Mathematics 581    Problem Set 8 (the last set)     Do not hand in.
Solutions will be posted Wednesday, June 5

Reminder.  Final exam in our classroom, Wednesday, June 12:  9:30 am - 11:18 am.

1. Let V be a vector space over a field F.  Suppose that  {v1, ... , vn}  spans  V.  Prove that there is a subset of  {v1, ... , vn} which is a basis for  V.
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If  V = {z}, then let the empty subset be the basis.  Otherwise construct a linearly independent subset  {w1, ... , wm} of {v1, ... , vn}  as folows.  For  w1 choose the first nonzero vi, reading from left to right.  For  w2 choose the first nonzero vi after  w1 which is not a linear combination of  w1.  Continue, at the jth stage choosing  wto be the first nonzero vi after
wj - 1 which is not a linear combination of  w1, w2, ..., wj - 1.  Then none of the w's chosen is a linear combination of the of  w's which precede it.  Hence, at every stage the w's so far chosen form a linearly independent set.  Do this until you have made one pass through all of the vi's. You have a linearly independent set  {w1, ... , wm}.  Those vi's which you skipped over while choosing the w's are linear combinations of the w's already chosen.  Hence, each vi is in the span of  {w1, ... , wm}.  Hence every  v  in  V  is in the span of  {w1, ... , wm}. (Why?)
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2.  Let  A  be a linear transformation of a vector space  V  over  F  into a vector space  W  over  F.

(a)  Let  z  be the zero vector in  V, and  zW  be the zero vector in  W.  Prove that
zA = zW.
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z = z + z.  Therefore,  zA = (z + z)A = zA + zA.  Adding -(zA) to both sides, we get
zA = zW.
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(b)  Prove that for all  v  in  V,  (-v)A = -(vA).
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v + (-v) = z.  Therefore, zW = zA = vA + (-v)A.  So, (-v)A  must be the inverse of  vA.
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3.  Considering the field of real numbers  R  as a vector space over its subfield of rational numbers  Q,  prove that  {1,  r}  is a linearly independent set in  R if and only if  r  is irrational.
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If  r  is irrational but  {1, r}  is linearly dependent, then there exist rational numbers  x, y, not both 0, such that  (*) x1 + yr = 0.  If  y = 0, then  x1 = 0, so that  x = 0 as well.  Therefore,  y is nonzero.  There are integers a, b, c, d, such that x = a/b  and  y = c/d.  The  c  and  d  must be nonzero, and we solve equation (*) for  r:  r = -x/y = -(a/b)(d/c), which is rational, a contradiction.  Hence,  {1, r} is linearly independent.

On the other hand, if  is rational, then  r = u/v for integers u  and  v, with  v  nonzero. Then
u1 + (-v)r = u - v(u/v) = 0, and  {1, r}  is linearly dependent.
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4.  Prove that if  F  is a field, then the set of two vectors {(a, b),  (c, d)}  is linearly independent if and only if  ad - bc is not zero.  [Caution: you may have to break up the proof into several cases.]
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Here is one way to organize the proof.

Suppose  ad - bc  is not zero.  If   x(a, b) + y(c, d) = (0, 0), then

(1)  xa + yc = 0, and
(2)  xb + yd = 0.

Multiply equation (1) by d  and equation (2) by  c  and subtract, obtaining x(ad - bc) = 0.
Hence, x = 0.  To get y = 0, multiply equation (1) by  b and equation (2) by  a  and subtract.

Suppose, on the other hand, that  ad - bc = 0.  We need to show that  {(a, b),  (c, d)} is linearly dependent.  It IS if  a = b = 0 or  c = d = 0,  so assume that  a  or  b  is nonzero and that  c  or  d  is nonzero.  If  a  is not zero, then d = bc/a, which implies that  if c = 0, then  d = 0.  So  c  is not zero.  Then   (c/a)(a, b) + (-1)(c, d) =  (c, cb/a) + (-1)(c, bc/a)  = (0, 0), and  {(a, b),  (c, d)} is linearly dependent.  If  a = 0, then  b  is not zero, and  ad - bc = 0  implies that  c = 0.  Then  d(a, b) + (-b)(c, d) = d(0, b) + (-b)(0, d) = (0, 0)  again.
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5.  Let  R[l]  be the vector space over the field of real numbers which consists of all polynomials with real coefficients.  Define a polynomial  p(l)  to be odd if  for every real number  a,  p(-a)= -p(a)  and define a polynomial  q(l)  to be even if for every real number  a,  q(-a)= q(a).

(a) Write down three odd polynomials of different degrees.
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l, l3, l5
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(b)  Write down three even polynomials of different degrees.
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1, l2, l4
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(c)  Prove that every polynomial is the sum of an odd polynomial and an even polynomial.
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If  p(l)  is any polynomial, then  (1/2)[p(l) + p(-l)]  is an even polynomial,
(1/2)[p(l) - p(-l)]  is an odd polynomial, and  p(l) = (1/2)[p(l) + p(-l)] + (1/2)[p(l) - p(-l)].
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(d)  Prove that if  p(l)  is both even and odd, then  p(l) = 0.
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p(-l)  would have to equal both  p(l)  and  -p(l).  So  p(l) = -p(l),  and adding  p(l)  to both sides,  2p(l) = 0, so that  p(l) = 0.
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(e)  Prove that if  s(l)  is a polynomial and  s(l) = p1(l) + q1(l) = p2(l) + q2(l),  where each  pi(l)  is odd and each  qi(l)  is even, then  p1(l) =  p2(l)  and  q1(l) = q2(l).
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If  s(l) = p1(l) + q1(l) = p2(l) + q2(l),  where each  pi(l)  is odd and each  qi(l)  is even, then  p1(l) - p2(l)  =  q2(l) - q1(l).  But  p1(l) - p2(l)  is even and  q2(l) - q1(l)  is odd,  Therefore by (d),  p1(l) - p2(l)  =  q2(l) - q1(l) = 0,  so that   p1(l) =  p2(l)  and  q1(l) = q2(l).
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In vector space notation, (c)  and  (d) show that if  O  is the subspace of odd polynomials and  E  is the subspace of even polynomials, then  O + E = R[l], and the intersection of O  and  E   is {0}.