patrick@nmtvax.UUCP (Patrick Madden) (03/11/86)
*** REPLACE THIS LINE WITH YOUR MATRIX *** Help! I'd like to get an A in linear, but I'm looking at a B.... However, there's a chance I can get a lot of brownie points for the following: Given matrix A of the form below, there are at most 4 matrices M1, M2, M3 and M4, such that Mi*Mi = I (M sub i squared is the identity matrix), and M1*M2*M3*M4 = A. The "A" matrix is: [K 0 0 .. 0 ] [2 0] [0 K 0 .. 0 ] and in particular [0 .5] [. 0 K 0. 0 ] [. . 0 K 0 ] [.....0 1/k^(n-1)] etc, and the nth row, nth column entry is 1/K^(n-1). The determinant of these beasts is 1. I have a feeling that these may have been discussed at a conference in Dallas on March 8th or thereabouts. The prof just got back from there with a smirk on his face.... Chances are this is a pretty rough one--he seemed pretty sure no one would get it. Anyway, if anyone has any ideas (or pointers, or knows anything about the confrence), I'd love to hear. My GPA would love to hear too. Thanx in advance, Patrick Madden, at a dusty terminal in a dusty town in a dusty state.... !cmcl2!lanl!unmc!nmtvax!patrick | "Mid the sagebrush and the thistle, !ucbvax!unmvax!nmtvax!patrick | I'll watch that guided missile"
pumphrey@ttidcb.UUCP (Larry Pumphrey) (03/12/86)
I think you need to supply more information as there appear to be
counter-examples as the problem is stated. The following points
should be clarified.
1. Over what field do the elements a of the matrices M
ij i
range (and hence k in the companion matrix A)? i.e., the
rationals, reals, complexes?
2. Any restrictions on the value of k?
3. Can any of the M be the identity matrix?
i
Counter-example:
----------------
If there are no restrictions on the value of k in the A matrix
then let k=1 and A becomes the identity matrix. Consider the 8
3X3 matrices, M as follows:
i
[1 0 0] [1 0 0] [1 0 0] [1 0 0]
M = [0 1 0] M = [0 1 0] M = [0 -1 0] M = [0 -1 0]
1 [0 0 1] 2 [0 0 -1] 3 [0 0 1] 4 [0 0 -1]
[-1 0 0] [-1 0 0] [-1 0 0] [-1 0 0]
M = [ 0 1 0] M = [ 0 1 0] M = [ 0 -1 0] M = [ 0 -1 0]
5 [ 0 0 1] 6 [ 0 0 -1] 7 [ 0 0 1] 8 [ 0 0 -1]
Note that all 8 matrices are distinct, each M squared is the
i
identity and their product is A = I. Also note that M = I.
1
It's an interesting problem, but something appears to be missing.
- Larryweemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (03/13/86)
In article <967@nmtvax.UUCP> patrick@nmtvax.UUCP writes: >Given matrix A of the form below, there are at most 4 matrices M1, M2, M3 and >M4, such that Mi*Mi = I (M sub i squared is the identity matrix), and >M1*M2*M3*M4 = A. The "A" matrix is: >[K 0 0 .. 0 ] [2 0] >[0 K 0 .. 0 ] and in particular [0 .5] >[. 0 K 0. 0 ] >[. . 0 K 0 ] >[.....0 1/k^(n-1)] > etc, and the nth row, nth column entry is 1/K^(n-1). >The determinant of these beasts is 1. The problem does not make sense as stated. "There are at most 4 matrices", for example, would make sense if the problem continued "M such that p(M) holds", where p(.) is some property. And if there is a quadruple of matrices with the asked for properties, then there are infinitely many quadruples, since any special orthogonal transformation, that is, replacing each Mi by O*Mi*inv(O), where O is a special orthogonal matrix, preserves the asked for property. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720
tim@ism780c.UUCP (Tim Smith) (03/14/86)
This sure sounds a lot like you want the net to help you get an undeserved A. Am I missing something? -- Tim Smith sdcrdcf!ism780c!tim || ima!ism780!tim || ihnp4!cithep!tim
weemba@brahms.BERKELEY.EDU (Matthew P. Wiener) (03/15/86)
In article <980@ism780c.UUCP> tim@ism780c.UUCP (Tim Smith) writes: >This sure sounds a lot like you want the net to help you get an >undeserved A. Am I missing something? First of all, if he can't state the problem correctly, he's going to have a hard time getting an A period, so don't worry. I don't know about the undeservedness issue. Sometimes understanding one good problem, no matter where the understanding comes from, is the most important part of the battle. Of course, if he then turns in his homework without accreditation, he's a jerk, but that's his problem. ucbvax!brahms!weemba Matthew P Wiener/UCB Math Dept/Berkeley CA 94720