Messages below were copied from the live Discussion for Math 5718 Spring 1999, see http://math.ucdenver.edu/~aknyazev/teaching/99/5718/ .

     Forum: Math 5718 S99
     Date: Wed, 27 Jan 1999 16:43:02 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 To make my recommendations clear:

 1. If you like Shilov, you do not have to read Axler at all.

 1a. If you like Shilov and want to read something else, I recommend Matrix Analysis by Roger A. Horn, Charles R. Johnson,
 see http://www.amazon.com/exec/obidos/ASIN/0521386322/qid=917476092/sr=1-1/002-4903982-5705246

 2. If you do not like Shilov, read Axler first, but then return to Shilov and try it again.

 Please let me know whether you like Shilov, or not.

     Forum: Math 5718 S99
     Date: Wed, 27 Jan 1999 19:56:22 MST
     From: Saulo <saulo@math.ucdenver.edu>

 Check on Problem 10, Chp1: the last row of the determinant on the right hand side of the equation has a strange pattern:

 x_1^(n-1) x_2^(n-2) ... x_n^(n-1)

typo in col 2

     Forum: Math 5718 S99
     Re:  Typo ? (Saulo)
     Keywords: typographical error
     Date: Wed, 27 Jan 1999 21:18:22 MST
     From: <frank.duca@ucdenver.edu>

 i believe, thru my great powers of deduction, that there is a misprint in the second determinant when the summation on the
 right is factored out of the first determinant.

 the last row, second column entry in the right determinant should be X-sub2 to the (n-1) power. this is the only way it makes
 sense to me.

 zat what u meant, saulo?

     Forum: Math 5718 S99
     Date: Thu, 28 Jan 1999 19:20:27 GMT
     From: <unknown>

 Is there another typo in the answer to problem two chapter 1?

 I have as an answer for the last product with a negative number the following:

 a14a31a23a42 and the answer in the back of the book says: a14a31a23a44 which I can't make work.

 Any comments?

 Thanks,

 Joe Kaiser

     Forum: Math 5718 S99
     Keywords: independent variable
     Date: Wed, 24 Feb 1999 16:58:58 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 A student asks:

 A quick question for you on Problem 2 of Quiz #1. Could you please clarify for me the proper interpretation of the lower
 case x please?

 1) One way to interpreted this problem is that the lower case x used in L=span{x}, 1+x+x**2, 1+x-x**2, and 2+ix is that
 these x do not have to be the same value. In other words, this can be phrased as L=span{e}, 1+m+m**2, 1+n-n**2, and
 2+i(p) where e,m,n, and p exist in the complex.

 2) The lower case x exist in the complex but once selected can not change and must be used in L=span{x}, 1+x+x**2,
 1+x-x**2, and 2+ix. Therefore,

 if x is selected as 1+i then L=span{1+i}, 1+(1+i)+(1+i)**2, 1+(1+i)-(1+i)**2, and 2+i(1+i).

 I am unsure what the correct interpretation of this lower case x is and it has a great impact on the solution.

 My answer:

 None of those is correct. The lower case x is the independent variable, like in functions sin(x) and cos(x). The value of x in
 sin(x) and cos(x) is the same, but it is not in your control when you use span. So, this would be incorrect to say that the

 span { sin(x), cos(x)} = span {sin (1) , cos (1)} when x=1.

 Such statement does not make any sense.

 However, e.g. sin(x)+cos(x) = sin(1)+cos(1). when x=1.

     Forum: Math 5718 S99
     Date: Sun, 07 Mar 1999 00:14:13 GMT
     From: <unknown>

 I believe the last two lines of the first paragraph of the proof should contain the phrases "no less than r-(n-k)" rather than "no
 more than r-(n-k)". Does anyone disagree?

     Forum: Math 5718 S99
     Date: Tue, 09 Mar 1999 23:45:05 GMT
     From: <unknown>

The 8th line of text of this subsection reads '...is the sum of k "elementary cloumns"...'.
I believe it should read '...is the sum of  n "elementary columns"...'.

 Two lines later, I think '...is the sum of k^k "elementary determinants"...' should read '...is the sum of n^k "elementary
 determinants"...'.

 Am I right?

     Forum: Math 5718 S99
     Date: Mon, 12 Apr 1999 19:37:15 GMT
     From: Don Romano <mor1@concentric.net>

 There is a typo on page 10 of Shilov, in the proof for the
 linear propetry of determinants theorem. Yes, the theorem
 from the midterm that we all know and love.

 In the second line of the proof you'll see: b_alpha_1
 (where "_" denotes subscript.) This should read: b_alpha_j.

 Our illustrious professor concurs.

     Forum: Math 5718 S99
     Re:  Problem 3.2 (Saulo)
     Date: Mon, 26 Apr 1999 16:04:37 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 I waited to see if other people answer, but this did not happen.

 In my opinion, the problem does not ask to find all possible (b_i)s explicitly. We only need to show that every vector y in the
 span can be found using minors of A of order k. And the hint does it. So, I think that you just need to explain/prove all the
 statements of the hint, and nothing more.

 >Andrew & Classmates,

 > Probably the most hard problem in the Chp 3 is the second >one. I've got an "ugly" solution that may not be written >clearly. In the
 > following I write its final conclusion, hoping that you can >solve the problem with your own approach.

 > From the question and the hint (that unfortunately doesn't >give us the final answer...), we need to determine the >coefficients b_i
 > (with respect to {e_1,....,e_n}) of any vector y in the span >L of {x_1,...,x_k} using ONLY the minors of A (A is formed by >the
 > coeffs. of x_j with respect to e_i). My final answer is that >the space of all possible (b_i)s (a subspace of R^n) is equal >to the
 > kernel of a matrix C with dimensions

 >         n
 >      (     )  x  n
 >        k+1

 > formed by the minors of order k of A, that corresponds to >the system of equations cited in the bottom of the hint.

 > Please let me know if you could finish it.

 > Saulo

     Forum: Math 5718 S99
     Date: Mon, 19 Apr 1999 12:14:46 MST
     From: cjameson <unknown>

 In subsection 6.11 page 134 line 3 the text defines height as being positive, then proceeds to call the zero vector a vector of
 height zero. Dr. Knyazev concurs that height should be defined as nonnegative.

 Also, 3/4 down the page at the end of a line the text is ...(p_2 = m_(r-1) - m_(r-2)). I believe this should read ...
(p_2 = m_(r) - m_(r-2)). Does anyone have any insights?

Yes, I agree

     Forum: Math 5718 S99
     Re:  text error (cjameson)
     Date: Mon, 26 Apr 1999 15:25:27 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 As we discussed in class last time, yes, I agreed that this was a typo.

     Forum: Math 5718 S99
     Date: Fri, 14 May 1999 01:39:15 GMT
     From: Saulo <saulo@math.ucdenver.edu>

 The claim in Ex. 18 / CHp 4 is wrong. The question will make sense if we replace the last sentence " (...) depends only on the
 colum indices of the minor M" by " (...) depends only on the row indices of the minor M".

     Forum: Math 5718 S99
     Date: Fri, 14 May 1999 01:39:15 GMT
     From: Saulo <saulo@math.ucdenver.edu>

 The claim in Ex. 18 / CHp 4 is wrong. The question will make sense if we replace the last sentence " (...) depends only on the
 colum indices of the minor M" by " (...) depends only on the row indices of the minor M".

     Forum: Math 5718 S99
     Date: Thu, 06 May 1999 11:14:32 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 I found yesterday night that some students for #4 tried to use a theorem, that was not in the book, namely, that a symmetric
 mapping leads to a symmetric matrix in any basis, and vice versa. As formulated above, the statement is NOT correct!

 It is correct, when the basis is orthonormal, though. Still, as usual, I suggest NOT to use any statements outside of the text,
 unless you provide complete proofs.

 Please, share this hint with those students who do not have a chance to read it.

 Hope this helps.

     Forum: Math 5718 S99
     Date: Fri, 14 May 1999 13:37:16 MST
     From: Andrew Knyazev <aknyazev@math.ucdenver.edu>

 For those of you who will continue studying linear
 algebra, e.g., for the prelim, I want to clarify
 that the algorithm for finding generalized eigenvectors,
 which we covered in class, is only a simplified
 version. It always works for eigenvalues
 with geometric multiplicity one. It often works
 for geometric multiplicity larger than one, as we see
 in our examples, see also

 http://bass.gmu.edu/ececourses/ece521/lecturenote/chap1/node3.html .

 However, in general, it may not work
 exactly the way we do it. For example, let u and v
 be two linearly independent eigenvectors for some fixed
 eigenvalue lambda,
 and geometric multiplicity of the eigenvalue
 is two. Then, what may happen is that none
 of u and v leads to a generalized eigenvector,
 but there may be generalized eigenvectors for some
 of their linear combinations. Thus, in general, there
 is nothing special with our choice of u and v,
 and when finding generalized eigenvectors
 on the first level one must attempt to solve
 the linear system for e_2 :

 (A - lambda I) e_2 = au+bv

 with arbitrary coefficients a and b.
 If you find such a' and b', you call e_1 = a'u+b'v.
 If you find another choice, a" and b", that
 gives you a linearly independent a"u+b"v, you call
 it f_1 and introduce another notation f_2 for the
 correspoding solution. In the latter case, thus,
 you would have e_1, f_1 and the corresponding e_2 and f_2.

 Try it for the matrix

 1 0 0 1
 0 1 0 1
 0 0 1 0
 0 0 0 1

 and use appropriate coordinate vectors u and v.

 A similar rule applies to next steps.
 When you find e_2, it will have two free
 parameters, say c and d,
 in it. When you solve for e_3 :

 (A - lambda I) e_3 = e_2,

 in general, one cannot just chose c and d in advance.
 Some choice of c and d may lead to e_3, for
 other choices of c and d the solution may not exist.
 To explain this, we notice that 1) all solutions e_2
 form a two-dimensional hyperplane, and
 2) solution e_3 exists if e_2 is in the range of
 A - lambda I, which is an n-2 dimensional subspace where n
 is the dimension of our space.
 There are several possibilities for an n-2 dimensional
 subspace and 2 dimensional hyperplane to intersect,
 and those possibilities characterize the choice
 of c and d we need.

 In general, one have to solve for e_3,
 keeping c and d in e_2 and trying to find c and d
 that lead to solutions e_3. One fixes those c and d
 in e_2, which give some e_3's. If there is no solution
 e_3 for any choice of c and d, that means there is
 no vector e_3 at all, and c and d can now be fixed
 any way you like. If there are several choices
 of c and d, than the big time starts. In any case, number
 of generalized eigenvectors on every level must not be
 larger than the number of vectors on the previous velev.

 I recommend you to try it for the matrix

 1 1 0 0 0 0
 0 1 1 0 0 0
 0 0 1 0 0 0
 0 0 0 1 1 0
 0 0 0 0 1 1
 0 0 0 0 0 1
 

 when you do not use coordinate vectors for
 the basis u and v of the eigenspace.

 This general form of the
 algorithm is quite messy, but, to my taste, is still better than the other one, from Shilov, see it also at

 http://forum.swarthmore.edu/dr.math/problems/russo4.29.98.html

 When you solve such problems, a good strategy is first to try to use
 our lite version of the algorithm with "natural" choice of
 basis vectors in a hope that you can guess the right path.
 It saves a lot of time.
 Then, if you see that you do not have enough vectors in the basis, you
 need to do it the hard way.

 Enjoy! Hope it is not too confusing.
 Let me know if you have any question.

Greetings Dr. Knyazev.

I meant to send this to you sooner, but I have been preoccupied with
setting up my computer.

There are a few things from Shilov's "Linear Algebra" that I think are
important for me to convey. Although the reading was slow at times, it was
not inordinately difficult until I reached chapter 6. Specifically,
the last sentence of Theorem 6.36. A clear grasp of the material
immediately following this theorem depends in part on a clear grasp of the
theorem. It took me a ridiculous amount of time to understand the theorem,
and I wanted to explain why.

On page 144, in the proof of Theorem 6.34, the text defines T_k as the
range of the operator Q_k(A). But, in Theorem 6.36 (page 145), T_k is the
nullspace of Q_k = (B_k)^(r_k). This material in this part of the text is
complex and demanding. This abrupt and unclarified change of definition
makes it difficult to grasp Theorem 6.36. In addition, because of the
change of definition, the proof of Theorem 6.36 is not as simple as it
would seem. The proof of Theorem 6.36 cites Theorem 6.34, and the two
theorems have different definitions of T_k. I think it is arguable whether
Theorem 6.34 applies to Theorem 6.36. Perhaps a corollary to Theorem 6.34
applicable to Theorem 6.36 could be formulated.

A couple of much more minor things.

1) In the last line of page 147 the symbol r_k appears twice. For each
root of the characteristic equation, this would require the rank of the
sub-matrix and the number of jordan blocks to be equal. This is, in
general, not true.

2) In the fourth to last line of page 148 (I mentioned this to you
previously) we find "(not containing lambda)". This would be correct if it
read "(whose determinants do not contain lambda)". S and T must contain
lambda, and their determinants must equal 1 if S and T only perform row and
column operations as described at the bottom of the page.

Best Wishes

Craig.