Math 110Professor Ken RibetLinear AlgebraSpring, 2005January 18, 2005See http://math.berkeley.edu/~ribet/110/ forinformation about:• The textbook• The discussion sections• Exam dates• Grading policy• Office hours (link)1Note especially that there is an online discussiongroup, Google’s Math 110. The URL for the groupis http://groups-beta.google.com/group/Math110/. Ifyou “join” the group, you can post comments andquestions. If you don’t join, you can still lurk andread what other people have written.2We have a fine lineup of experienced GSIs: Chu-Wee Lim, Scott Morrison, John Voight. Discussionsections are on Wednesdays ; see the class Web pagefor times and room numbers. Note that #103 is in433 Latimer, not 435 Latimer.Some of the sections are full. If you want tochange your s ec tion to one that is full, you need tospeak with Barbara Peavy in 967 Evans. The burdenwill be on you to convince her that your schedulerequires you to change into one of the full sections.3An initial question for feedback: Is lecturing witha laptop going to be effective? Mathematicianshave traditionally written on the board with railroadchalk (fat chalk) when giving large courses. ShouldI do that? Should I use transparencies? Somecombination? Post to the newsgroup with yourthoughts.Meanwhile, I will try to le cture today with a laptopand the projector.4Even though we’re in a large room, don’t hesitateto stop me to ask questions. One good thing aboutthe laptop is that I can see you while I’m speaking.You can interrupt—don’t be shy. If you would likesome clarification, so will your friends.5Vector Spaces and FieldsI imagine that you know more t han a bit aboutmatrices, systems of equations, linear maps,. . . .You have taken Math 54. In Math 110, westudy vector spaces and linear transformations moreabstractly. We will be interested a choosingbases for vector spaces in such a way that lineartransformations are represented by espe cially nicematrices.6What is a field? It’s a set with an addition anda multiplication; the system is required to satisfy alist of familiar-looking axioms (Appendix C of book).Some examples:• The field R of real numbers.• The field C of complex numbers.• The field {0, 1} with two elements.• The field Q of rational numbers.7Fix a field F . A vector space over F is a set Vtogether with two additional structures: an additionlaw (x, y) 7→ x + y on V and an operation of Fon V :F × V → V, (a, x) 7→ ax.These ope rations satisfy a whole bunch of axioms(VS 1–VS 8 in the book).We refer to V with its two additional structuressimply by the letter V .8ExamplesAn example that we all know: for each n ≥ 1,the set Fn= { (c1, . . . , cn) | ci∈ F for 1 ≤ i ≤ n }becomes an F -vector space with the operations(c1, . . . , cn) + (d1, . . . , dn) = (c1+ d1, . . . , cn+ dn)anda · (c1, . . . , cn) = (ac1, . . . , acn)(componentwise addition and multiplication byelements of F ).9The 0-vector space is the set V = {0} with theobvious operations: a·0 = 0 for all a ∈ F , 0+0 = 0.It is true somehow that {0} = Fnwhen n = 0.Warning: I sometimes write simply 0 for the 0-vector space. Our authors are careful to write {0}instead. I like my shorthand, but don’t want to forceit on anyone.10Another example: the set of m × n matricesa11a12· · · a1na21a22· · · a2n.........am1am2· · · amn,again with component-wise addition and the obviousmultplication by elements of F . This example isvisibly a re-labeling of Fmn. This set is calledMm×n(F ) in the book.11We can take V to be the space of all polynominalsover F (of any degree) or the space of polynomials ofdegree ≤ n for a fixed non-negative integer n . Thesespaces are called ℘(F ) and ℘n(F ), respectively.Since a polynomial of degree ≤ n,cnxn+ cn−1xn−1+ · · · + c1x + c0,is just a string of n + 1 numbe rs, ℘n(F ) is a suaveway of writing Fn+1.12If S is a set, we can take V to be the set offunctions f : S → F , and define the ope rations ina pointwise manner. Thus f + g is the function(f + g)(s) = f (s) + g(s), and af is the functiontaking s to af(s). If S = {1, 2, . . . , n}, we ge t Fn.An interesting variant is to take V instead to bethe set of functions S → F that have only a finitenumber of non-zero values. This is a genuine variantif S is infinite.13For e xample, if S is the set of natural numbers(i.e., non-negative integers), the “variant” V that wehave defined is the set of sequences c0, c1, c2, . . . suchthat cm= 0 for m sufficiently large. Such sequencesare really the same thing as polynomials. Hence theV that you get in this case is the F -vector space ofpolynomials over F .If you remove the requirement that cmbe zerofor large m, you get “formal power series” (non-terminating polynomials) such as 1+x+x2+x3+· · ·.14The axiomsFirst, V is an abelian group under addition:• We have x + y = y + x for x, y ∈ V ;• For x, y, z ∈ V , x + (y + z) = (x + y) + z;• There is a (unique) 0 ∈ V such that x + 0 = x forall x ∈ V ;• For each x ∈ V , there is a (unique) −x ∈ V suchthat x + (−x) = 0.15We should all know the proof that 0 is uniqueand that additive inverses are unique. For the first,assertion, suppose that 0 and 00both play the roleof zero. Then 0 + 00is both 0 and 00. Hence 0 = 00!If y and z are additive inverses for x, theny = y + 0 = y + (x + z) = (y + x) + z = 0 + z = z.16Next, two axioms about the action of F on V :• For all x ∈ V , 1 · x = x;• For a, b ∈ F and x ∈ V , a(bx) = (ab)x.Finally, there are two “distributive laws”: a(x +y) =ax + ay and (a + b)x = ax + bx, v alid for a, b ∈ F ,x, y ∈ V .17The deal about the axioms is that you want themto be easy to c heck but want them also to be powerfulenough to imply familiar-looking statements that hadbetter be true for us to preserve our sanity.For example, we want 0 · x = 0 and −(−x) = xfor all x ∈ V , (−a)x = a(−x) = −ax for a ∈ F ,x ∈ V , etc., etc.18Let’s prove something. Take a vector x in a vectorspace V . Consider 0·x. We’d be amazed if 0x turnedout to be anything other than 0 (the 0-vector).How do we actually prove that 0x = 0???19For x ∈ V , it’s true that 0x = (0 + 0)x = 0x + 0xbecause 0+0 = 0 in F and be cause of the distributivelaws. Let z = 0x. Then z = z+z. We can add −z toboth sides and get 0 = z + (−z) …