Math 115 Lab 8 Solutions

1. Find a basis for S where

S = span

1321

,1211

,1101

,1221

,3413

Answer: A quick way to find a basis for S is to place the vectors in a matrix A and find the RREF.For this problem

1 1 1 1 33 2 1 2 42 1 0 2 11 1 1 1 3

RREF1 0 1 0 20 1 2 0 50 0 0 1 00 0 0 0 0

Then the independent vectors that span S are the corresponding columns where the leading onesappear. Thus a basis of S consists of the first, second and fourth column of the set

B =

1321

,1211

,1221

2. Let the subset Q = {p(x) = a+ bx+ cx2 : p(0) = 0} of P2.(a) Prove that Q is a subspace of P2.

Answer: Let p(x) = a1 + b1x+ c1x2 and q(x) = a2 + b2x+ c2x2 in Q. Hence, a1 = p(0) = 0 anda2 = p(0) = 0. Then

(p+ q)(x) = p(x) + q(x) = (a1 + a2) + (b1 + b2)x+ (c1 + c2)x2

with (p+ q)(0) = p(0) + q(0) = a1 + a2 = 0. Thus (p+ q)(x) is in Q.Let p(x) = a+ bx+ cx2in Q and a scalar k R. Then

(kp)(x) = kp(x) = k(a+ bx+ cx2) = (ka) + (kb)x+ (kc)x2

with (kp)(0) = kp(0) = ka = 0.Finally, the zero polynomial satisfies p(0) = 0.Therefore Q is a linear subspace of P2.

(b) Find a basis for the subspace Q.Answer: Let p(x) = a+ bx+ cx2 in Q. Then a = p(0) = 0, hence p(x) = bx+ cx2. Thus the set{x, x2} spans Q. Moreover it is a linearly independent set, thus it defines a basis for Q.

3. Let H be the subset of the skew-symmetric 2 2 matrices.(a) Prove that H is a subspace of M22.

Answer: Let A,B be skew-symmetric, thus AT = A and BT = B. Therefore(A+B)T = AT +BT = AB = (A+B),

thus A+B is also skew symmetric.Let A be a skew-symmetric matrix and k R. Then

(kA)T = k(AT ) = k(A) = (kA),hence kA is also skew-symmetric.Finally, the zero matrix 0 is also skew-symmetric since 0T = 0.Therefore the subset of skew-symmetric matrices is a subspace.

(b) Find a basis for this subspace.

Answer: Let a skew-symmetric matrix A =[a bc d

]. Then

AT = A[a cb d

]=[a bc d

]

a = a,b = c,c = b,d = d

a = 0,b R,c = b,d = 0.

Hence every 2 2 skew-symmetric matrix can be written as

A =[0 bc d

]=[0 bb 0

]= b

[0 11 0

].

Therefore the matrix[0 11 0

]spans the subspace of the skew-symmetric matrices. Hence the set{[

0 11 0

]}forms a basis.

4. Prove that the set {1 + 2x2, 1 x} is an independent set of vectors in P2 and P3. Extend it to a basisof P3.Answer: Let a, b R such that

a(1 + 2x) + b(1 x) = 0,where 0 is the zero polynomial. Then

a+ 2ax2 + b bx = 0 (a+ b) bx+ 2ax2 = 0

a+ b = 0,b = 0,2a = 0

a = b = 0.

Thus the set is independent in P2 and in P3.First we will extend to a basis of P2. We know that {1, x, x2} is the standard basis for P2 and we notethat {1, 1 + 2x2, 1 x} is independent. Indeed, let a, b, c R such that

a1 + b(1 + 2x2) + c(1 x) = 0 (a+ b+ c) cx+ 2bx2 = 0

a+ b+ c = 0,c = 0,2b = 0

a = b = c = 0.

Moreover, it spans every element in P2 since,

x = 1 + 0(1 + 2x2) (1 x),x2 = 1

2 1 + 1

2(1 + 2x2) + 0(1 x).

(Alternatively, we can say that the subspace span{1, 1+2x2, 1x} has dimension 3 and it is a subspaceof P2 which also has dimension 3. Therefore span{1, 1 + 2x2, 1 x} = P2.)Finally, the set {1, 1 + 2x2, 1 x} extends to the basis {1, 1 + 2x2, 1 x, x3} of P3.

5. In each case either prove the assertion or give a counterexample showing that it is false

(a) Every set of four non-zero polynomials in P3 is a basis for P3.Answer: No. Pick your favourite polynomial p(x) in P3 (for example p(x) = x and let the fourpolynomials p1(x) = p(x), p2(x) =

2p(x), p3(x) = 1821p(x), p4(x) = pip(x). Then

1 p1(x) 12 p2(x) + 11821 p3(x)

1pi p4(x) = p(x) p(x) + p(x) p(x) = 0.

(b) P2 has a basis of polynomials p(x) such that p(0) = 1.Answer: Yes. We have shown that the set {1, 1 + x, 1 + x + x2} forms a basis for P2 and thesepolynomials satisfy p(x) = 1.

(c) If {~u,~v, ~w} is independent then a ~u+ b ~v + c ~w = ~0 for some a, b, c.Answer: Yes. If we let a = b = c = 0 then 0 u+ 0 v + 0 w = 0.

(d) The set {~u,~v, ~w} is independent if a ~u+ b ~v + c ~w = ~0 for some a, b, c.Answer: No. Let ~u,~v, ~w with ~v = 2~u, ~w = 3~u; then for a = 2, b = 1 and c = 0 we compute

a ~u+ b ~v + c ~w = 2~u ~v = ~0,

but {~u,~v, ~w} = {~u, 2~u, 3~u} is not independent.(e) If {~u,~v} is independent, then {~u,~v, ~u+ ~v} is independent.

Answer: No. Let a, b, c R such that

a ~u+ b ~v + c(~u+ ~v) = ~0.

Then

(a+ c) ~u+ (b+ c) ~v = ~0.

Since {~u,~v} is independent, then (a+ c) = 0 and (b+ c) = 0. Hence a = c, b = c and there isno restriction on c. So if we choose a = b = 1 and c = 1 we see that the set {~u,~v, ~u+ ~v} is notindependent.

(f) If {~u,~v, ~w} is independent then {~u+ ~w,~v + ~w} is independent.Answer: Yes. Let a, b R such that

a (~u+ ~w) + b (~v + ~w) = ~0.

Then

a ~u+ a ~w + b ~v + b ~w = ~0 a ~u+ b ~v + (a+ b) ~w = ~0.

Since {~u,~v, ~w} is independent we get that a = b = a+ b = 0, hence a = b = 0. So {~u+ ~w,~v + ~w}is independent.