − c tt .

(†)

The first expression of  shows  ∈ M and the second expression shows  ∈ N . Thus  ∈ M ∩ N . Hence we can write  as a linear combination of vectors in W :  = h  + h  +    + hr  r. In view of  = −c  − c −    − ct t, this gives h + h  +    + h r  r + c  + c  +    + c t t = . Since vectors in V

are linearly independent, we obtain h = h =    = hr = c = c  =    = ct = 0.

Thus (†) becomes a  + a  +    + ar r + b  + b  +    + b ss = . Since vectors in U

are linearly independent, we have a = a =    = ar = b = b =    = b s = 0.

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 indepenC dence 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 seen as follows. Write p(x) = cx n + q(x) with q(x) ∈ Pn−. Then p(x) = c((x + 1)n − xn) + (q(x + 1) − q(x)). Clearly q(x + 1) − q(x) ∈ Pn− . 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 p(x) + a p(x + 1) + a p(x + 2) +    + an p(x + n) = 0.

( ∗)

The nth derivative of the above identity gives us cn!(a + a  + a +    + a n) = 0, where c is the leading coefficient of p(x). Hence a  + a +    + an = 0

(∗∗)

We can rewrite the left hand side of (∗) as a telescoping sum (a + a +    + a n )p(x) + (a + a +    + a n)(p(x + 1) − p(x)) + (a +    + a n)(p(x + 2) − p(x + 1)) +    + + (an− + 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q(x) + b  q(x + 1) +    + bn− q(x + n − 1) = 0 where q(x) = p(x + 1) − p(x) is a polynomial of degree n − 1 and bk =

n

j k  a j .

By the induction hypothesis, we know b = b =    = bn− = 0. Now it is clear that a = a =    = an = 0....