Math H110, anonymous comments

Mon Jun 9 20:26:13 2003 : Question: The official prerequisite for 250A is 114, but some say 113 is enough. Is it? And do you recommend the class for undergraduates?

Reply: When I last taught the Math 250A (Fall, 2001), there were 41 students at the end of the class; 11 were undergraduates. The abstract algebra course is a good place for undergraduates to get their first taste of graduate mathematics. If you are contemplating graduate study in mathematics and want to see what it's like, then Math 250A is a good choice. If you haven't taken Math 114, you may find that the Galois theory in Math 250A is so slick that you won't get a feel for what's really going on. My view is that you can still take Math 250A, but you should work out lots of examples in a book like Stewart's "Galois theory". Obviously, you should ask your questions also to Bjorn Poonen, who'll be teaching the course.

Sun Jul 13 18:35:34 2003 : Hi--I'm curious to know why we're going to be using the FIS text (which seems rather computational) instead of Axler's. I took a look at Axler's in the bookstore, and his treatment seems somewhat more theoretical than FIS, which seems appropriate for an honors class. Are we going to follow the text closely, and more importantly, are you going to assign exercises from the text, or are you going to write up your own? thanks.

Reply: The main reason for using the text by Friedberg et. al. is that it's the most common choice for Math 110. I want students to be able to transfer out of the honors course without having to scramble to catch up with the work that has been done in their new section. In the other direction, a strong student from a non-honors section will have at least a fighting chance of being able to transfer into H110 without going completely crazy since the textbooks, at least, will be the same. The Friedberg text has a somewhat concrete perspective, but it has complete proofs of all the major results. I like the fact that it stresses elementary row and column operations, which are tremendously useful for practical applications but neglected by most purely theoretical books. My guess is that I will follow the book closely but will go fast when there are sections that are purely computational. For homework assignments, it'll be easy to include problems that are not taken from the book if the book's problems are too elementary for the class.

Mon Aug 4 08:04:57 2003 : Is the enrollment limit eventually going to be increased? What is the probability that people on the waitlist will be able to enroll in this class? Thanks.

Reply: The class limit was bumped up to 30, so there's currently no waiting list. I am guessing that some of the 30 students who are enrolled did not realize that they were signing up for an honors class. If so, there will be room for a few further students even if the class limit of 30 is not raised any further.

Fri Aug 8 11:44:50 2003 : for course planning purposes, how many hours of homework do you intend to assign every week?

Reply: It's hard to estimate long it will take students, on average, to do the weekly homework. If you look at the web sites from my two previous upper-division courses 110 and H113, you'll see how many questions I tend to assign each week. In H110, I imagine that I'll assign no more problems than in regular 110 but that the problems will be harder and more theoretical. When you plan out your courses, you should consider the more inclusive question: "How much time outside of class will I be spending on H110?" You will need to spend time before and after each lecture to internalize the concepts that are discussed in the lecture. I will follow the book, so you will know before each lecture what will be on the table. If you read the book carefully ahead of time, you'll be in a good position to focus on those aspects of the topic that seemed unclear when you first read over the material. For advice on issues like this, students new to Berkeley's math major might benefit from a chat with more seasoned students and with staff members Catherine Pauling and Alison Thompson. (Dexter Stewart has moved to another department; Alison is taking over Dexter's responsiblities.)

Wed Aug 20 20:55:41 2003 : Ken, I see we're both teaching linear algebra from Friedberg et al. this fall. Tom Weston had recommended the book to me last spring when we each found out we were both teaching linear algebra. Maybe he got the idea from you. There's a nice proof of the fundamental theorem of algebra based on ideas from linear algebra in the latest Monthly (the proof...). I'll try to include this as an application near the end of the course. The font scheme in the text is pretty crazy: italic R for {\mathbf R}? Let's see if you can guess who wrote this.

Reply: Keith, I have no clue as to who wrote the comment. About the "nice proof": I found it on the author's preprint page

Mon Aug 25 14:44:28 2003 : Final exam date is wrong on the web page - or perhaps you meant to provide an extended review period :)

Reply: Thanks for the report. Fixed now, I hope.

Mon Aug 25 20:52:37 2003 : Maybe it would be a good idea to post page numbers of suggested reading for each lecture before the lecture. How does that sound?

Reply: For some reason, I'd be happier if students tried to figure this out on their own. I think that it will help you organize the material more if you have to think about where we are, where we've been and where we might be going. On the other hand, if we're about to skip over some material or do things in a non-obvious order, I'll let you know by announcing it in class and/or sending e-mail to all students in H110.

Incidentally, I really think that attending class is of capital importance. The "course" includes all of our face to face interactions. Frequently there will be discussions in class that couldn't have been anticipated before the lecture.

Tue Aug 26 15:56:46 2003 : I'm slightly confused by problem 7 of section 1.1. It is asking for us to prove it for the generic case or only for vectors in 2 dimensional spaces? Also, on page 2, regarding the Parallelogram Law for Vector Addition, are there any restrictions on the vectors x and y? For example, do they have to be non-zero and non-parallel?

Reply: My impression is that problem 7 really was for vectors in n-space and not just for plane vectors. The parallelogram law on page 2 is surely true even in the degenerate cases that you cite if it's interpreted appropriately.

Tue Aug 26 20:39:26 2003 : This is partly to answer the above question and partly to make sure I'm doing the hw correctly. I think the first chapter is supposed to be motivational and vague (in the axiomatic sense), so I didnt worry too much about the about the nitty gritty stuff when doing number 7 and tried to make my reasoning similar to the examples from the chapter. Am I being too loose? Also, just for the sake of fruitless complaining, does the similar notation for vectors and scalars in the text bug anyone else?

Reply: As far as notation goes, it seems to me that there are books that try to distinguish vectors from scalars in various ways: some have vectors in bold letters and scalars in roman letters, while some have sclars in Greek letters and vectors from a to z. In this book, the authors don't have any general scheme, but I think that they do enough to make sure that the reader won't confuse vectors and scalars in any given situation. It gets more complicated when there are vector spaces and matrices around. As far as vagueness in this chapter is concerned, all I can say is that you should try to be as clear as possible. The concept of "vector space" is defined formally in the second section of the book, so you shouldn't have long to wait before the vagueness goes away.

Wed Aug 27 09:34:49 2003 : Assuming the simple 2-d case and that the two vectors are parallel, then wouldn't the 2 diagonals be collinear? How could collinear segments bisect each other? Unless of course I'm misintrepreting the definition of bisecting. Also is there a 113 way of doing #21 of 1.2? There is a direct product for VS, right?

Reply: OK, I give up: in problem #7 of § 1.1, it's a good idea to assume that the vectors that form your parallelogram or not actually parallel. (Thus, you assume that they span something planar and not just a line sigment.)

As far as #21 of § 1.2 goes, the Z that's being formed is indeed the direct product of V and W. From the point of view of Math 113, you could say that a vector space is an abelian group plus an extra structure, which you could characterize as an action of F on the abelian group. The abelian group that underlies Z is just the direct sum or product (they're the same thing here) of V and W as abelian groups. The problem defines an action of F on this direct sum: it's the componentwise action. You have to check that Z with this action is an F-vector space. You could plausibly begin your treatment by saying that we know from Math 113 that it's an abelian group; this means that you're claiming as already known the first handful of vector space axioms. Then you still have to check that more is true, namely that the action of F that's in the picture makes it so that the rest of the axioms hold.

Wed Aug 27 16:33:05 2003 : Dr Ribet, I like reading the sections prior to lecture, I don't know but maybe it would be possible to let us know what sections will be discussed in an upcoming meeting, or at least which sections of the book will be omitted. Thanks

Thu Aug 28 12:57:01 2003 : Woops, I didn't see that you had previouslyu responded to this.

Thu Aug 28 17:36:28 2003 : Is a single differentiable function from the reals to the reals infinitely differentiable? Can't find anything about it in my 104 book

Reply: Take a continuous function f that isn't differentiable and let F(x) be a definite integral of f, e.g., the integral of f from 0 to x. Then, by the fundamental theorem of calculus, F is differentiable with derivative f. Then F" is no better defined than f' is. If f is nowhere differentiable, then F" exists nowhere. If f'(0) doesn't exist, then F"(0) doesn't exist.

Sat Aug 30 21:33:44 2003 : Dr Ribet, I really appreciate that you post the HW assignments up promptly to allow me to get a head start over the weekend, when I have more time than during the week to look at things. Weekdays are always hectic. Just giving some positive feedback!

Reply: Thanks very much. I'll try to get homework assignments prepared sufficiently far in advance that students will have a full week to work on them. As I've said before, I think that the good strategy is for students to start thinking about problems as soon as they're assigned. Hard problems become a lot easier after you've had time to work on them for a while.

Sun Aug 31 10:03:07 2003 : how much more abstract algebra are we going to use in the class?

Reply: The question sounds as though it were written by someone wanting less abstract algebra! I'll do my best not to have extensive tangential discussions like the one relating a vector space structure on an abelian group to a ring homomorphism. On the other hand, I believe that it's a good idea to mention direct analogies with Math 113 when appropriate. Take the example of the construction of a quotient vector space, which I started to discuss on Friday. This is the exact same construction that one performs to make a quotient group. Not to say this would be almost negligent. If I knew that all my students would be taking Math 113 after my course, I wouldn't bother to mention the connection -- I'd leave it for the Math 113 instructor. It happens, though, that students can take 113 first and then go on to 110. Students who do this will benefit from the comment that they're seeing a specific construction for the second time. Students who take 110 first will get some insight into a construction that they'll see later on in 113.

Sun Aug 31 17:18:05 2003 : So far, I have found the abstract algebra to be pretty interesting, overall. However, maybe because I have no background in abstract algebra, I don't really see the significance of the relations between vector spaces and abelian groups, linear transformations and group homomorphisms, etc. I don't see the motivation behind studying the relationships, and I don't really see why studying the abstract algebra is necessarily important. Perhaps you could say a few words on this. Thanks.

Reply: Linear algebra is a part of abstract algebra. The majority of abstract algebra textbooks contain substantial discussions of linear algebra. Like most universities, UC Berkeley thinks that the material in Math 110 is important enough to deserve a course all its own. Moreover, we've decided that students can study abstract linear algebra before studying abstract algebra. This is all fine, but I think that students, and especially honors students, will benefit from extra perspective. The significance of the relation between vector spaces and abelian groups is that vector spaces are abelian groups -- they're abelian groups with some extra structure (an action of a field on the abelian group). Linear transformations between vector spaces are, in particular, homomorphisms between the vector spaces, thought of as abelian groups. They're homomorphisms with the extra property of "commuting" with field multiplication.

I can sense here that you'd rather hear less than more about abstract algebra. I'll steer away from abstract algebra for a while. Note that the construction of quotient spaces is done out in problem 31 on page 23 of the book.

Mon Sep 1 15:30:45 2003 : on problem twelve, does the author mean span(W) = V instead of W?

Reply: The book has it right: a set is a subspace if and only if it's equal to its own span. On the other hand, all textbooks have misprints, and most authors are eager to hear about misprints in their books. We should make a list of misprints that we find that are not noted already in the authors' list of errata. If you flag them by typing into the comment box, I'll make sure that the authors get the correction. I'll wait until December and then send them everything that we've found.

Mon Sep 1 15:48:34 2003 : I'm pretty sure the author means span(W)=W. And if you check the errata for the book,, there's nothing for that problem.

Tue Sep 2 15:23:39 2003 : On #14 in 1.5, it lists v as being in the set of "distinct" vectors of S. I am not convinced that under the assumption of S being lin dep, one can arrive at the conclusion that v itself be distinct. Certainly the vectors which combine to make v can be shown to be distict, but v itself? Am I right or can it indeed be shown?

Reply: To say that S is linearly dependent is to say that you can write 0 in a non-trivial way as a linear combination of vectors in S. The non-triviality means two things. The first is that you must use distinct vectors -- otherwise you could write 0 = v + (-1)v with v in the set. The second aspect of the non-triviality is the requirement that some coefficient in the linear combination be non-zero. (If the coefficients are all 0, you're just writing 0=0.) When S is linearly dependent, you can use high-school algebra to transform a non-trivial expression of 0 as a linear combination of distinct elements of S into an equation that writes one of those vectors in terms of the others. Conversely, if v is a linear combination of the u_i as in the statement of the problem, then you could write 0 in terms of v and the u_i in such a way that you exhibit the linear dependence of the set S. Conclusion: the problem seems fine as written.

Tue Sep 2 22:01:20 2003 : I have found the abstract algebra to be a refreshing way of looking at linear algebra. I would appreciate it if you do not completely remove this aspect from the course.

Wed Sep 3 00:25:11 2003 : I agree with the previous comment. Adding the abstract algebra portions do make the course more interesting.

Wed Sep 3 02:10:32 2003 : Hello, I was the one who posted the comment on -> Sun Aug 31 17:18:05 2003 I, too, am fine with the abstract algebra. What my previous comment was saying was that I didn't quite understand the motivation behind studying the abstract algebra.

Wed Sep 3 10:27:44 2003 : it might be a refreshing way of looking at linear algebra but some of us have never had any abstract algebra and it doesn't seem fair to all students to do so much abstract algebra

Wed Sep 3 16:14:40 2003 : This is Ribet -- I'm writing a comment on my own page. I want to complain about problem 17 in section 1.4. I'm doing it here because I hope that the comments page can be used to keep track of misprints in the book. The complaint here is that problem 17 has nothing to do with V -- it's all about W. I'd rephrase the problem by starting with a vector space W and then asking: under what conditions is it true that W has only a finite number of generating sets?

Some students asked me this question in office hours. It seems to me that one should prove that W has a finite number of generating sets if and only if W has only a finite number of elements. One should then establish necessary and sufficient conditions for W to have finite cardinality. Clearly, W is finite if it is the 0-vector space {0}. Clearly, W is infinite if the field F is infinite and W contains a non-zero vector (because it contains all scalar multplies of the vector). If the field F is finite, W has finitely many elements if and only if it has a generating set with only finitely many elements. I don't know if one can say anything more intelligent than this.

Wed Sep 3 23:08:23 2003 : I think that the assigned exercises are too routine for an honors class. They all are of "plug-in-the-definition" kind of problems. I expected more creative problems.

Reply: Well, I'll be happy to include some more difficult problems in the future. (I think that HW #2 had harder problems than HW #1.)

Meanwhile, be sure that you're doing the best possible job on the problems that have been assigned. The grader, John Voight, has finished the first assignment. You'll get back your papers on Friday. He had this to say: looks like most of the students are adequately prepared. There is, however, a lot of sloppy work.... About half of the class failed to check that addition and scalar multiplication were closed on the set of real-valued real differentiable functions (§ 1.2 #10) and another half of the class thought that § 1.2 #16 was asking to show that M_mn(Q) is a vector space over Q, not that M_mn(R) itself is a vector space over Q.
John told me that he finds it easier to write up model solutions than to make extensive marks on students' papers. You can download his solutions from John Voight's home page,

One more thing: Please staple your assignments together before handing them in. John is worried about losing pages.

Fri Sep 5 08:29:20 2003 : My friends and I actually debated on how to interpret #16 of section 1.2. Could we attribute the students' error to the poor phrasing of the question? Regardless, isn't this question not very honor-ish? I'm sure the students who missed the problem knew how to answer the intended question had their reading comprehension skills been adequate.

Reply: I hope that the assigned problems will begin to strike you as more honor-ish. This particular problem may have struck you as silly, but it did actually make a point, at least implicitly: If you have two fields k and K, with k stuck inside K, every K-vector space can be regarded as a k-vector space by remembering only the action of the smaller field k. Passing from K to k is called "restriction of scalars"; it's important in many contexts.

I thought that the problem was phrased unambiguously. In mathematics courses, one of the things that you're required to do is to read carefully and to express yourself precisely. This issue comes up quite often in Math 55, the discrete math class, in combinatorial questions, especially when probability is involved. You have to decide when two things are different, when order counts, when replacement is allowed, and so on. In our class, it's rare that there are questions of this subtlety, but you still have to pay attention sometimes.

Sun Sep 7 13:02:43 2003 : I really enjoy your lectures. Even though it all goes by quicker than I'm able to process -- but you suggestion to read ahead would most likely solve that problem.

Sun Sep 7 20:36:50 2003 : Will there be written solutions to some problems from hw?

Reply: Our grader, John Voight, wrote up the solutions to HW #1 last week. His H110 page offers them in three different formats! If I understand correctly, John will usually type up solutions as he grades papers. He says that it's easier for him to do this than to make marks on individual students' papers.

Tue Sep 9 21:34:35 2003 : Could you give maybe a clue for the first problem? I can't really think of a good way to approach it. I've tried quite a few things, but none seem to work... Also, for the second problem, are you sure you meant to ask "Show that the products st with s in S and t in T form a basis of V as an E-vector space"? I don't think that it is true... I don't think the products are necessarily linearly independent. Maybe you mean to ask that some subset of the set of products forms a basis?

Reply: For the first problem, suppose that the finite field is the field of integers mod p, where p is a prime. Think about the case of a vector space of dimensions 0, 1, 2, 3. Start with small examples and see if a pattern emerges. For the second problem, you can suppose that S and T are finite if you want to. To figure out what's going on, consider the case where the two fields are R asnd C. Also, there was a misprint, which I just corrected: the products st should be ts instead -- s is a vector and t is a field element; we usually write the scalars to the left of the vectors.

By the way, someone sent me e-mail to ask whether we'd be having a quiz today (Wednesday, September 10). We're not. I wasn't planning to have quizzes in this class. Also, I would never give a quiz without announcing it ahead of time. The person who wrote me the e-mail message had to miss class today and was worried about missing a quiz. Please don't miss class unless there is some overwhelming and compelling reason.

Thu Sep 11 16:01:53 2003 : John "the Reader" here: A few comments on HW #2. First, the nettlesome issue of whether to show 0 is in a subspace. The proposition in the book lists this one of three necessary and sufficient conditions. As many of you rightly noted, if w is in W and W is closed under scalar multiplication, then 0w=0 is in W. This presumes that W is nonempty! If W is nonempty and closed under linear combination, then yes, 0 is in W. This is important, though, because the empty set is not a vector space! Second, I will deduct points if you use a result but do not tell me where it is coming from. I'm not in class and so I can't know what Ken has proven or not. If you want to use a result from class, state it and say it was proven in class. If you are using a result from the book, please refer to it. Otherwise, linear algebra can start to get a little circular, because everything begins to imply everything else... Third, it is important to note that not every vector space is "finite-dimensional"! The vector space of polynomials over a field is an example of such--they can be of arbitrary degree. In particular, if you have a subset S of a vector space, that does not mean that it is finite (!), so span(S) may not be span({v1,...,vn}) for some vectors v1,...,vn. Cheers, John

Thu Sep 11 16:36:26 2003 : Is it ok for problem 21 to use the corollary from sec 1.7 that states that every vector space has a basis?

Reply: No. We didn't cover that section and the "corollary" is something of a fake. (It depends on an extra axiom of set theory.) I infer that you're talking about problem 21 in § 1.6. Please note that "infinite dimensional" does not mean "has a basis that happens to have an infinite number of elements". The term is defined on page 47, at the top. It means "has no finite basis." As far as we know from this definition, an infinite-dimensional vector space might have no basis at all.

Sun Sep 14 00:04:00 2003 : When you want to prove something like "the following are equivalent: P, Q, R", do you have to show that P iff Q, and Q iff R (or any other combination)?

Reply: There are all kinds of different strategies. A frequent one is to prove that P implies Q, that Q implies R and that R implies P. (It can be helpful to reorder the statements first.) If you end up with a series of implications that enable you to show, by following links in the series, that every statement implies every other one, then you've established the equivalence.

Sun Sep 14 01:01:23 2003 : I guess this is more of a comment for our reader John. I was wondering if it's possible in the future for the students to request certain problems being written up. Kind of like what Prof. Ribet did for h113. Thanks.

Reply: All problems are written up. Just look at John's 110 page.

Mon Sep 15 18:55:56 2003 : Can you post the averages of the first two homework assignments?

Reply: According to John, HW #1 had a mean of 16, standard deviation of 3; HW #2 had a mean of 18, standard deviation of 2.

Tue Sep 16 00:39:10 2003 : For the isomorphism problem, can we assume that (x+y)+W=(x+W)+(y+W) and c(x+W)=cx+W, as was noted in lecture? Also can we assume that if we have x+W=y+W, then y-x must be in W, as was noted in lecture?

Reply: The vector space structure on V/W is as you say; for example, the sum of x+W and y+W is (x+y)+W. If you use this without comment on your homework paper, no one will complain. If x+W = y+W, then it's true that x-y is in W. Again, if you use this without comment, no one will complain. I do want to stress, however, that this is not an assumption. Rather, it's an easy consequence of the definitions.

Thu Sep 18 15:59:51 2003 : I think the homeworks are too long and tedious. Many of the problems are similar but take a lot of time

Reply: The homework due on February 26 will be more theoretical than previous homeworks. Also, there are enough external problems, with mathematical notation, that I wrote this assignment in a separate .pdf file. If you would like me to print you out a copy on a sheet of paper, let me know by e-mail and I will do this for you. To some extent, I am making the homework more theoretical in response to students' comments. If you want more routine problems but have been afraid to say so, please express yourself by writing comments into the box.

Thu Sep 18 19:14:13 2003 : My study group thought the homework was actually pretty fun this week. And if you really hate the homework that much, you can choose not do it, ace a few midterms, and still get an A in the class.

Thu Sep 18 21:49:02 2003 : I do agree with the comment Thu Sep 18 15:59:51 2003 that the homework is trivial, long and tedious. Also I feel that the problems should motivate the theory. The definition of a vector space along with a long list of properties that can be derived from it is not a good motivation for studying linear algebra. Why not give problems that illustrate applications of linear algebra to the other branches of mathematics? As a reply to the previous comment: how can one do better than C grade without doing homeworks if the homework counts as 25% of the total grade?

Reply: I agree that students who don't do homework tend to hurt their final grades. I compute grades by adding all components together. In a situation where the mean scores have been quite close to the maximum possible score of 20, a student who gets a 0 for an assignment is not behaving rationally.

As for applications and motivation, I don't think that it would be a good idea for me to stress applications in a specific area (e.g., economics, electrical engineering, cryptography) because students have very varied backgrounds. Some of you are here (in H110) because you like the beauty of this abstract subject. Lots of you are here because you need the theorems of linear algebra as tools in your majors. (Several of you have come to my office hours to show me what you're doing with linear algebra in your other classes.) Most of you have some idea of the way that linear algebra can be applied because you've seen applications in Math 54. Students who want to see a variety of applications should look at the book Linear Algebra by Peter Lax or other books that stress applications. (One by Kenneth M. Hoffman and Ray Kunze is very well regarded, but I don't have a copy to refer to.)

Fri Sep 19 07:36:56 2003 : I am really enjoying the extra homework problems that are not from the textbook (such as the last one under assignment 4). Some of the textbook problems seem too easy, and these extra homework problems provide an opportunity to practice with some of the more challenging ideas from the lectures.

Fri Sep 19 20:09:31 2003 : Can anyone help me figure out how to get this LaTeX font that so many people use to type up their math homeworks? I heard it's a free program but I can't seem to find a place to download it.

Reply: Try the web page Getting Started with TeX, LaTeX, and friends.

Sun Sep 21 15:45:51 2003 : Isn't 2.3.13 the same as the first non-book problem in the HW?

Reply: The first non-book problem asks you to generalize "the first assertion of the problem." Here, "the problem" is 2.3.13. This got garbled because I inserted problem 10 of section 2.6. In 2.3.13, matrices are square. In the non-book generalization, they are allowed to be non-square.

Sun Sep 21 23:25:06 2003 : Did you mean to assign 2.3.13 and 2.5.10 rather than 2.6.10? Question 2.5.10 actually refers back to 2.3.13, while 2.6.10 is assigned again later in the problem set.

Reply: I intended to assign 2.6.10 and not 2.5.10. What must have happened is that the reference to 2.6.10 get left in at the beginning; I was probably intending to move it down to the bottom.

Mon Sep 22 01:21:35 2003 : I've been learning a LOT in the last few lectures. It takes me quite a bit of extra time and studying to fully absorb it and figure it out, but it's definitely been worth it. I especially like getting new perspectives and proofs of things from 54. I really am enjoying this class so far. Thanks, Professor Ribet!

Mon Sep 22 21:05:52 2003 : After multiplying out A and B two different ways (AB and BA) and computing the trace as one of the homework problem asked, I got the same answer. I am wondering if there is another "explanation" for this that explains the "meaning" of the trace. Thank you.

Reply: I'd prefer to think of this a bit more intrinsically. An endomorphism of a vector space V is a linear map from V to itself. When V has finite dimension, we can take the trace of an endomorphism of V and get a number. With this AB business, we should think that we are given a linear map T:V->W and a linear map U:W->V and that we are computing the traces of the endomorphisms UT of V and TU of W. We get the same answer both times and the question is why this is so.

The two functions "trace(UT)" and "trace(TU)" are linear functions of T and U, so we can replace T and U by simple linear transformations that are chosen from generating sets of the vector spaces L(V,W) and L(W,V). If you have a vector w in W and a functional f in V^*, you can make a linear transformation T by defining T(v):=f(v).w, the product of the scalar f(v) and the vector w. This T depends on f and w, of course. It's easy to see that these T generate L(V,W). In fact, if you choose a basis v_i of V and a basis w_j of W, and if you choose f=f_i (from the basis dual to the chosen basis of V) and select w=w_j, then you'll end up with the T whose matrix has a 1 in the ji-th place and 0's elsewhere. Hence, in studying this issue, we can take T to be the T determined by a pair (f,w) and take U to be the analogous element of L(W,V) that is defined by a pair (g,v) with g in W^* and v in V. Explicitly, U(w) is g(w).v.

What could trace(UT) be in this situation? There is an extremely natural answer, namely the product of the two numbers f(v) and g(w). If you choose a basis for V and compute the trace, you will actually get this answer. (I did this on a piece of paper but am reluctant to try to type the computation into this comments page.) In some sense, I think that I've now provided the explanation that you were seeking. Namely, the trace represents a bilinear (= linear in each argument) function on L(V,W) x L(W,V), and it turns out to be the most obvious such function. It happens also that this function has a nice symmetry: f(v)g(w) is unchanged if we exchange the two pairs (f,w) and (g,v). This symmetry is equivalent to the identity trace(AB) = trace(BA) that you verified by direct calculation. Does this help?

Tue Sep 23 22:15:41 2003 : The explanation of the trace that you gave is what I wanted to know. Thank you!

Wed Sep 24 22:21:11 2003 : Could we possibly spend some portion of Friday's lecture as midterm review? Also, is there anything you can tell us about what the midterm will be like? Do you think it would be a good thing, for example, for students to look over your Fall 2002 Math 110 class' midterms?

Reply: I don't think that it would be all that helpful to review things on Friday. Students should try to organize the material for themselves and distill essential things down to a single sheet of paper (2-sided). Doing this with a study group might be a good idea.

If you look at some of my old exams (110 last fall and H113 last spring), you'll get an idea of the style of my questions. An important thing to consider is that last year's exams were 80-minute exams and our midterms this semester are 50-minute exams. In 50 minutes, I can't ask you very much! Also, the midterm can't require too much reflection because you won't have much time to reflect! I might come up with some questions that are either recycled homework questions or variants of questions that you've had on the homework or things that we've done in class.

If you wonder whether a certain kind of question might appear on the exam, ask yourself what would happen if it were there. Would it be a fair question? Would it be reasonably easy to grade? Would only a few students get the answer or would everyone get it? I'd love to be able to make up an exam with three questions: one that almost everyone would get, one that 2/3 of the class would get, and one that few people would get. For various reasons, I doubt whether I can achieve this goal in our short exam. I think that the main purpose of this midterm will be to provide feedback to people who are unsure whether they should continue in this class or transfer into a regular 110 section. If your score on this exam is significantly below that of the main mass of students, then you should think about changing sections.

Fri Sep 26 10:10:47 2003 : Will the solutions to the problem set due today be on John's website before the midterm?

Reply: I've asked John whether he'll be able to make this happen, but I don't know his schedule. Check back a few times over the weekend.

Sat Sep 27 16:15:51 2003 : OK, I just put up HW #5 solutions. Have a great weekend! -John

Mon Sep 29 14:56:33 2003 : The midterm was very easy.

Reply: YMMV

What we have here is a classic bimodal distribution. There were 11 scores at 23 or above, no scores between 18 and 22, and 20 scores that were 17 or below. There were 10 students who got 10 or below and 10 students whose scores were between 11 and 17.

Tue Sep 30 11:40:40 2003 : Out of curosity what does YMMV exactly mean? I don't understand the idiom your mileage might vary.

Reply: "YMMV" is an expression that you see on the net; I understand it to mean "your experience may turn out to be different from mine." While some students found the midterm to be easy, only half the class got more than half the possible points. The goal of the exam was to enable students to decide whether or not they're losing control of the class material. If you find that you're no longer following what's going on in the class (meaning in lectures, in the reading and in the homework), then you should consider moving into a non-honors Math 110. Whether or not you actually decide to move is 100% up to you.

Tue Sep 30 21:15:22 2003 : For problem 4.2.28, can we use Theorem 4.7, even though it's in the following section?

Reply: Sounds like a bad idea because the proof of Theorem 4.7 involves consideration of elementary matrices. You want to avoid circular reasoning.

Thu Oct 2 23:41:13 2003 : "Down with Determinants" is available on Sheldon Axler's website:

Reply: I found this as well after class on Wednesday. I added a link to Axler's page on the main H110 page. You'll find the link in the general vicinity of his photo.

Wed Oct 8 19:12:45 2003 : Colored chalk is great!

Fri Oct 10 10:31:48 2003 : Yeah, coloured chalk looks good, but it takes a lot of time to write things in different colours.

Comment: I think that a big drawback of colored chalk is that it's hard to erase. I typically walk out of 2 Evans with un-erased boards; this is fairly normal for math courses. (Instructors begin their classes by erasing the boards left by previous instructors.) I'd feel terribly guilty if I left colored chalk on the board for Crystal to erase.

Sun Oct 12 13:58:27 2003 : Can you explain the notation used in Harm Dersken's article: P(K,d,Ar)?

Reply: In the article, there's P(K,d,r), which I suppose is what you mean. It's defined on page 2 of the article. The definition is sufficiently strange that it's helpful to consider the two examples that appear in Conrad's treatment. In Conrad's paper, one notes that every endomorphism of a vector space of odd dimension over the field of real numbers has an eigenvector. This is because odd-degree real polynomials have real roots (as follows from the intermediate value theorem of calculus). To see that every endomorphism on an odd dimensional vector space has an eigenvalue is to assert P(R,2,1). Conrad shows that every pair of commuting endomorphisms of such a vector space has a common eigenvector; this is P(R,2,2). Here, a pair of commuting endomorphisms basically is the same thing as a pair of square matrices of size n that commute with each other. Conrad's proof that P(R,2,2) follows from P(R,2,1) shows more generally that P(K,d,1) implies P(K,d,2) for every field K and every positive integer d. Conrad uses this implication when K is the field of complex numbers and d is a power of 2. Derksen treats also the assertion P(K,d,r) with r greater than 2, but this is not necessary for the proof of the fundamental theorem of algebra.

Mon Oct 13 16:10:47 2003 : Hi, it's Keith. In response to the comment about Derksen's notation, I didn't like the notation P(K,d,r) myself (among other things, I couldn't keep in my mind what it meant), and that's partly why I wrote up the proof which the class is now reading. As a side benefit, I found that the full strength of Derksen's induction isn't needed to get the Fundamental Theorem of Algebra, which is a point already noted in the previous reply.

Wed Oct 15 20:30:44 2003 : I do not understand 5.2 # 23. Why is the consequence not immediate?

Reply: Who says that it's not immediate? I think that the problem is easy if you view it in the right way, but it would be a mistake to get lulled by the notation and assert without justification that if W_1 and W_2 are each direct sums of subspaces of V, and if the sum of W_1 and W_2 is direct, then you can subsitute in blithly and say that W_1 + W_2 is the direct sum of the various subspaces that make up W_1 and W_2.

Thu Oct 16 20:54:22 2003 : Does there exist another person who doubts the validtiy of theorem 5.10 part d? How does the if part imply the then part? Also... What's preventing me from picking k = 2, W1 = W2 = V? This would mean d is not equivalent to a.

Reply: I just spoke with the student who posted this comment. The nub of the problem seems to be the union sign in the condition. The ordered bases gamma_i are intended to be lists of vectors. The union of these lists is the concatenation of them: you make a big string that lists the vectors in gamma_1, the vectors in gamma_2, and so on. If W_1=W_2 and gamma_1=gamma_2, and if W_1 is non-zero, then a list that begins with the vectors in gamma_1 and then repeats those vectors will not constitute an ordered basis of V.

Thu Oct 16 23:17:58 2003 : (d) is not true by itself. The theorem states that if (d) is true, then (a), (b), (c), and (e) follow. Or if any of (a), (b), (c), or (e) are true, then (d) must also be true.

Sun Oct 19 17:30:06 2003 : Are there only two problems on the homework this week?

The two posted problems are very good.

Reply: I added some stuff an hour or two ago.

Mon Oct 20 16:15:15 2003 : For the second part of the first problem on this week's homework, which definition of characteristic polynomial should we use? If we use the book's definition, that is, characteristic polynomial = f(t) = det(A-tI), doesn't the claim immediately follow?

Reply: Are you referring to the question of why the value of f(t)= det(A-tI) at lambda is the determinant of A-lambda.I? You have to know somehow that taking the determinant of a matrix with polynomials in it and then pluging in a number amounts to the same thing as plugging in the number and then taking the determinant. This is certainly true, but you need to provide a proof of some sort. It's not stated explicitly as a theorem in the textbook.

Sat Oct 25 21:28:40 2003 : For number 24 of 5.4, can we assume that the vector space is finite dimensional? The hint does say to apply exercise 23, where V was finite dimensional.

Reply: Sounds reasonable to me.

Fri Oct 31 19:29:47 2003 : Hello all! I will have the solutions for HW #10 posted on Saturday afternoon for your studying pleasure. Happy Halloween! -John

Sun Nov 2 01:05:07 2003 : Thank you!

Sun Nov 2 16:21:52 2003 : I'm not quite clear on the stuff you plan to cover on the midterm. If we could be tested on the most recent lecture material (into chapter 6), then we could be tested on stuff we havent had homework on. Should we be focusing on homework things, or is everything covered so far in lecture fair game?

Reply: My usual rule is that I don't ask questions about the lecture that was just before the exam but I can (and sometimes do) ask about material that was covered two lectures before the exam. In this situation, § 6.4 will not be on the exam, but the first three sections of Chapter 6 will be on the exam.

Mon Nov 3 11:31:20 2003 : Can we please not have HW due this friday?

Reply: The next homework assignment is due on Friday, November 7. However, papers that are turned in on November 10 will not be considered as late. (Mortgage payments are typically due on the first of the month but are considered as delinquent only if they are turned in after the 15th of the month.)

Tue Nov 4 15:58:27 2003 : in # 10, is V finite dimensional?

Reply: There are two #10s in this assignment. The second one, in § 6.4, starts off with the statement that V is finite-dimensional. I infer, therefore, that you're talking about § 6.2, #10. I'm pretty sure that the V here is not assumed to be finite-dimensional. If you can do the problem only when V is finite-dimensional, try to analyze how this assumption is used and then see if you can work around it.

Fri Nov 7 13:20:12 2003 : The second problem on the midterm asked to prove Schur's Lemma, which we discussed on Wednesday. However, I'm pretty sure that you said in class that the exam will cover themes up to the least squares approximation, which we studied on Monday. I would have appreciated you being more clear on what the exam would cover.

Reply: The second problem bears only a superficial resemblance to Schur's lemma. Schur's lemma is about inner product spaces and orthonormal bases. Its proof makes use of the adjoint of a linear transformation. The exam problem is algebraic: it has nothing to do with inner products and does not require that the field of scalars be the real or complex field. The idea of the proof is to pass from an n-dimensional space to a space of smaller dimension by using quotient spaces and to use one of the problems in the assignment that was due on October 17. It is true that proving Schur's lemma gave me the idea to ask the question. On the other hand, the best approach to solving the problem would have been to note at the outset that the solution couldn't involve Schur's lemma because that lemma wouldn't be covered by the exam.

Sat Nov 8 23:28:29 2003 : After reading the previous comment I wonder whether it was possible to "solve" the second problem on the midterm in the following way: Define an inner product on V as =v*(w) where v* denotes the the image of v under some fixed isomorphism from V to V* (as discussed in class, before introducing inner products on R and C). Then the proof of Schur's lemma works except that we cannot ensure that the square of norms of the vectors is 1 since the square of the norm can happen to be 0 in the field F or that the square of the norm is a non-residue.

Reply: The idea is very good, but I'd go further and say that there is no need to introduce an inner product. It's probably simplest just to keep V and V^* separate from each other and not identify them by bringing an inner product into the picture. The crux of the proof that we did in class is that there is a subspace of V of dimension dim(V) - 1 that is stable under T. Once you know this, the desired result follows by invoking the induction hypothesis: the restriction of T to the subspace is upper-triangular in some basis, and you get T upper-triangular on all of V just by completing the basis in any old way that you like. How to find the subspace? Look at the action of the transpose T^t of T on V^*. If you make a basis of V, then T is given by a matrix A. The matrix of T^t in the basis dual to the chosen basis is the transpose of A. Since we know that a matrix and its transpose have the same determinant, the characteristic polynomial of T^t is the same as the characteristic polynomial of T. It follows that T^t has an eigenvector f. Since f is a non-zero linear transformation V->F, the null space W of f is a subspace of V of dimension n-1, where n is the dimension of V. It is clear that W is stable under T. Indeed, if w is in W, then f(T(w)) = (T^t(f))(w). Now T^t(f) is a multiple of f because f is an eigenvector, so the expression (T^t(f))(w) is a multiple of f(w), which is 0. Hence T(w) lies in the kernel (= null space) of f, which is W.

Sun Nov 9 00:19:35 2003 : There is a typo in the defintion of S in the first question on the homework that is due November 14.

Reply: Thanks for pointing this out. I fixed the typo (while perhaps introducing others!) and also removed problem 6 from the list of problems in §6.5.

Wed Nov 12 23:27:34 2003 : What exactly is part b of the George Bergman problem asking us to do? Are we to rewrite "f'(0)=0"? It's not very clear.

Reply: The student who posted this comment came to my office and we had some discussion. If I recall correctly, f'(0) turns out to be some simple expression involving inner products. To say that it's 0 is to say that this simple expression vanishes. Accordingly, you get some identity involving inner products: two are equal, or maybe negatives of each other, or one of them is the sum of two others -- you get the idea.

Thu Nov 13 20:39:29 2003 : why did you remove #7???

Wed Nov 19 17:49:19 2003 : On the first question of this week's problem set, shouldn't the two conditions start with "for each _nonzero_ x/y..."?

Reply: Yes, thanks much for pointing this out!

Wed Nov 19 19:43:57 2003 : When defined bilinear forms, we did so only for VxV. If V,W are both vector spaces over F, is there something stopping us from defining a 'bilinear' form VxW-->F in the same way, even if we can't obtain a quadratic form from it?

Reply: There's nothing wrong with the concept, but the vector space of such functions might not be worthy of any special interest. It's the vector space Z*, where Z is the tensor product of V with W.

Thu Nov 20 13:02:46 2003 : On the first problem on the homework, it seems the statement would not be true unless H was a symmetric bilinear form. Should the question be about symmetric forms or am I making a mistake?

Reply: The latter, I believe.

Fri Nov 21 19:50:07 2003 : Are we going to have homework this week?

Reply: The next homework will be due after Thanksgiving. It'll probably be the last homework as well. I'll make it up over the weekend and expect to post it by the end of the day on Sunday.

Mon Nov 24 19:53:56 2003 : Some comments about homework #14. I did not expect to see computational questions (eg. 3), something like does A commute with A^t (question 4) in an honours course. In addition, we have done question 9 in an earlier homework, if I am not mistaken.

Reply: The homework problems were taken from old math department preliminary exams. What can make those exams hard is that there is no indication going in as to the nature of each problem. There can be very hard questions mixed in with very easy questions, silly questions with serious questions, and so on. The "computational" problem, for example, might be solvable by "pure thought" -- without any significant computation. Think about the characteristic polynomial and the generalized eigenspaces -- what can they be like? It's likely possible to find the canonical form by listing some a priori possibilities and then ruling out all but one of them. The commutation question is not one you'd expect, but there it was on the prelim, so now there it is on your homework! Sorry if it's silly. If you've done question #9 earlier, you can either do it again or cite the place where you did it before.

Thu Nov 27 17:56:22 2003 : For number 7, need V necessarily be over C? It seems like it doesnt.

Reply: The problem has nothing to do with the field of complex numbers, but you do need to assume that the characteristic polynomial of T splits as a product of linear factors. If you took T to be a non-trivial 2x2 rotation matrix over the field of real numbers, the vector space R^2 would have no invariant subspaces except itself and 0. Thus there would always be a W'. Nonetheless, T has no eigenvectors.

What's good about the problem as stated is that the optimal hypothesis is not presented. Thus the problem echos the sort of situation that you might have in honest applications of linear algebra: you have to figure out an argument to prove what you need, and only at the end do you have the luxury of trying to sort out how general your argument is.

Thu Dec 4 16:01:13 2003 : Hello everyone! John "the reader" Voight here. I graded the last HW, you should get it back forthwith. There were at total of 14 assignments, totalling 280 points; for what it's worth, the mean was 213/280, with a standard deviation of 49. If you would like to double-check your HW scores, you can e-mail me or talk to the professor. I hope you have enjoyed the course, and best of luck on the final exam! -John

Wed Dec 10 13:13:34 2003 : For #6 on the 'further review problems,' should the polynomials be distinct?

Wed Dec 10 17:35:30 2003 : Oops, sorry, I misread the question. Disregard my previous comment.

Sun Dec 14 17:19:03 2003 : When are your office hours before the final? Thanks!

Reply: How about 10:45 to 12:45 today (Monday)?

Mon Dec 15 12:22:29 2003 : um...I just woke up. Can you please have office hours tomorrow too?

Reply: If people come in between 2:30 and 3:30 today (Tuesday), I'll be happy to talk with them.

Mon Dec 15 22:23:40 2003 : do we like always get to bring a page of notes tomorrow?

Reply: Yes, of course.

Tue Dec 16 01:40:13 2003 : i think that there might be an error in one of the homework solutions: near the end of the solution for problem 6.6.5(b) in HW #13, the reader writes, "If !=0, then choose c < -||w2||^2/Re." But why can we divide by Re? Why can't be imaginary, in which case !=0 holds but Re=0? Or am I missing some easy observation?

Reply: I'm going to have to read a lot of solutions starting this evening. I'm not all that eager to read what the read wrote about 6.6.5(b). My attitude toward the problem, when I discussed it with students in office hours, is that I'd do out the real case on the board and leave the complex case for them to mull over by themselves. The complex case can't be seriously harder than the real case, right If I have some time today, I'll write down the solution in the real case.

OK, look: if there are w_1 and w_2 such that is non-zero, we can scale w_2 so that the inner product is a positive real number. Then there's no distinction between the inner product and its real part. It's true that the poor inner product might happen to be purely imaginary in the complex case, if w_1 and w_2 have chosen themselves maliciously. But we can correct this by multiplying by a scalar.

Tue Dec 16 01:44:16 2003 : oops, it ate my brackets. let me try again, this time with inner product denoted as [ , ]. i think that there might be an error in one of the homework solutions: near the end of the solution for problem 6.6.5(b) in HW #13, the reader writes, "If [w1,w2]!=0, then choose c < -||w2||^2/Re[w1,w2]." But why can we divide by Re[w1,w2]? Why can't [w1,w2] be imaginary, in which case [w1,w2]!=0 holds but Re[w1,w2]=0? Or am I missing some easy observation?

Tue Dec 16 14:02:22 2003 : We are dropping one homework grade, right?

Reply: Yes, because of the holiday season.

Tue Dec 16 21:12:20 2003 : according page 398 of the text, I don't think the statement in the solution to number 2 of the final is correct. "If T^2 = T, where T is a linear operator on a vector space V , then we know well that V is the direct sum of the null space of T and the space of vectors that are fixed by T." Don't we only know that T is a projection?

Reply: Suppose that T is the projection of V onto X along Y. Then V is the direct sum of X and Y; each v is uniquely of the form x+y. The map T takes x+y to x. The null space of T is Y. The space X is the set of vectors that are fixed by the projection, i.e., the ones that are mapped back to themselves by the projection. So, as I said, V is the direct sum of the null space of T and the null space of T-I.

I'm amazed that only two or three of you chose to do problem 8. It was very easy!

Tue Dec 16 21:14:45 2003 : also for number 6, I think one of them should be T[v - w]

Thu Dec 18 11:28:51 2003 : will you be posting our grades like you did for 113?

Reply: I posted the grades around 90 minutes after this comment was posted.